Is it rational or irrational?












2












$begingroup$


I am a mathematical putz - please be kind.



From what I know, a rational number is a non-imaginary number that can be written as $frac{p}{q}$. A repeating number can also be a rational number (i.e. $frac{1}{3} = 0.3333...$). Ok, so far, so good.



Ok, so now the question is to write $0.329999...$ (the 9 repeats) as a fraction and prove that it is rational, so here goes:



begin{align}
1000x &= 329.9999...\
100x &= 32.9999...\
end{align}



subtracting...



begin{align}
900x &= 297 \
x &= frac{297}{300} \
&= frac{33}{100} \
&= 0.33
end{align}



Herein my dilemma: I don't think that $0.32999...$ is the same as $0.33$ (except in the limiting case), but the math tells me it is. Based on the definition of rationality does that mean that $0.32999...$ is actually irrational, since it cannot be presented as $frac{p}{q}$? Does this therefore imply, that some irrational numbers can be written down with perfect clarity, like $0.32999...$?






rationality-testing







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The mistake is in the last step. $frac{33}{100} = 0.33(000 ...)$, not $0.3333...$. Also, $.3299999 ... = 0.33 (0000...)$. Why? The same reason why $0.999...= 1$.
    $endgroup$
    – Matti P.
    Feb 1 at 8:32










  • $begingroup$
    I found this as well: brilliant.org/wiki/is-0999-equal-1. It clears it up beautifully. You cannot have a number that is greater than $0.999...$ and less than 1 at the same time and the number line is continuous, so $0.999...$ must be 1. I am now richer, thank you.
    $endgroup$
    – Jaco Van Niekerk
    Feb 1 at 8:41












  • $begingroup$
    I won't recommend being too reliant on Brilliant, as it has subtle but serious mathematical errors. In this case, the linked article got it half right when it said "since 0.999... is defined to be that limit, it is defined to equal 1." The second half is wrong. 0.999... is defined to be the limit of the sequence (0.9,0.99,0.999,...), not the sequence nor any term in the sequence, there is no issue with it being equal to 1, and that can be proven. So it is not a matter of defining 0.999... to be 1, but simply a matter of proving that the limit of the sequence is 1.
    $endgroup$
    – user21820
    Feb 9 at 16:06










  • $begingroup$
    Similarly, 0.32999... is not a sequence, but the limit of the sequence of decimal expansions. It is also equal to the limit of other sequences such as (0.32923,0.329923,0.3299923,...).
    $endgroup$
    – user21820
    Feb 9 at 16:08


















2












$begingroup$


I am a mathematical putz - please be kind.



From what I know, a rational number is a non-imaginary number that can be written as $frac{p}{q}$. A repeating number can also be a rational number (i.e. $frac{1}{3} = 0.3333...$). Ok, so far, so good.



Ok, so now the question is to write $0.329999...$ (the 9 repeats) as a fraction and prove that it is rational, so here goes:



begin{align}
1000x &= 329.9999...\
100x &= 32.9999...\
end{align}



subtracting...



begin{align}
900x &= 297 \
x &= frac{297}{300} \
&= frac{33}{100} \
&= 0.33
end{align}



Herein my dilemma: I don't think that $0.32999...$ is the same as $0.33$ (except in the limiting case), but the math tells me it is. Based on the definition of rationality does that mean that $0.32999...$ is actually irrational, since it cannot be presented as $frac{p}{q}$? Does this therefore imply, that some irrational numbers can be written down with perfect clarity, like $0.32999...$?






rationality-testing







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The mistake is in the last step. $frac{33}{100} = 0.33(000 ...)$, not $0.3333...$. Also, $.3299999 ... = 0.33 (0000...)$. Why? The same reason why $0.999...= 1$.
    $endgroup$
    – Matti P.
    Feb 1 at 8:32










  • $begingroup$
    I found this as well: brilliant.org/wiki/is-0999-equal-1. It clears it up beautifully. You cannot have a number that is greater than $0.999...$ and less than 1 at the same time and the number line is continuous, so $0.999...$ must be 1. I am now richer, thank you.
    $endgroup$
    – Jaco Van Niekerk
    Feb 1 at 8:41












  • $begingroup$
    I won't recommend being too reliant on Brilliant, as it has subtle but serious mathematical errors. In this case, the linked article got it half right when it said "since 0.999... is defined to be that limit, it is defined to equal 1." The second half is wrong. 0.999... is defined to be the limit of the sequence (0.9,0.99,0.999,...), not the sequence nor any term in the sequence, there is no issue with it being equal to 1, and that can be proven. So it is not a matter of defining 0.999... to be 1, but simply a matter of proving that the limit of the sequence is 1.
    $endgroup$
    – user21820
    Feb 9 at 16:06










  • $begingroup$
    Similarly, 0.32999... is not a sequence, but the limit of the sequence of decimal expansions. It is also equal to the limit of other sequences such as (0.32923,0.329923,0.3299923,...).
    $endgroup$
    – user21820
    Feb 9 at 16:08
















2












2








2


1



$begingroup$


I am a mathematical putz - please be kind.



From what I know, a rational number is a non-imaginary number that can be written as $frac{p}{q}$. A repeating number can also be a rational number (i.e. $frac{1}{3} = 0.3333...$). Ok, so far, so good.



Ok, so now the question is to write $0.329999...$ (the 9 repeats) as a fraction and prove that it is rational, so here goes:



begin{align}
1000x &= 329.9999...\
100x &= 32.9999...\
end{align}



subtracting...



begin{align}
900x &= 297 \
x &= frac{297}{300} \
&= frac{33}{100} \
&= 0.33
end{align}



Herein my dilemma: I don't think that $0.32999...$ is the same as $0.33$ (except in the limiting case), but the math tells me it is. Based on the definition of rationality does that mean that $0.32999...$ is actually irrational, since it cannot be presented as $frac{p}{q}$? Does this therefore imply, that some irrational numbers can be written down with perfect clarity, like $0.32999...$?






rationality-testing







share|cite|improve this question











$endgroup$




I am a mathematical putz - please be kind.



From what I know, a rational number is a non-imaginary number that can be written as $frac{p}{q}$. A repeating number can also be a rational number (i.e. $frac{1}{3} = 0.3333...$). Ok, so far, so good.



Ok, so now the question is to write $0.329999...$ (the 9 repeats) as a fraction and prove that it is rational, so here goes:



begin{align}
1000x &= 329.9999...\
100x &= 32.9999...\
end{align}



subtracting...



begin{align}
900x &= 297 \
x &= frac{297}{300} \
&= frac{33}{100} \
&= 0.33
end{align}



Herein my dilemma: I don't think that $0.32999...$ is the same as $0.33$ (except in the limiting case), but the math tells me it is. Based on the definition of rationality does that mean that $0.32999...$ is actually irrational, since it cannot be presented as $frac{p}{q}$? Does this therefore imply, that some irrational numbers can be written down with perfect clarity, like $0.32999...$?






rationality-testing




rationality-testing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 1 at 8:55









Wouter

5,96921436




5,96921436










asked Feb 1 at 8:28









Jaco Van NiekerkJaco Van Niekerk

1204




1204








  • 2




    $begingroup$
    The mistake is in the last step. $frac{33}{100} = 0.33(000 ...)$, not $0.3333...$. Also, $.3299999 ... = 0.33 (0000...)$. Why? The same reason why $0.999...= 1$.
    $endgroup$
    – Matti P.
    Feb 1 at 8:32










  • $begingroup$
    I found this as well: brilliant.org/wiki/is-0999-equal-1. It clears it up beautifully. You cannot have a number that is greater than $0.999...$ and less than 1 at the same time and the number line is continuous, so $0.999...$ must be 1. I am now richer, thank you.
    $endgroup$
    – Jaco Van Niekerk
    Feb 1 at 8:41












  • $begingroup$
    I won't recommend being too reliant on Brilliant, as it has subtle but serious mathematical errors. In this case, the linked article got it half right when it said "since 0.999... is defined to be that limit, it is defined to equal 1." The second half is wrong. 0.999... is defined to be the limit of the sequence (0.9,0.99,0.999,...), not the sequence nor any term in the sequence, there is no issue with it being equal to 1, and that can be proven. So it is not a matter of defining 0.999... to be 1, but simply a matter of proving that the limit of the sequence is 1.
    $endgroup$
    – user21820
    Feb 9 at 16:06










  • $begingroup$
    Similarly, 0.32999... is not a sequence, but the limit of the sequence of decimal expansions. It is also equal to the limit of other sequences such as (0.32923,0.329923,0.3299923,...).
    $endgroup$
    – user21820
    Feb 9 at 16:08
















  • 2




    $begingroup$
    The mistake is in the last step. $frac{33}{100} = 0.33(000 ...)$, not $0.3333...$. Also, $.3299999 ... = 0.33 (0000...)$. Why? The same reason why $0.999...= 1$.
    $endgroup$
    – Matti P.
    Feb 1 at 8:32










  • $begingroup$
    I found this as well: brilliant.org/wiki/is-0999-equal-1. It clears it up beautifully. You cannot have a number that is greater than $0.999...$ and less than 1 at the same time and the number line is continuous, so $0.999...$ must be 1. I am now richer, thank you.
    $endgroup$
    – Jaco Van Niekerk
    Feb 1 at 8:41












  • $begingroup$
    I won't recommend being too reliant on Brilliant, as it has subtle but serious mathematical errors. In this case, the linked article got it half right when it said "since 0.999... is defined to be that limit, it is defined to equal 1." The second half is wrong. 0.999... is defined to be the limit of the sequence (0.9,0.99,0.999,...), not the sequence nor any term in the sequence, there is no issue with it being equal to 1, and that can be proven. So it is not a matter of defining 0.999... to be 1, but simply a matter of proving that the limit of the sequence is 1.
    $endgroup$
    – user21820
    Feb 9 at 16:06










  • $begingroup$
    Similarly, 0.32999... is not a sequence, but the limit of the sequence of decimal expansions. It is also equal to the limit of other sequences such as (0.32923,0.329923,0.3299923,...).
    $endgroup$
    – user21820
    Feb 9 at 16:08










2




2




$begingroup$
The mistake is in the last step. $frac{33}{100} = 0.33(000 ...)$, not $0.3333...$. Also, $.3299999 ... = 0.33 (0000...)$. Why? The same reason why $0.999...= 1$.
$endgroup$
– Matti P.
Feb 1 at 8:32




$begingroup$
The mistake is in the last step. $frac{33}{100} = 0.33(000 ...)$, not $0.3333...$. Also, $.3299999 ... = 0.33 (0000...)$. Why? The same reason why $0.999...= 1$.
$endgroup$
– Matti P.
Feb 1 at 8:32












$begingroup$
I found this as well: brilliant.org/wiki/is-0999-equal-1. It clears it up beautifully. You cannot have a number that is greater than $0.999...$ and less than 1 at the same time and the number line is continuous, so $0.999...$ must be 1. I am now richer, thank you.
$endgroup$
– Jaco Van Niekerk
Feb 1 at 8:41






$begingroup$
I found this as well: brilliant.org/wiki/is-0999-equal-1. It clears it up beautifully. You cannot have a number that is greater than $0.999...$ and less than 1 at the same time and the number line is continuous, so $0.999...$ must be 1. I am now richer, thank you.
$endgroup$
– Jaco Van Niekerk
Feb 1 at 8:41














$begingroup$
I won't recommend being too reliant on Brilliant, as it has subtle but serious mathematical errors. In this case, the linked article got it half right when it said "since 0.999... is defined to be that limit, it is defined to equal 1." The second half is wrong. 0.999... is defined to be the limit of the sequence (0.9,0.99,0.999,...), not the sequence nor any term in the sequence, there is no issue with it being equal to 1, and that can be proven. So it is not a matter of defining 0.999... to be 1, but simply a matter of proving that the limit of the sequence is 1.
$endgroup$
– user21820
Feb 9 at 16:06




$begingroup$
I won't recommend being too reliant on Brilliant, as it has subtle but serious mathematical errors. In this case, the linked article got it half right when it said "since 0.999... is defined to be that limit, it is defined to equal 1." The second half is wrong. 0.999... is defined to be the limit of the sequence (0.9,0.99,0.999,...), not the sequence nor any term in the sequence, there is no issue with it being equal to 1, and that can be proven. So it is not a matter of defining 0.999... to be 1, but simply a matter of proving that the limit of the sequence is 1.
$endgroup$
– user21820
Feb 9 at 16:06












$begingroup$
Similarly, 0.32999... is not a sequence, but the limit of the sequence of decimal expansions. It is also equal to the limit of other sequences such as (0.32923,0.329923,0.3299923,...).
$endgroup$
– user21820
Feb 9 at 16:08






$begingroup$
Similarly, 0.32999... is not a sequence, but the limit of the sequence of decimal expansions. It is also equal to the limit of other sequences such as (0.32923,0.329923,0.3299923,...).
$endgroup$
– user21820
Feb 9 at 16:08












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$begingroup$

That’s the problem when dealing with representation as decimal numbers. The point is that such representations are not unique but fractions are (up to equivalence). Although there is an error in your calculation, as decimal numbers, $0.33=0.32999dots$.






share|cite|improve this answer











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    1 Answer
    1






    active

    oldest

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    active

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    active

    oldest

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    2












    $begingroup$

    That’s the problem when dealing with representation as decimal numbers. The point is that such representations are not unique but fractions are (up to equivalence). Although there is an error in your calculation, as decimal numbers, $0.33=0.32999dots$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      That’s the problem when dealing with representation as decimal numbers. The point is that such representations are not unique but fractions are (up to equivalence). Although there is an error in your calculation, as decimal numbers, $0.33=0.32999dots$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        That’s the problem when dealing with representation as decimal numbers. The point is that such representations are not unique but fractions are (up to equivalence). Although there is an error in your calculation, as decimal numbers, $0.33=0.32999dots$.






        share|cite|improve this answer











        $endgroup$



        That’s the problem when dealing with representation as decimal numbers. The point is that such representations are not unique but fractions are (up to equivalence). Although there is an error in your calculation, as decimal numbers, $0.33=0.32999dots$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Feb 1 at 10:43









        Jaco Van Niekerk

        1204




        1204










        answered Feb 1 at 8:34









        JamesJames

        2,112422




        2,112422






























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