How irrational quantities physically exist in nature?












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We know that an irrational no has well defined decimal values upto infinite decimal places. These irrational quantities exist in nature in some kind of measurements. For an example, circumference of a circle is '2πr' , so if radius is rational then circumference will be irrational ,and this case is quite natural.
But I am unable to understand , how can a physical measurement exist with infinite precision (due to precision of infinite decimal values in an irrational no) ??
Isn't it in contrary with the laws of Physics??






irrational-numbers philosophy







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  • Comments are not for extended discussion; this conversation has been moved to chat.
    – Pedro Tamaroff♦
    Sep 17 '16 at 22:15










  • I seem to have learned that my answer wasn't a great answer. What did you really mean to ask? Did you mean "Can we determine $pi$ to arbitrary precision" using a Physics experiment? Would a better answer say something like "It's a mathematical constant so we can calculate a lot more digits than a simple experiment tells you, unlike the dimensionless fine structure constant which requires observations?" Did you define $pi$ to be a different value for Earth because of curved space inside it and want to know whether $pi$ for Earth can be measured that accurately?
    – Timothy
    Nov 22 '18 at 19:05
















22














We know that an irrational no has well defined decimal values upto infinite decimal places. These irrational quantities exist in nature in some kind of measurements. For an example, circumference of a circle is '2πr' , so if radius is rational then circumference will be irrational ,and this case is quite natural.
But I am unable to understand , how can a physical measurement exist with infinite precision (due to precision of infinite decimal values in an irrational no) ??
Isn't it in contrary with the laws of Physics??






irrational-numbers philosophy







share|cite|improve this question
























  • Comments are not for extended discussion; this conversation has been moved to chat.
    – Pedro Tamaroff♦
    Sep 17 '16 at 22:15










  • I seem to have learned that my answer wasn't a great answer. What did you really mean to ask? Did you mean "Can we determine $pi$ to arbitrary precision" using a Physics experiment? Would a better answer say something like "It's a mathematical constant so we can calculate a lot more digits than a simple experiment tells you, unlike the dimensionless fine structure constant which requires observations?" Did you define $pi$ to be a different value for Earth because of curved space inside it and want to know whether $pi$ for Earth can be measured that accurately?
    – Timothy
    Nov 22 '18 at 19:05














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We know that an irrational no has well defined decimal values upto infinite decimal places. These irrational quantities exist in nature in some kind of measurements. For an example, circumference of a circle is '2πr' , so if radius is rational then circumference will be irrational ,and this case is quite natural.
But I am unable to understand , how can a physical measurement exist with infinite precision (due to precision of infinite decimal values in an irrational no) ??
Isn't it in contrary with the laws of Physics??






irrational-numbers philosophy







share|cite|improve this question















We know that an irrational no has well defined decimal values upto infinite decimal places. These irrational quantities exist in nature in some kind of measurements. For an example, circumference of a circle is '2πr' , so if radius is rational then circumference will be irrational ,and this case is quite natural.
But I am unable to understand , how can a physical measurement exist with infinite precision (due to precision of infinite decimal values in an irrational no) ??
Isn't it in contrary with the laws of Physics??






irrational-numbers philosophy




irrational-numbers philosophy






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edited Aug 22 '16 at 11:40









Omnomnomnom

126k788176




126k788176










asked Aug 22 '16 at 11:36









Aman SinhaAman Sinha

12515




12515












  • Comments are not for extended discussion; this conversation has been moved to chat.
    – Pedro Tamaroff♦
    Sep 17 '16 at 22:15










  • I seem to have learned that my answer wasn't a great answer. What did you really mean to ask? Did you mean "Can we determine $pi$ to arbitrary precision" using a Physics experiment? Would a better answer say something like "It's a mathematical constant so we can calculate a lot more digits than a simple experiment tells you, unlike the dimensionless fine structure constant which requires observations?" Did you define $pi$ to be a different value for Earth because of curved space inside it and want to know whether $pi$ for Earth can be measured that accurately?
    – Timothy
    Nov 22 '18 at 19:05


















  • Comments are not for extended discussion; this conversation has been moved to chat.
    – Pedro Tamaroff♦
    Sep 17 '16 at 22:15










  • I seem to have learned that my answer wasn't a great answer. What did you really mean to ask? Did you mean "Can we determine $pi$ to arbitrary precision" using a Physics experiment? Would a better answer say something like "It's a mathematical constant so we can calculate a lot more digits than a simple experiment tells you, unlike the dimensionless fine structure constant which requires observations?" Did you define $pi$ to be a different value for Earth because of curved space inside it and want to know whether $pi$ for Earth can be measured that accurately?
    – Timothy
    Nov 22 '18 at 19:05
















Comments are not for extended discussion; this conversation has been moved to chat.
– Pedro Tamaroff♦
Sep 17 '16 at 22:15




Comments are not for extended discussion; this conversation has been moved to chat.
– Pedro Tamaroff♦
Sep 17 '16 at 22:15












I seem to have learned that my answer wasn't a great answer. What did you really mean to ask? Did you mean "Can we determine $pi$ to arbitrary precision" using a Physics experiment? Would a better answer say something like "It's a mathematical constant so we can calculate a lot more digits than a simple experiment tells you, unlike the dimensionless fine structure constant which requires observations?" Did you define $pi$ to be a different value for Earth because of curved space inside it and want to know whether $pi$ for Earth can be measured that accurately?
– Timothy
Nov 22 '18 at 19:05




I seem to have learned that my answer wasn't a great answer. What did you really mean to ask? Did you mean "Can we determine $pi$ to arbitrary precision" using a Physics experiment? Would a better answer say something like "It's a mathematical constant so we can calculate a lot more digits than a simple experiment tells you, unlike the dimensionless fine structure constant which requires observations?" Did you define $pi$ to be a different value for Earth because of curved space inside it and want to know whether $pi$ for Earth can be measured that accurately?
– Timothy
Nov 22 '18 at 19:05










8 Answers
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92














I've got bad news for you: Even rational quantities don't exist in nature. Numbers, of whatever type, are just a way to describe our observations. But when we measure, we always have a measurement error, so we cannot really say "this measured quantity has the value $x$", all we can say is "this measured quantity is close to the value $x$". Well, we can do a bit more: Namely quantify how close it is. This is why measurement results are normally written in the form $xpm y$. For example, a length might be measured to be $1.3pm 0.5~rm mm$.



Now we pretend that there is a "true value" of the measured quantity. For example, in the above example, we assume that there is a true value for the length, which is likely(!) less than $0.5~rm mm$ away from the value $1.3~rm mm$.



Exact numbers are therefore nothing we can measure, and nothing which we can even prove really exist in nature. They are abstract concepts we made up in our mind in order to describe what we observe. And they work very well for this purpose. But whether they really correspond to something "out there", nobody can really know.



But wait, you say, there are circles, and for those we can prove the circumference over diameter is $pi$. So $pi$ is part of our world, right? Wrong: You never have seen a circle. You've certainly seen lots of shapes that closely fit our concept of a circle. But likely if you look close enough, you'll find that it is not really a circle. And even if you cannot detect the difference, how do you know that if you had looked just a little bit closer, you'd have found a deviation?



And ultimately, those circle-lookalikes you've seen probably were all made of atoms anyway, which means they could not have been a circle, as they have volume, while the circle is an infinitely thin line.






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  • Comments are not for extended discussion; this conversation has been moved to chat.
    – Pedro Tamaroff♦
    Sep 17 '16 at 22:14










  • You seem to be mentioning only our inability of precision measuring the lengths which exist in nature; our inability of not measuring lengths exactly as they are, seems to not imply their non existence.
    – Immortal Player
    Dec 20 '18 at 13:39



















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One odd thing about the question is the apparent belief that the decimal notation of a number is somehow fundamental to its meaning, and so an unending decimal expansion is a sign that the number itself is somehow unending or imprecise.



First, a simple factual point: the question claims:




We know that an irrational no has well defined decimal values upto infinite decimal places




I'm not exactly sure what this means, but I suspect it's not true. For example, we do have formulae that will tell us what the $n^mathrm{th}$ decimal digit of $pi$ is – in other words, we have a perfectly precise specification of any digit of $pi$ you care to ask for. So there's no fundamental sense in which $pi$ is less well-specified than, say $4.6$, which after all has an infinite decimal expansion as well, namely $4.600000dots$



Anyway, this fixation on decimal expansions is unnecessary. Here's how I choose to imagine the "true nature" of a real number: the essence of a real number is that it measures some continuous quantity – for example, each real number is the length of some idealized line segment.



Then decimal expansions are a way to understand these lengths by comparing them to fixed rational lengths that we know about, namely $1$, $0.1$, $0.01$, and etc. When we say $pi = 3.141dots$, we can view this as a series of comparisons, comparing $pi$ to $3$ and saying it is larger, comparing it to $3.15$ and saying it is smaller, etc.



In this way, the decimal expansion isn't the "true nature" of a number, but simply as the view of a number you get when you look at it through the lens of powers of ten. This becomes more obvious when you realise that you can write the same number in different bases, view it through many different lenses, and indeed in some cases you get very different looking results, e.g. $1/3$ has an unending decimal expansion but a very simple ternary or nonary one.



Now ask again what it means for some real number, the length of an idealized line segment, to be a rational number. Well, to be rational is to be $p/q$ for some whole numbers $p$ and $q$, so a number is rational precisely if some multiple of it is a whole number (namely, $q$ times it is $p$). What that means is you can take your line segment and duplicate it $q$ times, you get a line segment that is exactly $p$ times a segment of length $1$.



Now the question becomes: why should it be rational? Why should that relationship ever hold exactly?






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  • "we do have formulae that will tell us what the nth decimal digit of π is " - I don't think this is possible in O(n)? (i.e. the only way is to compute all the digits up to the nth).
    – M.M
    Aug 22 '16 at 23:48






  • 3




    See Bailey–Borwein–Plouffe formula, although strictly speaking that's binary digits and not decimal digits. (In any case, computing all the digits up to the n$^text{th}$ suffices for the point I was trying to make).
    – Ben Millwood
    Aug 23 '16 at 6:39












  • Binary digits is no help in finding the decimal digits (without computing all digits up to the point)
    – M.M
    Aug 23 '16 at 7:06






  • 1




    Yeah, so you do need to compute all the digits up to the nth, I guess. I think that's fine.
    – Ben Millwood
    Aug 23 '16 at 8:07






  • 4




    @M.M does it matter that its not in O(n)? We do have a formulae for the nth decimal digit. Maybe not an super efficient formulae but still.
    – Jakob
    Aug 23 '16 at 8:14



















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Philosophically, you could consider an "irrational quantity" as the limit of a sequences of "rational quantities", so e.g. 2pi is what the circumference of a unit circle approaches as you make it more and more perfect.






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  • 2




    :-) The completeness of it all.
    – Simply Beautiful Art
    Aug 23 '16 at 21:57










  • More than philosophically, you can define real numbers as sets of sequences of rationals.
    – YoTengoUnLCD
    Aug 24 '16 at 1:32










  • @YoTengoUnLCD right but here I'm talking about sequences of rational measurements in the physical world
    – aabeshou
    Aug 25 '16 at 9:49



















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Relations between phenomena in Nature (φύσις, the origin of physics) existed probably before anyone thought about integers, rationals, and laws of physics. There is a legend about Kronecker saying integers were god-made, and the rest was a creation of mankind, but we leave that to the history of mathematics for the moment.



These relations therefore accommodate a variety of unit systems. The celerity of light, if constant, does not care about being measured in meter per second, or foot per lunar month.



Any instrument has finite precision, no phenomenon is "pure" enough to be measured precisely.



What matters, if laws are accurate enough, is to have sufficient precision. So, How Much Pi Do You Need?




NASA scientists keep the space station operational with only 15 or 16
significant digits of $pi$, and the fundamental constants of the
universe only require 32.




So, with "quadruple precision" you can almost work with "decimals", with respect to your floating point system. Which is much simple than with certain long-period rationals: $1/97$ has a period of 96 digits.



Finally, for $pi$, even if you cannot work with infinite precision, you can still use indefinite precision, adding a decimal when you want, using the Bailey-Borwein-Plouffe formula, to directly calculate the value of any given digit (base 16) of $pi$ without calculating the preceding digits:



$$pi =sum _{k=0}^{infty }left[{frac {1}{16^{k}}}left({frac {4}{8k+1}}-{frac {2}{8k+4}}-{frac {1}{8k+5}}-{frac {1}{8k+6}}right)right],.$$






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    2














    An interesting question. It relates mathematical, i.e. mental concepts to "reality". Let us assume for the purpose of this discussion that there is a reality, and that it exists outside of our mind. Then the relation of the well-understood mental concept of irrational numbers and that reality depends on the properties of that reality (which, in all reality, are probably not completely known). I give three examples.





    1. A "continuous paradigm" reality



      If the underlying structure of space and matter were completely continuous (i.e. there were no Planck constant, no atoms etc.), all properties of natural objects like length or weight were irrational: Because there are infinitely more irrational than rational numbers, it is virtually impossible to properly hit, say, 2 cm when cutting a piece of wood. It would always be 2cm + "random delta" (however small), and that delta will "always" be irrational. The same goes for the International Prototype Kilogram and all other artifacts and natural phenomena.





    2. A blocky reality



      If reality were like Minecraft, and all physical properties were "blocky" (albeit at an atomic scale), all properties would be rational: Everything is a multiple of block lengths, weights, times etc. and thus all properties can be expressed as fractions of integers.





    3. Our quantum world



      Our reality is a bit like Minecraft, except that things get fuzzy if you look too close. This means that no property of an object we want to measure can be measured exactly; on a fundamental level there is always a, well, uncertainty. (This is not a deficiency in our way of measuring -- those errors would come on top --, but a property of our universe.) Because exactness is limited one needs not be afraid of running out of decimal places.



      Any interval produced by the amount of uncertainty has many rational numbers in them; so in a way you can probably get away without using irrational numbers in describing reality. (But you could also get away without ever using any rational numbers, mind you). Any existing circle can have both: A rational diameter and a rational circumference, as far as we can tell (+/- the Planck length).








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      1














      Take a line segment of length 1. With compass and straight edge it is possible to construct a line segment starting from its endpoint that is perpendicular to it and also of length 1. Draw that line segment. Now draw the line connecting the two end points.



      By the Pythagorean Theorem you have just drawn a line segment of length $sqrt{2}$.



      It is a fairly straightforward proof to show that this is an irrational number.



      So it exists.



      Now the question is - what law of physics you would believe this to violate. Sure - if you try to measure it, you won't be able to measure it to that accuracy (and okay, you probably didn't draw it to that accuracy), but assuming that you had the ability to exactly use a compass and straightedge then you would have drawn something with exactly length $sqrt{2}$.



      This is why the Pythagorean Theorem was a bit of a scandal. Up until this point it was assumed that for any two line segments we could find a smaller line segment that went into both of them evenly. This assumption made proofs about similar triangles quite easy. Mathematicians knew that $sqrt{2}$ wasn't rational, but they could just argue no such length exists (much like $-1$ used to have no square root). The Pythagorean Theorem showed that $sqrt{2}$ does too exist as a length.






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        0














        Certain natural phenomena in probability theory such as Buffon's needle experiment have answers containing an irrational constant, namely $pi$. Therefore such constants are arguably "in nature" (or any reasonable idealisation thereof) even if you are not particularly interested in the circle itself. The fact that there is an exact formula for the probability that involves an irrational number is precisely an argument against limiting precision in this fashion. Mathematicians have great respect for quantum theory but that theory involves a different type of idealisation of nature.






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        • 1




          The OP is arguing against infinite precision, and your example is exactly an infinite precision type of example. What happens if you limit your probability to something around the volume of the observable universe in Planck units. Using a uniform distribution, do $pi$ and other irrational numbers ever come up then?
          – Asaf Karagila♦
          Aug 27 '16 at 19:05










        • The fact that there is an exact formula for the probability that involves an irrational number is precisely an argument against limiting precision in this fashion. I have great respect for quantum theory and even published a paper in a quantum studies journal but you are confusing apples and oranges.
          – Mikhail Katz
          Aug 28 '16 at 11:11










        • If you don't want to read my comment, don't expect me to answer yours properly. Have a nice day.
          – Asaf Karagila♦
          Aug 28 '16 at 11:55



















        -1














        Technically, irrational numbers certainly exist. Just because we can't measure $pi$ to arbitrary precision doesn't mean not all of its digits can be calculated mathematically. The fine structure constant is built into the laws of Physics and we need to make observations to determine what its digits are. $pi$ on the other hand is a mathematical constant which is not built into the laws of Euclidean geometry and rather is derived from them and all of its digits can be calculated entirely mathematically and that constant can be proven to be irrational.



        $pi$ can't be measured exactly in our universe. According to https://physics.stackexchange.com/questions/395266/has-the-zeroth-law-of-thermodynamics-ever-been-proven/395270#395270, temperature probably can't be defined so I will instead describe how hot something is by its internal energy which can also be added or lost from the material. Here's one attempted way to measure $pi$. Take a large diamond sphere and drill a thin hole through the center and create a thin diamond chain that fits in that hole and measure its circumference and diameter while regulating its internal energy. There might be factors limiting how accurately you can measure it. For example, it has the highest shear modulus of all at about 5,000,000 atm but with a large enough sphere, the stretching of the chain from tension will limit the number of digits you can measure more than does the number of links. If your measurment is limited entirely my the number of links in the chain, you still can't measure $pi$ to arbitrary precision in that way because once the links get down to the size of a few atoms, Brownian motion of its atoms probably becomes significant which can sublimate away part of one of the links causing the chain to break. Despite that, Euclidean geometry still exists mathematically and all the digits of $pi$ can be calculated from it.






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        • How could you tell if $x$ is irrational or not, if you can only compute it to the billionth trillion decimal digit?
          – Asaf Karagila♦
          Nov 21 '18 at 23:32










        • @AsafKaragila Just because we don't know what any of the later digits are doesn't mean we don't know a certain property of the set of all digits after that place, the property of not repeating any finite string of digits over and over. To prove that $pi$ is irrational, it suffices to show that for any positive rational number less than 1, the $sin$ of it is not exactly $frac{1}{2}$. Any positive rational less than 1 can be expressed as $frac{p}{q}$ where the greatest common factor of p and q is 1 and p < q. The $sin$ of it is the limit of its Taylor series. The nth term must be a multiple
          – Timothy
          Nov 22 '18 at 1:46










        • of $frac{1}{n!q^n}$. Eventually you will get to the point where the absolute value of the sum of all the remaining terms is less than $frac{1}{n!q^n}$. I believe I can prove that the sum of all the terms after the nth terms adds up to something that is not a multiple of $frac{1}{n!q^n}$ but I'm not quite sure how to do that.
          – Timothy
          Nov 22 '18 at 3:21












        • How can you prove that $pi$ exists? As long as you can only measure finitely many digits, what makes you so sure that you're not just wrong? Or that your assumption in the proof that the value should be $pi$ include non-physical assumptions (e.g. that space is continuous and not discrete)?
          – Asaf Karagila♦
          Nov 22 '18 at 7:54










        • @AsafKaragila 2-dimensional Euclidean geometry is a well defined thing and sets of points with the property of being a circle exist. I guess we can call it an imaginary circle if we don't add an additional assumption that each point either has matter in it or it doesn't and the set of all points with matter in it is a circle. I don't yet know how to write a formal proof that $pi$ exists and is the same for all imaginary circles in Euclidean geometry but a complete formal proof would probably render my answer unreadable for a lot of people and a expert can probably figure out how to write a
          – Timothy
          Nov 22 '18 at 18:45











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        8 Answers
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        92














        I've got bad news for you: Even rational quantities don't exist in nature. Numbers, of whatever type, are just a way to describe our observations. But when we measure, we always have a measurement error, so we cannot really say "this measured quantity has the value $x$", all we can say is "this measured quantity is close to the value $x$". Well, we can do a bit more: Namely quantify how close it is. This is why measurement results are normally written in the form $xpm y$. For example, a length might be measured to be $1.3pm 0.5~rm mm$.



        Now we pretend that there is a "true value" of the measured quantity. For example, in the above example, we assume that there is a true value for the length, which is likely(!) less than $0.5~rm mm$ away from the value $1.3~rm mm$.



        Exact numbers are therefore nothing we can measure, and nothing which we can even prove really exist in nature. They are abstract concepts we made up in our mind in order to describe what we observe. And they work very well for this purpose. But whether they really correspond to something "out there", nobody can really know.



        But wait, you say, there are circles, and for those we can prove the circumference over diameter is $pi$. So $pi$ is part of our world, right? Wrong: You never have seen a circle. You've certainly seen lots of shapes that closely fit our concept of a circle. But likely if you look close enough, you'll find that it is not really a circle. And even if you cannot detect the difference, how do you know that if you had looked just a little bit closer, you'd have found a deviation?



        And ultimately, those circle-lookalikes you've seen probably were all made of atoms anyway, which means they could not have been a circle, as they have volume, while the circle is an infinitely thin line.






        share|cite|improve this answer





















        • Comments are not for extended discussion; this conversation has been moved to chat.
          – Pedro Tamaroff♦
          Sep 17 '16 at 22:14










        • You seem to be mentioning only our inability of precision measuring the lengths which exist in nature; our inability of not measuring lengths exactly as they are, seems to not imply their non existence.
          – Immortal Player
          Dec 20 '18 at 13:39
















        92














        I've got bad news for you: Even rational quantities don't exist in nature. Numbers, of whatever type, are just a way to describe our observations. But when we measure, we always have a measurement error, so we cannot really say "this measured quantity has the value $x$", all we can say is "this measured quantity is close to the value $x$". Well, we can do a bit more: Namely quantify how close it is. This is why measurement results are normally written in the form $xpm y$. For example, a length might be measured to be $1.3pm 0.5~rm mm$.



        Now we pretend that there is a "true value" of the measured quantity. For example, in the above example, we assume that there is a true value for the length, which is likely(!) less than $0.5~rm mm$ away from the value $1.3~rm mm$.



        Exact numbers are therefore nothing we can measure, and nothing which we can even prove really exist in nature. They are abstract concepts we made up in our mind in order to describe what we observe. And they work very well for this purpose. But whether they really correspond to something "out there", nobody can really know.



        But wait, you say, there are circles, and for those we can prove the circumference over diameter is $pi$. So $pi$ is part of our world, right? Wrong: You never have seen a circle. You've certainly seen lots of shapes that closely fit our concept of a circle. But likely if you look close enough, you'll find that it is not really a circle. And even if you cannot detect the difference, how do you know that if you had looked just a little bit closer, you'd have found a deviation?



        And ultimately, those circle-lookalikes you've seen probably were all made of atoms anyway, which means they could not have been a circle, as they have volume, while the circle is an infinitely thin line.






        share|cite|improve this answer





















        • Comments are not for extended discussion; this conversation has been moved to chat.
          – Pedro Tamaroff♦
          Sep 17 '16 at 22:14










        • You seem to be mentioning only our inability of precision measuring the lengths which exist in nature; our inability of not measuring lengths exactly as they are, seems to not imply their non existence.
          – Immortal Player
          Dec 20 '18 at 13:39














        92












        92








        92






        I've got bad news for you: Even rational quantities don't exist in nature. Numbers, of whatever type, are just a way to describe our observations. But when we measure, we always have a measurement error, so we cannot really say "this measured quantity has the value $x$", all we can say is "this measured quantity is close to the value $x$". Well, we can do a bit more: Namely quantify how close it is. This is why measurement results are normally written in the form $xpm y$. For example, a length might be measured to be $1.3pm 0.5~rm mm$.



        Now we pretend that there is a "true value" of the measured quantity. For example, in the above example, we assume that there is a true value for the length, which is likely(!) less than $0.5~rm mm$ away from the value $1.3~rm mm$.



        Exact numbers are therefore nothing we can measure, and nothing which we can even prove really exist in nature. They are abstract concepts we made up in our mind in order to describe what we observe. And they work very well for this purpose. But whether they really correspond to something "out there", nobody can really know.



        But wait, you say, there are circles, and for those we can prove the circumference over diameter is $pi$. So $pi$ is part of our world, right? Wrong: You never have seen a circle. You've certainly seen lots of shapes that closely fit our concept of a circle. But likely if you look close enough, you'll find that it is not really a circle. And even if you cannot detect the difference, how do you know that if you had looked just a little bit closer, you'd have found a deviation?



        And ultimately, those circle-lookalikes you've seen probably were all made of atoms anyway, which means they could not have been a circle, as they have volume, while the circle is an infinitely thin line.






        share|cite|improve this answer












        I've got bad news for you: Even rational quantities don't exist in nature. Numbers, of whatever type, are just a way to describe our observations. But when we measure, we always have a measurement error, so we cannot really say "this measured quantity has the value $x$", all we can say is "this measured quantity is close to the value $x$". Well, we can do a bit more: Namely quantify how close it is. This is why measurement results are normally written in the form $xpm y$. For example, a length might be measured to be $1.3pm 0.5~rm mm$.



        Now we pretend that there is a "true value" of the measured quantity. For example, in the above example, we assume that there is a true value for the length, which is likely(!) less than $0.5~rm mm$ away from the value $1.3~rm mm$.



        Exact numbers are therefore nothing we can measure, and nothing which we can even prove really exist in nature. They are abstract concepts we made up in our mind in order to describe what we observe. And they work very well for this purpose. But whether they really correspond to something "out there", nobody can really know.



        But wait, you say, there are circles, and for those we can prove the circumference over diameter is $pi$. So $pi$ is part of our world, right? Wrong: You never have seen a circle. You've certainly seen lots of shapes that closely fit our concept of a circle. But likely if you look close enough, you'll find that it is not really a circle. And even if you cannot detect the difference, how do you know that if you had looked just a little bit closer, you'd have found a deviation?



        And ultimately, those circle-lookalikes you've seen probably were all made of atoms anyway, which means they could not have been a circle, as they have volume, while the circle is an infinitely thin line.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 22 '16 at 12:05









        celtschkceltschk

        29.8k75599




        29.8k75599












        • Comments are not for extended discussion; this conversation has been moved to chat.
          – Pedro Tamaroff♦
          Sep 17 '16 at 22:14










        • You seem to be mentioning only our inability of precision measuring the lengths which exist in nature; our inability of not measuring lengths exactly as they are, seems to not imply their non existence.
          – Immortal Player
          Dec 20 '18 at 13:39


















        • Comments are not for extended discussion; this conversation has been moved to chat.
          – Pedro Tamaroff♦
          Sep 17 '16 at 22:14










        • You seem to be mentioning only our inability of precision measuring the lengths which exist in nature; our inability of not measuring lengths exactly as they are, seems to not imply their non existence.
          – Immortal Player
          Dec 20 '18 at 13:39
















        Comments are not for extended discussion; this conversation has been moved to chat.
        – Pedro Tamaroff♦
        Sep 17 '16 at 22:14




        Comments are not for extended discussion; this conversation has been moved to chat.
        – Pedro Tamaroff♦
        Sep 17 '16 at 22:14












        You seem to be mentioning only our inability of precision measuring the lengths which exist in nature; our inability of not measuring lengths exactly as they are, seems to not imply their non existence.
        – Immortal Player
        Dec 20 '18 at 13:39




        You seem to be mentioning only our inability of precision measuring the lengths which exist in nature; our inability of not measuring lengths exactly as they are, seems to not imply their non existence.
        – Immortal Player
        Dec 20 '18 at 13:39











        20














        One odd thing about the question is the apparent belief that the decimal notation of a number is somehow fundamental to its meaning, and so an unending decimal expansion is a sign that the number itself is somehow unending or imprecise.



        First, a simple factual point: the question claims:




        We know that an irrational no has well defined decimal values upto infinite decimal places




        I'm not exactly sure what this means, but I suspect it's not true. For example, we do have formulae that will tell us what the $n^mathrm{th}$ decimal digit of $pi$ is – in other words, we have a perfectly precise specification of any digit of $pi$ you care to ask for. So there's no fundamental sense in which $pi$ is less well-specified than, say $4.6$, which after all has an infinite decimal expansion as well, namely $4.600000dots$



        Anyway, this fixation on decimal expansions is unnecessary. Here's how I choose to imagine the "true nature" of a real number: the essence of a real number is that it measures some continuous quantity – for example, each real number is the length of some idealized line segment.



        Then decimal expansions are a way to understand these lengths by comparing them to fixed rational lengths that we know about, namely $1$, $0.1$, $0.01$, and etc. When we say $pi = 3.141dots$, we can view this as a series of comparisons, comparing $pi$ to $3$ and saying it is larger, comparing it to $3.15$ and saying it is smaller, etc.



        In this way, the decimal expansion isn't the "true nature" of a number, but simply as the view of a number you get when you look at it through the lens of powers of ten. This becomes more obvious when you realise that you can write the same number in different bases, view it through many different lenses, and indeed in some cases you get very different looking results, e.g. $1/3$ has an unending decimal expansion but a very simple ternary or nonary one.



        Now ask again what it means for some real number, the length of an idealized line segment, to be a rational number. Well, to be rational is to be $p/q$ for some whole numbers $p$ and $q$, so a number is rational precisely if some multiple of it is a whole number (namely, $q$ times it is $p$). What that means is you can take your line segment and duplicate it $q$ times, you get a line segment that is exactly $p$ times a segment of length $1$.



        Now the question becomes: why should it be rational? Why should that relationship ever hold exactly?






        share|cite|improve this answer





















        • "we do have formulae that will tell us what the nth decimal digit of π is " - I don't think this is possible in O(n)? (i.e. the only way is to compute all the digits up to the nth).
          – M.M
          Aug 22 '16 at 23:48






        • 3




          See Bailey–Borwein–Plouffe formula, although strictly speaking that's binary digits and not decimal digits. (In any case, computing all the digits up to the n$^text{th}$ suffices for the point I was trying to make).
          – Ben Millwood
          Aug 23 '16 at 6:39












        • Binary digits is no help in finding the decimal digits (without computing all digits up to the point)
          – M.M
          Aug 23 '16 at 7:06






        • 1




          Yeah, so you do need to compute all the digits up to the nth, I guess. I think that's fine.
          – Ben Millwood
          Aug 23 '16 at 8:07






        • 4




          @M.M does it matter that its not in O(n)? We do have a formulae for the nth decimal digit. Maybe not an super efficient formulae but still.
          – Jakob
          Aug 23 '16 at 8:14
















        20














        One odd thing about the question is the apparent belief that the decimal notation of a number is somehow fundamental to its meaning, and so an unending decimal expansion is a sign that the number itself is somehow unending or imprecise.



        First, a simple factual point: the question claims:




        We know that an irrational no has well defined decimal values upto infinite decimal places




        I'm not exactly sure what this means, but I suspect it's not true. For example, we do have formulae that will tell us what the $n^mathrm{th}$ decimal digit of $pi$ is – in other words, we have a perfectly precise specification of any digit of $pi$ you care to ask for. So there's no fundamental sense in which $pi$ is less well-specified than, say $4.6$, which after all has an infinite decimal expansion as well, namely $4.600000dots$



        Anyway, this fixation on decimal expansions is unnecessary. Here's how I choose to imagine the "true nature" of a real number: the essence of a real number is that it measures some continuous quantity – for example, each real number is the length of some idealized line segment.



        Then decimal expansions are a way to understand these lengths by comparing them to fixed rational lengths that we know about, namely $1$, $0.1$, $0.01$, and etc. When we say $pi = 3.141dots$, we can view this as a series of comparisons, comparing $pi$ to $3$ and saying it is larger, comparing it to $3.15$ and saying it is smaller, etc.



        In this way, the decimal expansion isn't the "true nature" of a number, but simply as the view of a number you get when you look at it through the lens of powers of ten. This becomes more obvious when you realise that you can write the same number in different bases, view it through many different lenses, and indeed in some cases you get very different looking results, e.g. $1/3$ has an unending decimal expansion but a very simple ternary or nonary one.



        Now ask again what it means for some real number, the length of an idealized line segment, to be a rational number. Well, to be rational is to be $p/q$ for some whole numbers $p$ and $q$, so a number is rational precisely if some multiple of it is a whole number (namely, $q$ times it is $p$). What that means is you can take your line segment and duplicate it $q$ times, you get a line segment that is exactly $p$ times a segment of length $1$.



        Now the question becomes: why should it be rational? Why should that relationship ever hold exactly?






        share|cite|improve this answer





















        • "we do have formulae that will tell us what the nth decimal digit of π is " - I don't think this is possible in O(n)? (i.e. the only way is to compute all the digits up to the nth).
          – M.M
          Aug 22 '16 at 23:48






        • 3




          See Bailey–Borwein–Plouffe formula, although strictly speaking that's binary digits and not decimal digits. (In any case, computing all the digits up to the n$^text{th}$ suffices for the point I was trying to make).
          – Ben Millwood
          Aug 23 '16 at 6:39












        • Binary digits is no help in finding the decimal digits (without computing all digits up to the point)
          – M.M
          Aug 23 '16 at 7:06






        • 1




          Yeah, so you do need to compute all the digits up to the nth, I guess. I think that's fine.
          – Ben Millwood
          Aug 23 '16 at 8:07






        • 4




          @M.M does it matter that its not in O(n)? We do have a formulae for the nth decimal digit. Maybe not an super efficient formulae but still.
          – Jakob
          Aug 23 '16 at 8:14














        20












        20








        20






        One odd thing about the question is the apparent belief that the decimal notation of a number is somehow fundamental to its meaning, and so an unending decimal expansion is a sign that the number itself is somehow unending or imprecise.



        First, a simple factual point: the question claims:




        We know that an irrational no has well defined decimal values upto infinite decimal places




        I'm not exactly sure what this means, but I suspect it's not true. For example, we do have formulae that will tell us what the $n^mathrm{th}$ decimal digit of $pi$ is – in other words, we have a perfectly precise specification of any digit of $pi$ you care to ask for. So there's no fundamental sense in which $pi$ is less well-specified than, say $4.6$, which after all has an infinite decimal expansion as well, namely $4.600000dots$



        Anyway, this fixation on decimal expansions is unnecessary. Here's how I choose to imagine the "true nature" of a real number: the essence of a real number is that it measures some continuous quantity – for example, each real number is the length of some idealized line segment.



        Then decimal expansions are a way to understand these lengths by comparing them to fixed rational lengths that we know about, namely $1$, $0.1$, $0.01$, and etc. When we say $pi = 3.141dots$, we can view this as a series of comparisons, comparing $pi$ to $3$ and saying it is larger, comparing it to $3.15$ and saying it is smaller, etc.



        In this way, the decimal expansion isn't the "true nature" of a number, but simply as the view of a number you get when you look at it through the lens of powers of ten. This becomes more obvious when you realise that you can write the same number in different bases, view it through many different lenses, and indeed in some cases you get very different looking results, e.g. $1/3$ has an unending decimal expansion but a very simple ternary or nonary one.



        Now ask again what it means for some real number, the length of an idealized line segment, to be a rational number. Well, to be rational is to be $p/q$ for some whole numbers $p$ and $q$, so a number is rational precisely if some multiple of it is a whole number (namely, $q$ times it is $p$). What that means is you can take your line segment and duplicate it $q$ times, you get a line segment that is exactly $p$ times a segment of length $1$.



        Now the question becomes: why should it be rational? Why should that relationship ever hold exactly?






        share|cite|improve this answer












        One odd thing about the question is the apparent belief that the decimal notation of a number is somehow fundamental to its meaning, and so an unending decimal expansion is a sign that the number itself is somehow unending or imprecise.



        First, a simple factual point: the question claims:




        We know that an irrational no has well defined decimal values upto infinite decimal places




        I'm not exactly sure what this means, but I suspect it's not true. For example, we do have formulae that will tell us what the $n^mathrm{th}$ decimal digit of $pi$ is – in other words, we have a perfectly precise specification of any digit of $pi$ you care to ask for. So there's no fundamental sense in which $pi$ is less well-specified than, say $4.6$, which after all has an infinite decimal expansion as well, namely $4.600000dots$



        Anyway, this fixation on decimal expansions is unnecessary. Here's how I choose to imagine the "true nature" of a real number: the essence of a real number is that it measures some continuous quantity – for example, each real number is the length of some idealized line segment.



        Then decimal expansions are a way to understand these lengths by comparing them to fixed rational lengths that we know about, namely $1$, $0.1$, $0.01$, and etc. When we say $pi = 3.141dots$, we can view this as a series of comparisons, comparing $pi$ to $3$ and saying it is larger, comparing it to $3.15$ and saying it is smaller, etc.



        In this way, the decimal expansion isn't the "true nature" of a number, but simply as the view of a number you get when you look at it through the lens of powers of ten. This becomes more obvious when you realise that you can write the same number in different bases, view it through many different lenses, and indeed in some cases you get very different looking results, e.g. $1/3$ has an unending decimal expansion but a very simple ternary or nonary one.



        Now ask again what it means for some real number, the length of an idealized line segment, to be a rational number. Well, to be rational is to be $p/q$ for some whole numbers $p$ and $q$, so a number is rational precisely if some multiple of it is a whole number (namely, $q$ times it is $p$). What that means is you can take your line segment and duplicate it $q$ times, you get a line segment that is exactly $p$ times a segment of length $1$.



        Now the question becomes: why should it be rational? Why should that relationship ever hold exactly?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 22 '16 at 17:09









        Ben MillwoodBen Millwood

        11.2k32049




        11.2k32049












        • "we do have formulae that will tell us what the nth decimal digit of π is " - I don't think this is possible in O(n)? (i.e. the only way is to compute all the digits up to the nth).
          – M.M
          Aug 22 '16 at 23:48






        • 3




          See Bailey–Borwein–Plouffe formula, although strictly speaking that's binary digits and not decimal digits. (In any case, computing all the digits up to the n$^text{th}$ suffices for the point I was trying to make).
          – Ben Millwood
          Aug 23 '16 at 6:39












        • Binary digits is no help in finding the decimal digits (without computing all digits up to the point)
          – M.M
          Aug 23 '16 at 7:06






        • 1




          Yeah, so you do need to compute all the digits up to the nth, I guess. I think that's fine.
          – Ben Millwood
          Aug 23 '16 at 8:07






        • 4




          @M.M does it matter that its not in O(n)? We do have a formulae for the nth decimal digit. Maybe not an super efficient formulae but still.
          – Jakob
          Aug 23 '16 at 8:14


















        • "we do have formulae that will tell us what the nth decimal digit of π is " - I don't think this is possible in O(n)? (i.e. the only way is to compute all the digits up to the nth).
          – M.M
          Aug 22 '16 at 23:48






        • 3




          See Bailey–Borwein–Plouffe formula, although strictly speaking that's binary digits and not decimal digits. (In any case, computing all the digits up to the n$^text{th}$ suffices for the point I was trying to make).
          – Ben Millwood
          Aug 23 '16 at 6:39












        • Binary digits is no help in finding the decimal digits (without computing all digits up to the point)
          – M.M
          Aug 23 '16 at 7:06






        • 1




          Yeah, so you do need to compute all the digits up to the nth, I guess. I think that's fine.
          – Ben Millwood
          Aug 23 '16 at 8:07






        • 4




          @M.M does it matter that its not in O(n)? We do have a formulae for the nth decimal digit. Maybe not an super efficient formulae but still.
          – Jakob
          Aug 23 '16 at 8:14
















        "we do have formulae that will tell us what the nth decimal digit of π is " - I don't think this is possible in O(n)? (i.e. the only way is to compute all the digits up to the nth).
        – M.M
        Aug 22 '16 at 23:48




        "we do have formulae that will tell us what the nth decimal digit of π is " - I don't think this is possible in O(n)? (i.e. the only way is to compute all the digits up to the nth).
        – M.M
        Aug 22 '16 at 23:48




        3




        3




        See Bailey–Borwein–Plouffe formula, although strictly speaking that's binary digits and not decimal digits. (In any case, computing all the digits up to the n$^text{th}$ suffices for the point I was trying to make).
        – Ben Millwood
        Aug 23 '16 at 6:39






        See Bailey–Borwein–Plouffe formula, although strictly speaking that's binary digits and not decimal digits. (In any case, computing all the digits up to the n$^text{th}$ suffices for the point I was trying to make).
        – Ben Millwood
        Aug 23 '16 at 6:39














        Binary digits is no help in finding the decimal digits (without computing all digits up to the point)
        – M.M
        Aug 23 '16 at 7:06




        Binary digits is no help in finding the decimal digits (without computing all digits up to the point)
        – M.M
        Aug 23 '16 at 7:06




        1




        1




        Yeah, so you do need to compute all the digits up to the nth, I guess. I think that's fine.
        – Ben Millwood
        Aug 23 '16 at 8:07




        Yeah, so you do need to compute all the digits up to the nth, I guess. I think that's fine.
        – Ben Millwood
        Aug 23 '16 at 8:07




        4




        4




        @M.M does it matter that its not in O(n)? We do have a formulae for the nth decimal digit. Maybe not an super efficient formulae but still.
        – Jakob
        Aug 23 '16 at 8:14




        @M.M does it matter that its not in O(n)? We do have a formulae for the nth decimal digit. Maybe not an super efficient formulae but still.
        – Jakob
        Aug 23 '16 at 8:14











        5














        Philosophically, you could consider an "irrational quantity" as the limit of a sequences of "rational quantities", so e.g. 2pi is what the circumference of a unit circle approaches as you make it more and more perfect.






        share|cite|improve this answer

















        • 2




          :-) The completeness of it all.
          – Simply Beautiful Art
          Aug 23 '16 at 21:57










        • More than philosophically, you can define real numbers as sets of sequences of rationals.
          – YoTengoUnLCD
          Aug 24 '16 at 1:32










        • @YoTengoUnLCD right but here I'm talking about sequences of rational measurements in the physical world
          – aabeshou
          Aug 25 '16 at 9:49
















        5














        Philosophically, you could consider an "irrational quantity" as the limit of a sequences of "rational quantities", so e.g. 2pi is what the circumference of a unit circle approaches as you make it more and more perfect.






        share|cite|improve this answer

















        • 2




          :-) The completeness of it all.
          – Simply Beautiful Art
          Aug 23 '16 at 21:57










        • More than philosophically, you can define real numbers as sets of sequences of rationals.
          – YoTengoUnLCD
          Aug 24 '16 at 1:32










        • @YoTengoUnLCD right but here I'm talking about sequences of rational measurements in the physical world
          – aabeshou
          Aug 25 '16 at 9:49














        5












        5








        5






        Philosophically, you could consider an "irrational quantity" as the limit of a sequences of "rational quantities", so e.g. 2pi is what the circumference of a unit circle approaches as you make it more and more perfect.






        share|cite|improve this answer












        Philosophically, you could consider an "irrational quantity" as the limit of a sequences of "rational quantities", so e.g. 2pi is what the circumference of a unit circle approaches as you make it more and more perfect.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 23 '16 at 0:12









        aabeshouaabeshou

        275110




        275110








        • 2




          :-) The completeness of it all.
          – Simply Beautiful Art
          Aug 23 '16 at 21:57










        • More than philosophically, you can define real numbers as sets of sequences of rationals.
          – YoTengoUnLCD
          Aug 24 '16 at 1:32










        • @YoTengoUnLCD right but here I'm talking about sequences of rational measurements in the physical world
          – aabeshou
          Aug 25 '16 at 9:49














        • 2




          :-) The completeness of it all.
          – Simply Beautiful Art
          Aug 23 '16 at 21:57










        • More than philosophically, you can define real numbers as sets of sequences of rationals.
          – YoTengoUnLCD
          Aug 24 '16 at 1:32










        • @YoTengoUnLCD right but here I'm talking about sequences of rational measurements in the physical world
          – aabeshou
          Aug 25 '16 at 9:49








        2




        2




        :-) The completeness of it all.
        – Simply Beautiful Art
        Aug 23 '16 at 21:57




        :-) The completeness of it all.
        – Simply Beautiful Art
        Aug 23 '16 at 21:57












        More than philosophically, you can define real numbers as sets of sequences of rationals.
        – YoTengoUnLCD
        Aug 24 '16 at 1:32




        More than philosophically, you can define real numbers as sets of sequences of rationals.
        – YoTengoUnLCD
        Aug 24 '16 at 1:32












        @YoTengoUnLCD right but here I'm talking about sequences of rational measurements in the physical world
        – aabeshou
        Aug 25 '16 at 9:49




        @YoTengoUnLCD right but here I'm talking about sequences of rational measurements in the physical world
        – aabeshou
        Aug 25 '16 at 9:49











        2














        Relations between phenomena in Nature (φύσις, the origin of physics) existed probably before anyone thought about integers, rationals, and laws of physics. There is a legend about Kronecker saying integers were god-made, and the rest was a creation of mankind, but we leave that to the history of mathematics for the moment.



        These relations therefore accommodate a variety of unit systems. The celerity of light, if constant, does not care about being measured in meter per second, or foot per lunar month.



        Any instrument has finite precision, no phenomenon is "pure" enough to be measured precisely.



        What matters, if laws are accurate enough, is to have sufficient precision. So, How Much Pi Do You Need?




        NASA scientists keep the space station operational with only 15 or 16
        significant digits of $pi$, and the fundamental constants of the
        universe only require 32.




        So, with "quadruple precision" you can almost work with "decimals", with respect to your floating point system. Which is much simple than with certain long-period rationals: $1/97$ has a period of 96 digits.



        Finally, for $pi$, even if you cannot work with infinite precision, you can still use indefinite precision, adding a decimal when you want, using the Bailey-Borwein-Plouffe formula, to directly calculate the value of any given digit (base 16) of $pi$ without calculating the preceding digits:



        $$pi =sum _{k=0}^{infty }left[{frac {1}{16^{k}}}left({frac {4}{8k+1}}-{frac {2}{8k+4}}-{frac {1}{8k+5}}-{frac {1}{8k+6}}right)right],.$$






        share|cite|improve this answer




























          2














          Relations between phenomena in Nature (φύσις, the origin of physics) existed probably before anyone thought about integers, rationals, and laws of physics. There is a legend about Kronecker saying integers were god-made, and the rest was a creation of mankind, but we leave that to the history of mathematics for the moment.



          These relations therefore accommodate a variety of unit systems. The celerity of light, if constant, does not care about being measured in meter per second, or foot per lunar month.



          Any instrument has finite precision, no phenomenon is "pure" enough to be measured precisely.



          What matters, if laws are accurate enough, is to have sufficient precision. So, How Much Pi Do You Need?




          NASA scientists keep the space station operational with only 15 or 16
          significant digits of $pi$, and the fundamental constants of the
          universe only require 32.




          So, with "quadruple precision" you can almost work with "decimals", with respect to your floating point system. Which is much simple than with certain long-period rationals: $1/97$ has a period of 96 digits.



          Finally, for $pi$, even if you cannot work with infinite precision, you can still use indefinite precision, adding a decimal when you want, using the Bailey-Borwein-Plouffe formula, to directly calculate the value of any given digit (base 16) of $pi$ without calculating the preceding digits:



          $$pi =sum _{k=0}^{infty }left[{frac {1}{16^{k}}}left({frac {4}{8k+1}}-{frac {2}{8k+4}}-{frac {1}{8k+5}}-{frac {1}{8k+6}}right)right],.$$






          share|cite|improve this answer


























            2












            2








            2






            Relations between phenomena in Nature (φύσις, the origin of physics) existed probably before anyone thought about integers, rationals, and laws of physics. There is a legend about Kronecker saying integers were god-made, and the rest was a creation of mankind, but we leave that to the history of mathematics for the moment.



            These relations therefore accommodate a variety of unit systems. The celerity of light, if constant, does not care about being measured in meter per second, or foot per lunar month.



            Any instrument has finite precision, no phenomenon is "pure" enough to be measured precisely.



            What matters, if laws are accurate enough, is to have sufficient precision. So, How Much Pi Do You Need?




            NASA scientists keep the space station operational with only 15 or 16
            significant digits of $pi$, and the fundamental constants of the
            universe only require 32.




            So, with "quadruple precision" you can almost work with "decimals", with respect to your floating point system. Which is much simple than with certain long-period rationals: $1/97$ has a period of 96 digits.



            Finally, for $pi$, even if you cannot work with infinite precision, you can still use indefinite precision, adding a decimal when you want, using the Bailey-Borwein-Plouffe formula, to directly calculate the value of any given digit (base 16) of $pi$ without calculating the preceding digits:



            $$pi =sum _{k=0}^{infty }left[{frac {1}{16^{k}}}left({frac {4}{8k+1}}-{frac {2}{8k+4}}-{frac {1}{8k+5}}-{frac {1}{8k+6}}right)right],.$$






            share|cite|improve this answer














            Relations between phenomena in Nature (φύσις, the origin of physics) existed probably before anyone thought about integers, rationals, and laws of physics. There is a legend about Kronecker saying integers were god-made, and the rest was a creation of mankind, but we leave that to the history of mathematics for the moment.



            These relations therefore accommodate a variety of unit systems. The celerity of light, if constant, does not care about being measured in meter per second, or foot per lunar month.



            Any instrument has finite precision, no phenomenon is "pure" enough to be measured precisely.



            What matters, if laws are accurate enough, is to have sufficient precision. So, How Much Pi Do You Need?




            NASA scientists keep the space station operational with only 15 or 16
            significant digits of $pi$, and the fundamental constants of the
            universe only require 32.




            So, with "quadruple precision" you can almost work with "decimals", with respect to your floating point system. Which is much simple than with certain long-period rationals: $1/97$ has a period of 96 digits.



            Finally, for $pi$, even if you cannot work with infinite precision, you can still use indefinite precision, adding a decimal when you want, using the Bailey-Borwein-Plouffe formula, to directly calculate the value of any given digit (base 16) of $pi$ without calculating the preceding digits:



            $$pi =sum _{k=0}^{infty }left[{frac {1}{16^{k}}}left({frac {4}{8k+1}}-{frac {2}{8k+4}}-{frac {1}{8k+5}}-{frac {1}{8k+6}}right)right],.$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Apr 13 '17 at 12:51









            Community♦

            1




            1










            answered Aug 23 '16 at 15:47









            Laurent DuvalLaurent Duval

            5,29811239




            5,29811239























                2














                An interesting question. It relates mathematical, i.e. mental concepts to "reality". Let us assume for the purpose of this discussion that there is a reality, and that it exists outside of our mind. Then the relation of the well-understood mental concept of irrational numbers and that reality depends on the properties of that reality (which, in all reality, are probably not completely known). I give three examples.





                1. A "continuous paradigm" reality



                  If the underlying structure of space and matter were completely continuous (i.e. there were no Planck constant, no atoms etc.), all properties of natural objects like length or weight were irrational: Because there are infinitely more irrational than rational numbers, it is virtually impossible to properly hit, say, 2 cm when cutting a piece of wood. It would always be 2cm + "random delta" (however small), and that delta will "always" be irrational. The same goes for the International Prototype Kilogram and all other artifacts and natural phenomena.





                2. A blocky reality



                  If reality were like Minecraft, and all physical properties were "blocky" (albeit at an atomic scale), all properties would be rational: Everything is a multiple of block lengths, weights, times etc. and thus all properties can be expressed as fractions of integers.





                3. Our quantum world



                  Our reality is a bit like Minecraft, except that things get fuzzy if you look too close. This means that no property of an object we want to measure can be measured exactly; on a fundamental level there is always a, well, uncertainty. (This is not a deficiency in our way of measuring -- those errors would come on top --, but a property of our universe.) Because exactness is limited one needs not be afraid of running out of decimal places.



                  Any interval produced by the amount of uncertainty has many rational numbers in them; so in a way you can probably get away without using irrational numbers in describing reality. (But you could also get away without ever using any rational numbers, mind you). Any existing circle can have both: A rational diameter and a rational circumference, as far as we can tell (+/- the Planck length).








                share|cite|improve this answer




























                  2














                  An interesting question. It relates mathematical, i.e. mental concepts to "reality". Let us assume for the purpose of this discussion that there is a reality, and that it exists outside of our mind. Then the relation of the well-understood mental concept of irrational numbers and that reality depends on the properties of that reality (which, in all reality, are probably not completely known). I give three examples.





                  1. A "continuous paradigm" reality



                    If the underlying structure of space and matter were completely continuous (i.e. there were no Planck constant, no atoms etc.), all properties of natural objects like length or weight were irrational: Because there are infinitely more irrational than rational numbers, it is virtually impossible to properly hit, say, 2 cm when cutting a piece of wood. It would always be 2cm + "random delta" (however small), and that delta will "always" be irrational. The same goes for the International Prototype Kilogram and all other artifacts and natural phenomena.





                  2. A blocky reality



                    If reality were like Minecraft, and all physical properties were "blocky" (albeit at an atomic scale), all properties would be rational: Everything is a multiple of block lengths, weights, times etc. and thus all properties can be expressed as fractions of integers.





                  3. Our quantum world



                    Our reality is a bit like Minecraft, except that things get fuzzy if you look too close. This means that no property of an object we want to measure can be measured exactly; on a fundamental level there is always a, well, uncertainty. (This is not a deficiency in our way of measuring -- those errors would come on top --, but a property of our universe.) Because exactness is limited one needs not be afraid of running out of decimal places.



                    Any interval produced by the amount of uncertainty has many rational numbers in them; so in a way you can probably get away without using irrational numbers in describing reality. (But you could also get away without ever using any rational numbers, mind you). Any existing circle can have both: A rational diameter and a rational circumference, as far as we can tell (+/- the Planck length).








                  share|cite|improve this answer


























                    2












                    2








                    2






                    An interesting question. It relates mathematical, i.e. mental concepts to "reality". Let us assume for the purpose of this discussion that there is a reality, and that it exists outside of our mind. Then the relation of the well-understood mental concept of irrational numbers and that reality depends on the properties of that reality (which, in all reality, are probably not completely known). I give three examples.





                    1. A "continuous paradigm" reality



                      If the underlying structure of space and matter were completely continuous (i.e. there were no Planck constant, no atoms etc.), all properties of natural objects like length or weight were irrational: Because there are infinitely more irrational than rational numbers, it is virtually impossible to properly hit, say, 2 cm when cutting a piece of wood. It would always be 2cm + "random delta" (however small), and that delta will "always" be irrational. The same goes for the International Prototype Kilogram and all other artifacts and natural phenomena.





                    2. A blocky reality



                      If reality were like Minecraft, and all physical properties were "blocky" (albeit at an atomic scale), all properties would be rational: Everything is a multiple of block lengths, weights, times etc. and thus all properties can be expressed as fractions of integers.





                    3. Our quantum world



                      Our reality is a bit like Minecraft, except that things get fuzzy if you look too close. This means that no property of an object we want to measure can be measured exactly; on a fundamental level there is always a, well, uncertainty. (This is not a deficiency in our way of measuring -- those errors would come on top --, but a property of our universe.) Because exactness is limited one needs not be afraid of running out of decimal places.



                      Any interval produced by the amount of uncertainty has many rational numbers in them; so in a way you can probably get away without using irrational numbers in describing reality. (But you could also get away without ever using any rational numbers, mind you). Any existing circle can have both: A rational diameter and a rational circumference, as far as we can tell (+/- the Planck length).








                    share|cite|improve this answer














                    An interesting question. It relates mathematical, i.e. mental concepts to "reality". Let us assume for the purpose of this discussion that there is a reality, and that it exists outside of our mind. Then the relation of the well-understood mental concept of irrational numbers and that reality depends on the properties of that reality (which, in all reality, are probably not completely known). I give three examples.





                    1. A "continuous paradigm" reality



                      If the underlying structure of space and matter were completely continuous (i.e. there were no Planck constant, no atoms etc.), all properties of natural objects like length or weight were irrational: Because there are infinitely more irrational than rational numbers, it is virtually impossible to properly hit, say, 2 cm when cutting a piece of wood. It would always be 2cm + "random delta" (however small), and that delta will "always" be irrational. The same goes for the International Prototype Kilogram and all other artifacts and natural phenomena.





                    2. A blocky reality



                      If reality were like Minecraft, and all physical properties were "blocky" (albeit at an atomic scale), all properties would be rational: Everything is a multiple of block lengths, weights, times etc. and thus all properties can be expressed as fractions of integers.





                    3. Our quantum world



                      Our reality is a bit like Minecraft, except that things get fuzzy if you look too close. This means that no property of an object we want to measure can be measured exactly; on a fundamental level there is always a, well, uncertainty. (This is not a deficiency in our way of measuring -- those errors would come on top --, but a property of our universe.) Because exactness is limited one needs not be afraid of running out of decimal places.



                      Any interval produced by the amount of uncertainty has many rational numbers in them; so in a way you can probably get away without using irrational numbers in describing reality. (But you could also get away without ever using any rational numbers, mind you). Any existing circle can have both: A rational diameter and a rational circumference, as far as we can tell (+/- the Planck length).









                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Aug 23 '16 at 23:12

























                    answered Aug 23 '16 at 15:05









                    Peter A. SchneiderPeter A. Schneider

                    228110




                    228110























                        1














                        Take a line segment of length 1. With compass and straight edge it is possible to construct a line segment starting from its endpoint that is perpendicular to it and also of length 1. Draw that line segment. Now draw the line connecting the two end points.



                        By the Pythagorean Theorem you have just drawn a line segment of length $sqrt{2}$.



                        It is a fairly straightforward proof to show that this is an irrational number.



                        So it exists.



                        Now the question is - what law of physics you would believe this to violate. Sure - if you try to measure it, you won't be able to measure it to that accuracy (and okay, you probably didn't draw it to that accuracy), but assuming that you had the ability to exactly use a compass and straightedge then you would have drawn something with exactly length $sqrt{2}$.



                        This is why the Pythagorean Theorem was a bit of a scandal. Up until this point it was assumed that for any two line segments we could find a smaller line segment that went into both of them evenly. This assumption made proofs about similar triangles quite easy. Mathematicians knew that $sqrt{2}$ wasn't rational, but they could just argue no such length exists (much like $-1$ used to have no square root). The Pythagorean Theorem showed that $sqrt{2}$ does too exist as a length.






                        share|cite|improve this answer


























                          1














                          Take a line segment of length 1. With compass and straight edge it is possible to construct a line segment starting from its endpoint that is perpendicular to it and also of length 1. Draw that line segment. Now draw the line connecting the two end points.



                          By the Pythagorean Theorem you have just drawn a line segment of length $sqrt{2}$.



                          It is a fairly straightforward proof to show that this is an irrational number.



                          So it exists.



                          Now the question is - what law of physics you would believe this to violate. Sure - if you try to measure it, you won't be able to measure it to that accuracy (and okay, you probably didn't draw it to that accuracy), but assuming that you had the ability to exactly use a compass and straightedge then you would have drawn something with exactly length $sqrt{2}$.



                          This is why the Pythagorean Theorem was a bit of a scandal. Up until this point it was assumed that for any two line segments we could find a smaller line segment that went into both of them evenly. This assumption made proofs about similar triangles quite easy. Mathematicians knew that $sqrt{2}$ wasn't rational, but they could just argue no such length exists (much like $-1$ used to have no square root). The Pythagorean Theorem showed that $sqrt{2}$ does too exist as a length.






                          share|cite|improve this answer
























                            1












                            1








                            1






                            Take a line segment of length 1. With compass and straight edge it is possible to construct a line segment starting from its endpoint that is perpendicular to it and also of length 1. Draw that line segment. Now draw the line connecting the two end points.



                            By the Pythagorean Theorem you have just drawn a line segment of length $sqrt{2}$.



                            It is a fairly straightforward proof to show that this is an irrational number.



                            So it exists.



                            Now the question is - what law of physics you would believe this to violate. Sure - if you try to measure it, you won't be able to measure it to that accuracy (and okay, you probably didn't draw it to that accuracy), but assuming that you had the ability to exactly use a compass and straightedge then you would have drawn something with exactly length $sqrt{2}$.



                            This is why the Pythagorean Theorem was a bit of a scandal. Up until this point it was assumed that for any two line segments we could find a smaller line segment that went into both of them evenly. This assumption made proofs about similar triangles quite easy. Mathematicians knew that $sqrt{2}$ wasn't rational, but they could just argue no such length exists (much like $-1$ used to have no square root). The Pythagorean Theorem showed that $sqrt{2}$ does too exist as a length.






                            share|cite|improve this answer












                            Take a line segment of length 1. With compass and straight edge it is possible to construct a line segment starting from its endpoint that is perpendicular to it and also of length 1. Draw that line segment. Now draw the line connecting the two end points.



                            By the Pythagorean Theorem you have just drawn a line segment of length $sqrt{2}$.



                            It is a fairly straightforward proof to show that this is an irrational number.



                            So it exists.



                            Now the question is - what law of physics you would believe this to violate. Sure - if you try to measure it, you won't be able to measure it to that accuracy (and okay, you probably didn't draw it to that accuracy), but assuming that you had the ability to exactly use a compass and straightedge then you would have drawn something with exactly length $sqrt{2}$.



                            This is why the Pythagorean Theorem was a bit of a scandal. Up until this point it was assumed that for any two line segments we could find a smaller line segment that went into both of them evenly. This assumption made proofs about similar triangles quite easy. Mathematicians knew that $sqrt{2}$ wasn't rational, but they could just argue no such length exists (much like $-1$ used to have no square root). The Pythagorean Theorem showed that $sqrt{2}$ does too exist as a length.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Aug 24 '16 at 4:35









                            JoelJoel

                            434210




                            434210























                                0














                                Certain natural phenomena in probability theory such as Buffon's needle experiment have answers containing an irrational constant, namely $pi$. Therefore such constants are arguably "in nature" (or any reasonable idealisation thereof) even if you are not particularly interested in the circle itself. The fact that there is an exact formula for the probability that involves an irrational number is precisely an argument against limiting precision in this fashion. Mathematicians have great respect for quantum theory but that theory involves a different type of idealisation of nature.






                                share|cite|improve this answer



















                                • 1




                                  The OP is arguing against infinite precision, and your example is exactly an infinite precision type of example. What happens if you limit your probability to something around the volume of the observable universe in Planck units. Using a uniform distribution, do $pi$ and other irrational numbers ever come up then?
                                  – Asaf Karagila♦
                                  Aug 27 '16 at 19:05










                                • The fact that there is an exact formula for the probability that involves an irrational number is precisely an argument against limiting precision in this fashion. I have great respect for quantum theory and even published a paper in a quantum studies journal but you are confusing apples and oranges.
                                  – Mikhail Katz
                                  Aug 28 '16 at 11:11










                                • If you don't want to read my comment, don't expect me to answer yours properly. Have a nice day.
                                  – Asaf Karagila♦
                                  Aug 28 '16 at 11:55
















                                0














                                Certain natural phenomena in probability theory such as Buffon's needle experiment have answers containing an irrational constant, namely $pi$. Therefore such constants are arguably "in nature" (or any reasonable idealisation thereof) even if you are not particularly interested in the circle itself. The fact that there is an exact formula for the probability that involves an irrational number is precisely an argument against limiting precision in this fashion. Mathematicians have great respect for quantum theory but that theory involves a different type of idealisation of nature.






                                share|cite|improve this answer



















                                • 1




                                  The OP is arguing against infinite precision, and your example is exactly an infinite precision type of example. What happens if you limit your probability to something around the volume of the observable universe in Planck units. Using a uniform distribution, do $pi$ and other irrational numbers ever come up then?
                                  – Asaf Karagila♦
                                  Aug 27 '16 at 19:05










                                • The fact that there is an exact formula for the probability that involves an irrational number is precisely an argument against limiting precision in this fashion. I have great respect for quantum theory and even published a paper in a quantum studies journal but you are confusing apples and oranges.
                                  – Mikhail Katz
                                  Aug 28 '16 at 11:11










                                • If you don't want to read my comment, don't expect me to answer yours properly. Have a nice day.
                                  – Asaf Karagila♦
                                  Aug 28 '16 at 11:55














                                0












                                0








                                0






                                Certain natural phenomena in probability theory such as Buffon's needle experiment have answers containing an irrational constant, namely $pi$. Therefore such constants are arguably "in nature" (or any reasonable idealisation thereof) even if you are not particularly interested in the circle itself. The fact that there is an exact formula for the probability that involves an irrational number is precisely an argument against limiting precision in this fashion. Mathematicians have great respect for quantum theory but that theory involves a different type of idealisation of nature.






                                share|cite|improve this answer














                                Certain natural phenomena in probability theory such as Buffon's needle experiment have answers containing an irrational constant, namely $pi$. Therefore such constants are arguably "in nature" (or any reasonable idealisation thereof) even if you are not particularly interested in the circle itself. The fact that there is an exact formula for the probability that involves an irrational number is precisely an argument against limiting precision in this fashion. Mathematicians have great respect for quantum theory but that theory involves a different type of idealisation of nature.







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Aug 28 '16 at 11:52

























                                answered Aug 27 '16 at 18:59









                                Mikhail KatzMikhail Katz

                                30.6k14298




                                30.6k14298








                                • 1




                                  The OP is arguing against infinite precision, and your example is exactly an infinite precision type of example. What happens if you limit your probability to something around the volume of the observable universe in Planck units. Using a uniform distribution, do $pi$ and other irrational numbers ever come up then?
                                  – Asaf Karagila♦
                                  Aug 27 '16 at 19:05










                                • The fact that there is an exact formula for the probability that involves an irrational number is precisely an argument against limiting precision in this fashion. I have great respect for quantum theory and even published a paper in a quantum studies journal but you are confusing apples and oranges.
                                  – Mikhail Katz
                                  Aug 28 '16 at 11:11










                                • If you don't want to read my comment, don't expect me to answer yours properly. Have a nice day.
                                  – Asaf Karagila♦
                                  Aug 28 '16 at 11:55














                                • 1




                                  The OP is arguing against infinite precision, and your example is exactly an infinite precision type of example. What happens if you limit your probability to something around the volume of the observable universe in Planck units. Using a uniform distribution, do $pi$ and other irrational numbers ever come up then?
                                  – Asaf Karagila♦
                                  Aug 27 '16 at 19:05










                                • The fact that there is an exact formula for the probability that involves an irrational number is precisely an argument against limiting precision in this fashion. I have great respect for quantum theory and even published a paper in a quantum studies journal but you are confusing apples and oranges.
                                  – Mikhail Katz
                                  Aug 28 '16 at 11:11










                                • If you don't want to read my comment, don't expect me to answer yours properly. Have a nice day.
                                  – Asaf Karagila♦
                                  Aug 28 '16 at 11:55








                                1




                                1




                                The OP is arguing against infinite precision, and your example is exactly an infinite precision type of example. What happens if you limit your probability to something around the volume of the observable universe in Planck units. Using a uniform distribution, do $pi$ and other irrational numbers ever come up then?
                                – Asaf Karagila♦
                                Aug 27 '16 at 19:05




                                The OP is arguing against infinite precision, and your example is exactly an infinite precision type of example. What happens if you limit your probability to something around the volume of the observable universe in Planck units. Using a uniform distribution, do $pi$ and other irrational numbers ever come up then?
                                – Asaf Karagila♦
                                Aug 27 '16 at 19:05












                                The fact that there is an exact formula for the probability that involves an irrational number is precisely an argument against limiting precision in this fashion. I have great respect for quantum theory and even published a paper in a quantum studies journal but you are confusing apples and oranges.
                                – Mikhail Katz
                                Aug 28 '16 at 11:11




                                The fact that there is an exact formula for the probability that involves an irrational number is precisely an argument against limiting precision in this fashion. I have great respect for quantum theory and even published a paper in a quantum studies journal but you are confusing apples and oranges.
                                – Mikhail Katz
                                Aug 28 '16 at 11:11












                                If you don't want to read my comment, don't expect me to answer yours properly. Have a nice day.
                                – Asaf Karagila♦
                                Aug 28 '16 at 11:55




                                If you don't want to read my comment, don't expect me to answer yours properly. Have a nice day.
                                – Asaf Karagila♦
                                Aug 28 '16 at 11:55











                                -1














                                Technically, irrational numbers certainly exist. Just because we can't measure $pi$ to arbitrary precision doesn't mean not all of its digits can be calculated mathematically. The fine structure constant is built into the laws of Physics and we need to make observations to determine what its digits are. $pi$ on the other hand is a mathematical constant which is not built into the laws of Euclidean geometry and rather is derived from them and all of its digits can be calculated entirely mathematically and that constant can be proven to be irrational.



                                $pi$ can't be measured exactly in our universe. According to https://physics.stackexchange.com/questions/395266/has-the-zeroth-law-of-thermodynamics-ever-been-proven/395270#395270, temperature probably can't be defined so I will instead describe how hot something is by its internal energy which can also be added or lost from the material. Here's one attempted way to measure $pi$. Take a large diamond sphere and drill a thin hole through the center and create a thin diamond chain that fits in that hole and measure its circumference and diameter while regulating its internal energy. There might be factors limiting how accurately you can measure it. For example, it has the highest shear modulus of all at about 5,000,000 atm but with a large enough sphere, the stretching of the chain from tension will limit the number of digits you can measure more than does the number of links. If your measurment is limited entirely my the number of links in the chain, you still can't measure $pi$ to arbitrary precision in that way because once the links get down to the size of a few atoms, Brownian motion of its atoms probably becomes significant which can sublimate away part of one of the links causing the chain to break. Despite that, Euclidean geometry still exists mathematically and all the digits of $pi$ can be calculated from it.






                                share|cite|improve this answer























                                • How could you tell if $x$ is irrational or not, if you can only compute it to the billionth trillion decimal digit?
                                  – Asaf Karagila♦
                                  Nov 21 '18 at 23:32










                                • @AsafKaragila Just because we don't know what any of the later digits are doesn't mean we don't know a certain property of the set of all digits after that place, the property of not repeating any finite string of digits over and over. To prove that $pi$ is irrational, it suffices to show that for any positive rational number less than 1, the $sin$ of it is not exactly $frac{1}{2}$. Any positive rational less than 1 can be expressed as $frac{p}{q}$ where the greatest common factor of p and q is 1 and p < q. The $sin$ of it is the limit of its Taylor series. The nth term must be a multiple
                                  – Timothy
                                  Nov 22 '18 at 1:46










                                • of $frac{1}{n!q^n}$. Eventually you will get to the point where the absolute value of the sum of all the remaining terms is less than $frac{1}{n!q^n}$. I believe I can prove that the sum of all the terms after the nth terms adds up to something that is not a multiple of $frac{1}{n!q^n}$ but I'm not quite sure how to do that.
                                  – Timothy
                                  Nov 22 '18 at 3:21












                                • How can you prove that $pi$ exists? As long as you can only measure finitely many digits, what makes you so sure that you're not just wrong? Or that your assumption in the proof that the value should be $pi$ include non-physical assumptions (e.g. that space is continuous and not discrete)?
                                  – Asaf Karagila♦
                                  Nov 22 '18 at 7:54










                                • @AsafKaragila 2-dimensional Euclidean geometry is a well defined thing and sets of points with the property of being a circle exist. I guess we can call it an imaginary circle if we don't add an additional assumption that each point either has matter in it or it doesn't and the set of all points with matter in it is a circle. I don't yet know how to write a formal proof that $pi$ exists and is the same for all imaginary circles in Euclidean geometry but a complete formal proof would probably render my answer unreadable for a lot of people and a expert can probably figure out how to write a
                                  – Timothy
                                  Nov 22 '18 at 18:45
















                                -1














                                Technically, irrational numbers certainly exist. Just because we can't measure $pi$ to arbitrary precision doesn't mean not all of its digits can be calculated mathematically. The fine structure constant is built into the laws of Physics and we need to make observations to determine what its digits are. $pi$ on the other hand is a mathematical constant which is not built into the laws of Euclidean geometry and rather is derived from them and all of its digits can be calculated entirely mathematically and that constant can be proven to be irrational.



                                $pi$ can't be measured exactly in our universe. According to https://physics.stackexchange.com/questions/395266/has-the-zeroth-law-of-thermodynamics-ever-been-proven/395270#395270, temperature probably can't be defined so I will instead describe how hot something is by its internal energy which can also be added or lost from the material. Here's one attempted way to measure $pi$. Take a large diamond sphere and drill a thin hole through the center and create a thin diamond chain that fits in that hole and measure its circumference and diameter while regulating its internal energy. There might be factors limiting how accurately you can measure it. For example, it has the highest shear modulus of all at about 5,000,000 atm but with a large enough sphere, the stretching of the chain from tension will limit the number of digits you can measure more than does the number of links. If your measurment is limited entirely my the number of links in the chain, you still can't measure $pi$ to arbitrary precision in that way because once the links get down to the size of a few atoms, Brownian motion of its atoms probably becomes significant which can sublimate away part of one of the links causing the chain to break. Despite that, Euclidean geometry still exists mathematically and all the digits of $pi$ can be calculated from it.






                                share|cite|improve this answer























                                • How could you tell if $x$ is irrational or not, if you can only compute it to the billionth trillion decimal digit?
                                  – Asaf Karagila♦
                                  Nov 21 '18 at 23:32










                                • @AsafKaragila Just because we don't know what any of the later digits are doesn't mean we don't know a certain property of the set of all digits after that place, the property of not repeating any finite string of digits over and over. To prove that $pi$ is irrational, it suffices to show that for any positive rational number less than 1, the $sin$ of it is not exactly $frac{1}{2}$. Any positive rational less than 1 can be expressed as $frac{p}{q}$ where the greatest common factor of p and q is 1 and p < q. The $sin$ of it is the limit of its Taylor series. The nth term must be a multiple
                                  – Timothy
                                  Nov 22 '18 at 1:46










                                • of $frac{1}{n!q^n}$. Eventually you will get to the point where the absolute value of the sum of all the remaining terms is less than $frac{1}{n!q^n}$. I believe I can prove that the sum of all the terms after the nth terms adds up to something that is not a multiple of $frac{1}{n!q^n}$ but I'm not quite sure how to do that.
                                  – Timothy
                                  Nov 22 '18 at 3:21












                                • How can you prove that $pi$ exists? As long as you can only measure finitely many digits, what makes you so sure that you're not just wrong? Or that your assumption in the proof that the value should be $pi$ include non-physical assumptions (e.g. that space is continuous and not discrete)?
                                  – Asaf Karagila♦
                                  Nov 22 '18 at 7:54










                                • @AsafKaragila 2-dimensional Euclidean geometry is a well defined thing and sets of points with the property of being a circle exist. I guess we can call it an imaginary circle if we don't add an additional assumption that each point either has matter in it or it doesn't and the set of all points with matter in it is a circle. I don't yet know how to write a formal proof that $pi$ exists and is the same for all imaginary circles in Euclidean geometry but a complete formal proof would probably render my answer unreadable for a lot of people and a expert can probably figure out how to write a
                                  – Timothy
                                  Nov 22 '18 at 18:45














                                -1












                                -1








                                -1






                                Technically, irrational numbers certainly exist. Just because we can't measure $pi$ to arbitrary precision doesn't mean not all of its digits can be calculated mathematically. The fine structure constant is built into the laws of Physics and we need to make observations to determine what its digits are. $pi$ on the other hand is a mathematical constant which is not built into the laws of Euclidean geometry and rather is derived from them and all of its digits can be calculated entirely mathematically and that constant can be proven to be irrational.



                                $pi$ can't be measured exactly in our universe. According to https://physics.stackexchange.com/questions/395266/has-the-zeroth-law-of-thermodynamics-ever-been-proven/395270#395270, temperature probably can't be defined so I will instead describe how hot something is by its internal energy which can also be added or lost from the material. Here's one attempted way to measure $pi$. Take a large diamond sphere and drill a thin hole through the center and create a thin diamond chain that fits in that hole and measure its circumference and diameter while regulating its internal energy. There might be factors limiting how accurately you can measure it. For example, it has the highest shear modulus of all at about 5,000,000 atm but with a large enough sphere, the stretching of the chain from tension will limit the number of digits you can measure more than does the number of links. If your measurment is limited entirely my the number of links in the chain, you still can't measure $pi$ to arbitrary precision in that way because once the links get down to the size of a few atoms, Brownian motion of its atoms probably becomes significant which can sublimate away part of one of the links causing the chain to break. Despite that, Euclidean geometry still exists mathematically and all the digits of $pi$ can be calculated from it.






                                share|cite|improve this answer














                                Technically, irrational numbers certainly exist. Just because we can't measure $pi$ to arbitrary precision doesn't mean not all of its digits can be calculated mathematically. The fine structure constant is built into the laws of Physics and we need to make observations to determine what its digits are. $pi$ on the other hand is a mathematical constant which is not built into the laws of Euclidean geometry and rather is derived from them and all of its digits can be calculated entirely mathematically and that constant can be proven to be irrational.



                                $pi$ can't be measured exactly in our universe. According to https://physics.stackexchange.com/questions/395266/has-the-zeroth-law-of-thermodynamics-ever-been-proven/395270#395270, temperature probably can't be defined so I will instead describe how hot something is by its internal energy which can also be added or lost from the material. Here's one attempted way to measure $pi$. Take a large diamond sphere and drill a thin hole through the center and create a thin diamond chain that fits in that hole and measure its circumference and diameter while regulating its internal energy. There might be factors limiting how accurately you can measure it. For example, it has the highest shear modulus of all at about 5,000,000 atm but with a large enough sphere, the stretching of the chain from tension will limit the number of digits you can measure more than does the number of links. If your measurment is limited entirely my the number of links in the chain, you still can't measure $pi$ to arbitrary precision in that way because once the links get down to the size of a few atoms, Brownian motion of its atoms probably becomes significant which can sublimate away part of one of the links causing the chain to break. Despite that, Euclidean geometry still exists mathematically and all the digits of $pi$ can be calculated from it.







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                                edited Dec 9 '18 at 5:21

























                                answered Nov 21 '18 at 18:30









                                TimothyTimothy

                                288211




                                288211












                                • How could you tell if $x$ is irrational or not, if you can only compute it to the billionth trillion decimal digit?
                                  – Asaf Karagila♦
                                  Nov 21 '18 at 23:32










                                • @AsafKaragila Just because we don't know what any of the later digits are doesn't mean we don't know a certain property of the set of all digits after that place, the property of not repeating any finite string of digits over and over. To prove that $pi$ is irrational, it suffices to show that for any positive rational number less than 1, the $sin$ of it is not exactly $frac{1}{2}$. Any positive rational less than 1 can be expressed as $frac{p}{q}$ where the greatest common factor of p and q is 1 and p < q. The $sin$ of it is the limit of its Taylor series. The nth term must be a multiple
                                  – Timothy
                                  Nov 22 '18 at 1:46










                                • of $frac{1}{n!q^n}$. Eventually you will get to the point where the absolute value of the sum of all the remaining terms is less than $frac{1}{n!q^n}$. I believe I can prove that the sum of all the terms after the nth terms adds up to something that is not a multiple of $frac{1}{n!q^n}$ but I'm not quite sure how to do that.
                                  – Timothy
                                  Nov 22 '18 at 3:21












                                • How can you prove that $pi$ exists? As long as you can only measure finitely many digits, what makes you so sure that you're not just wrong? Or that your assumption in the proof that the value should be $pi$ include non-physical assumptions (e.g. that space is continuous and not discrete)?
                                  – Asaf Karagila♦
                                  Nov 22 '18 at 7:54










                                • @AsafKaragila 2-dimensional Euclidean geometry is a well defined thing and sets of points with the property of being a circle exist. I guess we can call it an imaginary circle if we don't add an additional assumption that each point either has matter in it or it doesn't and the set of all points with matter in it is a circle. I don't yet know how to write a formal proof that $pi$ exists and is the same for all imaginary circles in Euclidean geometry but a complete formal proof would probably render my answer unreadable for a lot of people and a expert can probably figure out how to write a
                                  – Timothy
                                  Nov 22 '18 at 18:45


















                                • How could you tell if $x$ is irrational or not, if you can only compute it to the billionth trillion decimal digit?
                                  – Asaf Karagila♦
                                  Nov 21 '18 at 23:32










                                • @AsafKaragila Just because we don't know what any of the later digits are doesn't mean we don't know a certain property of the set of all digits after that place, the property of not repeating any finite string of digits over and over. To prove that $pi$ is irrational, it suffices to show that for any positive rational number less than 1, the $sin$ of it is not exactly $frac{1}{2}$. Any positive rational less than 1 can be expressed as $frac{p}{q}$ where the greatest common factor of p and q is 1 and p < q. The $sin$ of it is the limit of its Taylor series. The nth term must be a multiple
                                  – Timothy
                                  Nov 22 '18 at 1:46










                                • of $frac{1}{n!q^n}$. Eventually you will get to the point where the absolute value of the sum of all the remaining terms is less than $frac{1}{n!q^n}$. I believe I can prove that the sum of all the terms after the nth terms adds up to something that is not a multiple of $frac{1}{n!q^n}$ but I'm not quite sure how to do that.
                                  – Timothy
                                  Nov 22 '18 at 3:21












                                • How can you prove that $pi$ exists? As long as you can only measure finitely many digits, what makes you so sure that you're not just wrong? Or that your assumption in the proof that the value should be $pi$ include non-physical assumptions (e.g. that space is continuous and not discrete)?
                                  – Asaf Karagila♦
                                  Nov 22 '18 at 7:54










                                • @AsafKaragila 2-dimensional Euclidean geometry is a well defined thing and sets of points with the property of being a circle exist. I guess we can call it an imaginary circle if we don't add an additional assumption that each point either has matter in it or it doesn't and the set of all points with matter in it is a circle. I don't yet know how to write a formal proof that $pi$ exists and is the same for all imaginary circles in Euclidean geometry but a complete formal proof would probably render my answer unreadable for a lot of people and a expert can probably figure out how to write a
                                  – Timothy
                                  Nov 22 '18 at 18:45
















                                How could you tell if $x$ is irrational or not, if you can only compute it to the billionth trillion decimal digit?
                                – Asaf Karagila♦
                                Nov 21 '18 at 23:32




                                How could you tell if $x$ is irrational or not, if you can only compute it to the billionth trillion decimal digit?
                                – Asaf Karagila♦
                                Nov 21 '18 at 23:32












                                @AsafKaragila Just because we don't know what any of the later digits are doesn't mean we don't know a certain property of the set of all digits after that place, the property of not repeating any finite string of digits over and over. To prove that $pi$ is irrational, it suffices to show that for any positive rational number less than 1, the $sin$ of it is not exactly $frac{1}{2}$. Any positive rational less than 1 can be expressed as $frac{p}{q}$ where the greatest common factor of p and q is 1 and p < q. The $sin$ of it is the limit of its Taylor series. The nth term must be a multiple
                                – Timothy
                                Nov 22 '18 at 1:46




                                @AsafKaragila Just because we don't know what any of the later digits are doesn't mean we don't know a certain property of the set of all digits after that place, the property of not repeating any finite string of digits over and over. To prove that $pi$ is irrational, it suffices to show that for any positive rational number less than 1, the $sin$ of it is not exactly $frac{1}{2}$. Any positive rational less than 1 can be expressed as $frac{p}{q}$ where the greatest common factor of p and q is 1 and p < q. The $sin$ of it is the limit of its Taylor series. The nth term must be a multiple
                                – Timothy
                                Nov 22 '18 at 1:46












                                of $frac{1}{n!q^n}$. Eventually you will get to the point where the absolute value of the sum of all the remaining terms is less than $frac{1}{n!q^n}$. I believe I can prove that the sum of all the terms after the nth terms adds up to something that is not a multiple of $frac{1}{n!q^n}$ but I'm not quite sure how to do that.
                                – Timothy
                                Nov 22 '18 at 3:21






                                of $frac{1}{n!q^n}$. Eventually you will get to the point where the absolute value of the sum of all the remaining terms is less than $frac{1}{n!q^n}$. I believe I can prove that the sum of all the terms after the nth terms adds up to something that is not a multiple of $frac{1}{n!q^n}$ but I'm not quite sure how to do that.
                                – Timothy
                                Nov 22 '18 at 3:21














                                How can you prove that $pi$ exists? As long as you can only measure finitely many digits, what makes you so sure that you're not just wrong? Or that your assumption in the proof that the value should be $pi$ include non-physical assumptions (e.g. that space is continuous and not discrete)?
                                – Asaf Karagila♦
                                Nov 22 '18 at 7:54




                                How can you prove that $pi$ exists? As long as you can only measure finitely many digits, what makes you so sure that you're not just wrong? Or that your assumption in the proof that the value should be $pi$ include non-physical assumptions (e.g. that space is continuous and not discrete)?
                                – Asaf Karagila♦
                                Nov 22 '18 at 7:54












                                @AsafKaragila 2-dimensional Euclidean geometry is a well defined thing and sets of points with the property of being a circle exist. I guess we can call it an imaginary circle if we don't add an additional assumption that each point either has matter in it or it doesn't and the set of all points with matter in it is a circle. I don't yet know how to write a formal proof that $pi$ exists and is the same for all imaginary circles in Euclidean geometry but a complete formal proof would probably render my answer unreadable for a lot of people and a expert can probably figure out how to write a
                                – Timothy
                                Nov 22 '18 at 18:45




                                @AsafKaragila 2-dimensional Euclidean geometry is a well defined thing and sets of points with the property of being a circle exist. I guess we can call it an imaginary circle if we don't add an additional assumption that each point either has matter in it or it doesn't and the set of all points with matter in it is a circle. I don't yet know how to write a formal proof that $pi$ exists and is the same for all imaginary circles in Euclidean geometry but a complete formal proof would probably render my answer unreadable for a lot of people and a expert can probably figure out how to write a
                                – Timothy
                                Nov 22 '18 at 18:45


















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