Mixing
The validity of the proofs of the Pythagorean Theorem and the concept of area
$begingroup$
this might be a very elemental question but it has been bothering me for a while. Must of the proofs I've seen of the Pythagorean Theorem involve showing that the areas of the squares with side length $a$ and $b$ add up to the area of the square with side length $c$. This is generally done by rearranging triangles.
My problem with this type of proofs is that they only show that the areas must be the same but don't show that $a^2+b^2=c^2$.
Why must the area of a square with side $a$ be defined as $a^2$. Say for example that you had another way of measuring the surface of a square with a given side length (and it behaves as we would intuitively want area to behave). If this function is called $A$ then the visual proofs of the theorem would only show that $A(a)+A(b)=A(c)$.
So, does this type of proof works because we just happen to define area as we do, or does $A(a)+A(b)=A(c)$ must imply $a^2+b^2=c^2$?
Now, if $A(a)+A(b)=A(c)$ does imply $a^2+b^2=c^2$ that would mean that our function $A$ (which behaves as area does) must include the square of the side in its formula. For example $A(x)=kx^2, k>0$ (which does imply the pythagorean theorem). Are there other ways to define the surface of a square such that it behaves as it physically does? Would the visual proofs still be valid?
Thank you!
geometry measure-theory euclidean-geometry
asked Jun 20 '17 at 3:14
MathUser123MathUser123
17210
$endgroup$
add a comment |
$begingroup$
this might be a very elemental question but it has been bothering me for a while. Must of the proofs I've seen of the Pythagorean Theorem involve showing that the areas of the squares with side length $a$ and $b$ add up to the area of the square with side length $c$. This is generally done by rearranging triangles.
My problem with this type of proofs is that they only show that the areas must be the same but don't show that $a^2+b^2=c^2$.
Why must the area of a square with side $a$ be defined as $a^2$. Say for example that you had another way of measuring the surface of a square with a given side length (and it behaves as we would intuitively want area to behave). If this function is called $A$ then the visual proofs of the theorem would only show that $A(a)+A(b)=A(c)$.
So, does this type of proof works because we just happen to define area as we do, or does $A(a)+A(b)=A(c)$ must imply $a^2+b^2=c^2$?
Now, if $A(a)+A(b)=A(c)$ does imply $a^2+b^2=c^2$ that would mean that our function $A$ (which behaves as area does) must include the square of the side in its formula. For example $A(x)=kx^2, k>0$ (which does imply the pythagorean theorem). Are there other ways to define the surface of a square such that it behaves as it physically does? Would the visual proofs still be valid?
Thank you!
geometry measure-theory euclidean-geometry
asked Jun 20 '17 at 3:14
MathUser123MathUser123
17210
$endgroup$
2
$begingroup$
Related: See my answer to this question math.stackexchange.com/questions/675522/…. This also provides an answer to your question: Most proofs of the PT use existence of a certain function called "area" satisfying certain properties. One can prove that there is no function satisfying all but one of these axioms and also, as you suggested, with $A(b)=b$.
$endgroup$
– Moishe Cohen
Jun 20 '17 at 3:25
$begingroup$
proofwiki.org/wiki/Area_of_Square
$endgroup$
– Ryan A
Jun 20 '17 at 3:27
add a comment |
$begingroup$
this might be a very elemental question but it has been bothering me for a while. Must of the proofs I've seen of the Pythagorean Theorem involve showing that the areas of the squares with side length $a$ and $b$ add up to the area of the square with side length $c$. This is generally done by rearranging triangles.
My problem with this type of proofs is that they only show that the areas must be the same but don't show that $a^2+b^2=c^2$.
Why must the area of a square with side $a$ be defined as $a^2$. Say for example that you had another way of measuring the surface of a square with a given side length (and it behaves as we would intuitively want area to behave). If this function is called $A$ then the visual proofs of the theorem would only show that $A(a)+A(b)=A(c)$.
So, does this type of proof works because we just happen to define area as we do, or does $A(a)+A(b)=A(c)$ must imply $a^2+b^2=c^2$?
Now, if $A(a)+A(b)=A(c)$ does imply $a^2+b^2=c^2$ that would mean that our function $A$ (which behaves as area does) must include the square of the side in its formula. For example $A(x)=kx^2, k>0$ (which does imply the pythagorean theorem). Are there other ways to define the surface of a square such that it behaves as it physically does? Would the visual proofs still be valid?
Thank you!
geometry measure-theory euclidean-geometry
asked Jun 20 '17 at 3:14
MathUser123MathUser123
17210
$endgroup$
this might be a very elemental question but it has been bothering me for a while. Must of the proofs I've seen of the Pythagorean Theorem involve showing that the areas of the squares with side length $a$ and $b$ add up to the area of the square with side length $c$. This is generally done by rearranging triangles.
My problem with this type of proofs is that they only show that the areas must be the same but don't show that $a^2+b^2=c^2$.
Why must the area of a square with side $a$ be defined as $a^2$. Say for example that you had another way of measuring the surface of a square with a given side length (and it behaves as we would intuitively want area to behave). If this function is called $A$ then the visual proofs of the theorem would only show that $A(a)+A(b)=A(c)$.
So, does this type of proof works because we just happen to define area as we do, or does $A(a)+A(b)=A(c)$ must imply $a^2+b^2=c^2$?
Now, if $A(a)+A(b)=A(c)$ does imply $a^2+b^2=c^2$ that would mean that our function $A$ (which behaves as area does) must include the square of the side in its formula. For example $A(x)=kx^2, k>0$ (which does imply the pythagorean theorem). Are there other ways to define the surface of a square such that it behaves as it physically does? Would the visual proofs still be valid?
Thank you!
geometry measure-theory euclidean-geometry
geometry measure-theory euclidean-geometry
asked Jun 20 '17 at 3:14
MathUser123MathUser123
17210
asked Jun 20 '17 at 3:14
MathUser123MathUser123
17210
asked Jun 20 '17 at 3:14
MathUser123MathUser123
17210
asked Jun 20 '17 at 3:14
MathUser123MathUser123
17210
asked Jun 20 '17 at 3:14
MathUser123MathUser123
17210
17210
2
$begingroup$
Related: See my answer to this question math.stackexchange.com/questions/675522/…. This also provides an answer to your question: Most proofs of the PT use existence of a certain function called "area" satisfying certain properties. One can prove that there is no function satisfying all but one of these axioms and also, as you suggested, with $A(b)=b$.
$endgroup$
– Moishe Cohen
Jun 20 '17 at 3:25
$begingroup$
proofwiki.org/wiki/Area_of_Square
$endgroup$
– Ryan A
Jun 20 '17 at 3:27
add a comment |
2
$begingroup$
Related: See my answer to this question math.stackexchange.com/questions/675522/…. This also provides an answer to your question: Most proofs of the PT use existence of a certain function called "area" satisfying certain properties. One can prove that there is no function satisfying all but one of these axioms and also, as you suggested, with $A(b)=b$.
$endgroup$
– Moishe Cohen
Jun 20 '17 at 3:25
$begingroup$
proofwiki.org/wiki/Area_of_Square
$endgroup$
– Ryan A
Jun 20 '17 at 3:27
2
2
$begingroup$
Related: See my answer to this question math.stackexchange.com/questions/675522/…. This also provides an answer to your question: Most proofs of the PT use existence of a certain function called "area" satisfying certain properties. One can prove that there is no function satisfying all but one of these axioms and also, as you suggested, with $A(b)=b$.
$endgroup$
– Moishe Cohen
Jun 20 '17 at 3:25
$begingroup$
Related: See my answer to this question math.stackexchange.com/questions/675522/…. This also provides an answer to your question: Most proofs of the PT use existence of a certain function called "area" satisfying certain properties. One can prove that there is no function satisfying all but one of these axioms and also, as you suggested, with $A(b)=b$.
$endgroup$
– Moishe Cohen
Jun 20 '17 at 3:25
$begingroup$
proofwiki.org/wiki/Area_of_Square
$endgroup$
– Ryan A
Jun 20 '17 at 3:27
$begingroup$
proofwiki.org/wiki/Area_of_Square
$endgroup$
– Ryan A
Jun 20 '17 at 3:27
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Fix an angle $theta$, i.e. a number between 0 and $2pi$. Let A,B and C be the sides of a right-angled triangle, with AB the hypotenuse. Draw three circular arcs with these 3 sides as end points and all subtending angle $theta$ at their respective centres P, Q,R
Now the sum of areas of the circular sectors satisfy similar additive law:
area (ABP) = area (BCQ) + area(CAR)
It might be easier to see the simpler case of semi-circular arcs (the three sides would be diameters of respective circles)
answered Jun 20 '17 at 4:30
P VanchinathanP Vanchinathan
15.3k12136
$endgroup$
$begingroup$
I don't understand what you're saying in your answer. Would it be possible for you to more clearly explain it?
$endgroup$
– Timothy
Jan 22 at 5:37
add a comment |
$begingroup$
Pythagora's theorem would imply $a^2+b^2=c^2$ even if you should choose a different definition for the area of a square.
That happens because the ratio of the areas of similar figures is the square of the ratio of any two corresponding lengths. From that it follows that the area of a square of side $a$ is $ka^2$, with $k$ some positive constant which is the same for all squares (all squares are similar among them).
Pythagora's theorem then implies $ka^2+kb^2=kc^2$, and dividing by $k$ we recover the usual relation.
answered Jun 20 '17 at 21:53
AretinoAretino
24.9k21444
$endgroup$
$begingroup$
I made a mistake downvoting this answer. I was about to say it didn't answer the question for the following reason. Although it can be proven, people don't feel the need to prove it and just accept that there exists a function satisfying these conditions. 1. For any real numbers x, y, z, w, d((x, y), (x + z, y + w)) = d((0, 0), (z, w)) 2. For any real number x, y, d((0, 0), (x, y)) is a nonnegative real number. 3. For any nonnegative real number x, d((0, 0), (x, 0)) = x 4. For any real numbers x, y, d((0, 0), (x, -y)) = d((0, 0), (x, y)). 5. For any real numbers x, y, z, w, d((0, 0), (xz - yw,
$endgroup$
– Timothy
Jan 22 at 5:44
$begingroup$
xw + yz)) = d((0, 0), (x, y)) × d((0, 0), (z, w)). So once they show that any function satisfying those conditions must be $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$, then they're fine with assuming that that function does satisfy those conditions. However, why should you assume that when you redefine distance using area, that new definition also satisfies those properties? Maybe because it can be shown that that new definition is in fact also $sqrt{x^2 + y^2}$ which is known to satisfy them. However, the task was to prove the Pythagorean theorem that way without reliance on a result
$endgroup$
– Timothy
Jan 22 at 5:52
$begingroup$
that wasn't proven under those other assumptions of distance which you didn't prove the distance formula from. Maybe it is however obvious that the composition of two operations on the unit square gets you a square with an area that's the product of the area of the squares each of those operations gets you because as described at math.stackexchange.com/questions/675522/…, any linear transformation multiplies the area of any shape by the same number.
$endgroup$
– Timothy
Jan 22 at 5:59
add a comment |
$begingroup$
None of the other answers give a rigourous proof that the Pythagorean theorem holds for all right angle triangles in $mathbb{R}^2$, even ones whose legs are not parallel to the axes so I will give my own proof of that in this answer. It is technically not a valid proof of the Pythagorean theorem. However, once you give a precise definition of distance, you can deduce from the distance formula the Pythagorean theorem and the statement that for any square, its area is the square of the length of its edge.
We seek a definition of distance from any point in $mathbb{R}^2$ to $mathbb{R}^2$, a function from $(mathbb{R}^2)^2$ to $mathbb{R}$ that satisfies the following properties.
- For any points $(x, y)$ and $(z, w)$, $d((x, y), (x + z, y + w)) = d((0, 0), (z, w))$
- For any point $(x, y)$, $d((0, 0), (x, y))$ is nonnegative
- For any nonnegative real number $x$, $d((0, 0), (x, 0)) = x$
- For any point $(x, y)$, $d((0, 0), (x, -y)) = d((0, 0), (x, y))$
- For any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$
Suppose a function $d$ from $(mathbb{R}^2)^2$ to $mathbb{R}$ satisfies those conditions, then for any point $(x, y)$, $d((0, 0), (x, y))^2 = d((0, 0), (x, y))d((0, 0), (x, y)) = d((0, 0), (x, y))d((0, 0), (x, -y)) = d((0, 0), (x^2 + y^2, 0)) = x^2 + y^2$ so $d((0, 0), (x, y)) = sqrt{x^2 + y^2}$ so for any points $(x, y)$ and $(z, w)$, $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ Now I will show that $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ actually satisfies those properties. It's trivial to show that it satisfies the first 4 conditions. It also satisfies the fifth condition because for any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = sqrt{(xz - yw)^2 + (xw + yz)^2} = sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = sqrt{(x^2 + y^2)(z^2 + w^2)} = sqrt{x^2 + y^2}sqrt{z^2 + w^2} = d((0, 0), (x, y))d((0, 0), (z, w))$
As a result of this, from now on, I will define the distance from any point $(x, y)$ to any point $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$ and denote it as $d((x, y), (z, w))$. I will also use $d(x, y)$ as shorthand for $d((0, 0), (x, y))$ For any square, for any displacement along one of its edges, let $x$ be the $x$ component of its displacement and $y$ be the $y$ component of its displacement.
This image shows that the area of that square is $(x - y)^2 + 2xy = (x^2 - 2xy + y^2) + 2xy = x^2 + y^2 = (d(x, y))^2$
Source:
- https://www.maa.org/press/periodicals/convergence/proportionality-in-similar-triangles-a-cross-cultural-comparison-the-student-module
answered Jan 22 at 5:29
TimothyTimothy
326214
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Fix an angle $theta$, i.e. a number between 0 and $2pi$. Let A,B and C be the sides of a right-angled triangle, with AB the hypotenuse. Draw three circular arcs with these 3 sides as end points and all subtending angle $theta$ at their respective centres P, Q,R
Now the sum of areas of the circular sectors satisfy similar additive law:
area (ABP) = area (BCQ) + area(CAR)
It might be easier to see the simpler case of semi-circular arcs (the three sides would be diameters of respective circles)
answered Jun 20 '17 at 4:30
P VanchinathanP Vanchinathan
15.3k12136
$endgroup$
$begingroup$
I don't understand what you're saying in your answer. Would it be possible for you to more clearly explain it?
$endgroup$
– Timothy
Jan 22 at 5:37
add a comment |
$begingroup$
Fix an angle $theta$, i.e. a number between 0 and $2pi$. Let A,B and C be the sides of a right-angled triangle, with AB the hypotenuse. Draw three circular arcs with these 3 sides as end points and all subtending angle $theta$ at their respective centres P, Q,R
Now the sum of areas of the circular sectors satisfy similar additive law:
area (ABP) = area (BCQ) + area(CAR)
It might be easier to see the simpler case of semi-circular arcs (the three sides would be diameters of respective circles)
answered Jun 20 '17 at 4:30
P VanchinathanP Vanchinathan
15.3k12136
$endgroup$
$begingroup$
I don't understand what you're saying in your answer. Would it be possible for you to more clearly explain it?
$endgroup$
– Timothy
Jan 22 at 5:37
add a comment |
$begingroup$
Fix an angle $theta$, i.e. a number between 0 and $2pi$. Let A,B and C be the sides of a right-angled triangle, with AB the hypotenuse. Draw three circular arcs with these 3 sides as end points and all subtending angle $theta$ at their respective centres P, Q,R
Now the sum of areas of the circular sectors satisfy similar additive law:
area (ABP) = area (BCQ) + area(CAR)
It might be easier to see the simpler case of semi-circular arcs (the three sides would be diameters of respective circles)
answered Jun 20 '17 at 4:30
P VanchinathanP Vanchinathan
15.3k12136
$endgroup$
Fix an angle $theta$, i.e. a number between 0 and $2pi$. Let A,B and C be the sides of a right-angled triangle, with AB the hypotenuse. Draw three circular arcs with these 3 sides as end points and all subtending angle $theta$ at their respective centres P, Q,R
Now the sum of areas of the circular sectors satisfy similar additive law:
area (ABP) = area (BCQ) + area(CAR)
It might be easier to see the simpler case of semi-circular arcs (the three sides would be diameters of respective circles)
answered Jun 20 '17 at 4:30
P VanchinathanP Vanchinathan
15.3k12136
answered Jun 20 '17 at 4:30
P VanchinathanP Vanchinathan
15.3k12136
answered Jun 20 '17 at 4:30
P VanchinathanP Vanchinathan
15.3k12136
answered Jun 20 '17 at 4:30
P VanchinathanP Vanchinathan
15.3k12136
15.3k12136
$begingroup$
I don't understand what you're saying in your answer. Would it be possible for you to more clearly explain it?
$endgroup$
– Timothy
Jan 22 at 5:37
add a comment |
$begingroup$
I don't understand what you're saying in your answer. Would it be possible for you to more clearly explain it?
$endgroup$
– Timothy
Jan 22 at 5:37
$begingroup$
I don't understand what you're saying in your answer. Would it be possible for you to more clearly explain it?
$endgroup$
– Timothy
Jan 22 at 5:37
$begingroup$
I don't understand what you're saying in your answer. Would it be possible for you to more clearly explain it?
$endgroup$
– Timothy
Jan 22 at 5:37
add a comment |
$begingroup$
Pythagora's theorem would imply $a^2+b^2=c^2$ even if you should choose a different definition for the area of a square.
That happens because the ratio of the areas of similar figures is the square of the ratio of any two corresponding lengths. From that it follows that the area of a square of side $a$ is $ka^2$, with $k$ some positive constant which is the same for all squares (all squares are similar among them).
Pythagora's theorem then implies $ka^2+kb^2=kc^2$, and dividing by $k$ we recover the usual relation.
answered Jun 20 '17 at 21:53
AretinoAretino
24.9k21444
$endgroup$
$begingroup$
I made a mistake downvoting this answer. I was about to say it didn't answer the question for the following reason. Although it can be proven, people don't feel the need to prove it and just accept that there exists a function satisfying these conditions. 1. For any real numbers x, y, z, w, d((x, y), (x + z, y + w)) = d((0, 0), (z, w)) 2. For any real number x, y, d((0, 0), (x, y)) is a nonnegative real number. 3. For any nonnegative real number x, d((0, 0), (x, 0)) = x 4. For any real numbers x, y, d((0, 0), (x, -y)) = d((0, 0), (x, y)). 5. For any real numbers x, y, z, w, d((0, 0), (xz - yw,
$endgroup$
– Timothy
Jan 22 at 5:44
$begingroup$
xw + yz)) = d((0, 0), (x, y)) × d((0, 0), (z, w)). So once they show that any function satisfying those conditions must be $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$, then they're fine with assuming that that function does satisfy those conditions. However, why should you assume that when you redefine distance using area, that new definition also satisfies those properties? Maybe because it can be shown that that new definition is in fact also $sqrt{x^2 + y^2}$ which is known to satisfy them. However, the task was to prove the Pythagorean theorem that way without reliance on a result
$endgroup$
– Timothy
Jan 22 at 5:52
$begingroup$
that wasn't proven under those other assumptions of distance which you didn't prove the distance formula from. Maybe it is however obvious that the composition of two operations on the unit square gets you a square with an area that's the product of the area of the squares each of those operations gets you because as described at math.stackexchange.com/questions/675522/…, any linear transformation multiplies the area of any shape by the same number.
$endgroup$
– Timothy
Jan 22 at 5:59
add a comment |
$begingroup$
Pythagora's theorem would imply $a^2+b^2=c^2$ even if you should choose a different definition for the area of a square.
That happens because the ratio of the areas of similar figures is the square of the ratio of any two corresponding lengths. From that it follows that the area of a square of side $a$ is $ka^2$, with $k$ some positive constant which is the same for all squares (all squares are similar among them).
Pythagora's theorem then implies $ka^2+kb^2=kc^2$, and dividing by $k$ we recover the usual relation.
answered Jun 20 '17 at 21:53
AretinoAretino
24.9k21444
$endgroup$
$begingroup$
I made a mistake downvoting this answer. I was about to say it didn't answer the question for the following reason. Although it can be proven, people don't feel the need to prove it and just accept that there exists a function satisfying these conditions. 1. For any real numbers x, y, z, w, d((x, y), (x + z, y + w)) = d((0, 0), (z, w)) 2. For any real number x, y, d((0, 0), (x, y)) is a nonnegative real number. 3. For any nonnegative real number x, d((0, 0), (x, 0)) = x 4. For any real numbers x, y, d((0, 0), (x, -y)) = d((0, 0), (x, y)). 5. For any real numbers x, y, z, w, d((0, 0), (xz - yw,
$endgroup$
– Timothy
Jan 22 at 5:44
$begingroup$
xw + yz)) = d((0, 0), (x, y)) × d((0, 0), (z, w)). So once they show that any function satisfying those conditions must be $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$, then they're fine with assuming that that function does satisfy those conditions. However, why should you assume that when you redefine distance using area, that new definition also satisfies those properties? Maybe because it can be shown that that new definition is in fact also $sqrt{x^2 + y^2}$ which is known to satisfy them. However, the task was to prove the Pythagorean theorem that way without reliance on a result
$endgroup$
– Timothy
Jan 22 at 5:52
$begingroup$
that wasn't proven under those other assumptions of distance which you didn't prove the distance formula from. Maybe it is however obvious that the composition of two operations on the unit square gets you a square with an area that's the product of the area of the squares each of those operations gets you because as described at math.stackexchange.com/questions/675522/…, any linear transformation multiplies the area of any shape by the same number.
$endgroup$
– Timothy
Jan 22 at 5:59
add a comment |
$begingroup$
Pythagora's theorem would imply $a^2+b^2=c^2$ even if you should choose a different definition for the area of a square.
That happens because the ratio of the areas of similar figures is the square of the ratio of any two corresponding lengths. From that it follows that the area of a square of side $a$ is $ka^2$, with $k$ some positive constant which is the same for all squares (all squares are similar among them).
Pythagora's theorem then implies $ka^2+kb^2=kc^2$, and dividing by $k$ we recover the usual relation.
answered Jun 20 '17 at 21:53
AretinoAretino
24.9k21444
$endgroup$
Pythagora's theorem would imply $a^2+b^2=c^2$ even if you should choose a different definition for the area of a square.
That happens because the ratio of the areas of similar figures is the square of the ratio of any two corresponding lengths. From that it follows that the area of a square of side $a$ is $ka^2$, with $k$ some positive constant which is the same for all squares (all squares are similar among them).
Pythagora's theorem then implies $ka^2+kb^2=kc^2$, and dividing by $k$ we recover the usual relation.
answered Jun 20 '17 at 21:53
AretinoAretino
24.9k21444
answered Jun 20 '17 at 21:53
AretinoAretino
24.9k21444
answered Jun 20 '17 at 21:53
AretinoAretino
24.9k21444
answered Jun 20 '17 at 21:53
AretinoAretino
24.9k21444
24.9k21444
$begingroup$
I made a mistake downvoting this answer. I was about to say it didn't answer the question for the following reason. Although it can be proven, people don't feel the need to prove it and just accept that there exists a function satisfying these conditions. 1. For any real numbers x, y, z, w, d((x, y), (x + z, y + w)) = d((0, 0), (z, w)) 2. For any real number x, y, d((0, 0), (x, y)) is a nonnegative real number. 3. For any nonnegative real number x, d((0, 0), (x, 0)) = x 4. For any real numbers x, y, d((0, 0), (x, -y)) = d((0, 0), (x, y)). 5. For any real numbers x, y, z, w, d((0, 0), (xz - yw,
$endgroup$
– Timothy
Jan 22 at 5:44
$begingroup$
xw + yz)) = d((0, 0), (x, y)) × d((0, 0), (z, w)). So once they show that any function satisfying those conditions must be $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$, then they're fine with assuming that that function does satisfy those conditions. However, why should you assume that when you redefine distance using area, that new definition also satisfies those properties? Maybe because it can be shown that that new definition is in fact also $sqrt{x^2 + y^2}$ which is known to satisfy them. However, the task was to prove the Pythagorean theorem that way without reliance on a result
$endgroup$
– Timothy
Jan 22 at 5:52
$begingroup$
that wasn't proven under those other assumptions of distance which you didn't prove the distance formula from. Maybe it is however obvious that the composition of two operations on the unit square gets you a square with an area that's the product of the area of the squares each of those operations gets you because as described at math.stackexchange.com/questions/675522/…, any linear transformation multiplies the area of any shape by the same number.
$endgroup$
– Timothy
Jan 22 at 5:59
add a comment |
$begingroup$
I made a mistake downvoting this answer. I was about to say it didn't answer the question for the following reason. Although it can be proven, people don't feel the need to prove it and just accept that there exists a function satisfying these conditions. 1. For any real numbers x, y, z, w, d((x, y), (x + z, y + w)) = d((0, 0), (z, w)) 2. For any real number x, y, d((0, 0), (x, y)) is a nonnegative real number. 3. For any nonnegative real number x, d((0, 0), (x, 0)) = x 4. For any real numbers x, y, d((0, 0), (x, -y)) = d((0, 0), (x, y)). 5. For any real numbers x, y, z, w, d((0, 0), (xz - yw,
$endgroup$
– Timothy
Jan 22 at 5:44
$begingroup$
xw + yz)) = d((0, 0), (x, y)) × d((0, 0), (z, w)). So once they show that any function satisfying those conditions must be $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$, then they're fine with assuming that that function does satisfy those conditions. However, why should you assume that when you redefine distance using area, that new definition also satisfies those properties? Maybe because it can be shown that that new definition is in fact also $sqrt{x^2 + y^2}$ which is known to satisfy them. However, the task was to prove the Pythagorean theorem that way without reliance on a result
$endgroup$
– Timothy
Jan 22 at 5:52
$begingroup$
that wasn't proven under those other assumptions of distance which you didn't prove the distance formula from. Maybe it is however obvious that the composition of two operations on the unit square gets you a square with an area that's the product of the area of the squares each of those operations gets you because as described at math.stackexchange.com/questions/675522/…, any linear transformation multiplies the area of any shape by the same number.
$endgroup$
– Timothy
Jan 22 at 5:59
$begingroup$
I made a mistake downvoting this answer. I was about to say it didn't answer the question for the following reason. Although it can be proven, people don't feel the need to prove it and just accept that there exists a function satisfying these conditions. 1. For any real numbers x, y, z, w, d((x, y), (x + z, y + w)) = d((0, 0), (z, w)) 2. For any real number x, y, d((0, 0), (x, y)) is a nonnegative real number. 3. For any nonnegative real number x, d((0, 0), (x, 0)) = x 4. For any real numbers x, y, d((0, 0), (x, -y)) = d((0, 0), (x, y)). 5. For any real numbers x, y, z, w, d((0, 0), (xz - yw,
$endgroup$
– Timothy
Jan 22 at 5:44
$begingroup$
I made a mistake downvoting this answer. I was about to say it didn't answer the question for the following reason. Although it can be proven, people don't feel the need to prove it and just accept that there exists a function satisfying these conditions. 1. For any real numbers x, y, z, w, d((x, y), (x + z, y + w)) = d((0, 0), (z, w)) 2. For any real number x, y, d((0, 0), (x, y)) is a nonnegative real number. 3. For any nonnegative real number x, d((0, 0), (x, 0)) = x 4. For any real numbers x, y, d((0, 0), (x, -y)) = d((0, 0), (x, y)). 5. For any real numbers x, y, z, w, d((0, 0), (xz - yw,
$endgroup$
– Timothy
Jan 22 at 5:44
$begingroup$
xw + yz)) = d((0, 0), (x, y)) × d((0, 0), (z, w)). So once they show that any function satisfying those conditions must be $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$, then they're fine with assuming that that function does satisfy those conditions. However, why should you assume that when you redefine distance using area, that new definition also satisfies those properties? Maybe because it can be shown that that new definition is in fact also $sqrt{x^2 + y^2}$ which is known to satisfy them. However, the task was to prove the Pythagorean theorem that way without reliance on a result
$endgroup$
– Timothy
Jan 22 at 5:52
$begingroup$
xw + yz)) = d((0, 0), (x, y)) × d((0, 0), (z, w)). So once they show that any function satisfying those conditions must be $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$, then they're fine with assuming that that function does satisfy those conditions. However, why should you assume that when you redefine distance using area, that new definition also satisfies those properties? Maybe because it can be shown that that new definition is in fact also $sqrt{x^2 + y^2}$ which is known to satisfy them. However, the task was to prove the Pythagorean theorem that way without reliance on a result
$endgroup$
– Timothy
Jan 22 at 5:52
$begingroup$
that wasn't proven under those other assumptions of distance which you didn't prove the distance formula from. Maybe it is however obvious that the composition of two operations on the unit square gets you a square with an area that's the product of the area of the squares each of those operations gets you because as described at math.stackexchange.com/questions/675522/…, any linear transformation multiplies the area of any shape by the same number.
$endgroup$
– Timothy
Jan 22 at 5:59
$begingroup$
that wasn't proven under those other assumptions of distance which you didn't prove the distance formula from. Maybe it is however obvious that the composition of two operations on the unit square gets you a square with an area that's the product of the area of the squares each of those operations gets you because as described at math.stackexchange.com/questions/675522/…, any linear transformation multiplies the area of any shape by the same number.
$endgroup$
– Timothy
Jan 22 at 5:59
add a comment |
$begingroup$
None of the other answers give a rigourous proof that the Pythagorean theorem holds for all right angle triangles in $mathbb{R}^2$, even ones whose legs are not parallel to the axes so I will give my own proof of that in this answer. It is technically not a valid proof of the Pythagorean theorem. However, once you give a precise definition of distance, you can deduce from the distance formula the Pythagorean theorem and the statement that for any square, its area is the square of the length of its edge.
We seek a definition of distance from any point in $mathbb{R}^2$ to $mathbb{R}^2$, a function from $(mathbb{R}^2)^2$ to $mathbb{R}$ that satisfies the following properties.
- For any points $(x, y)$ and $(z, w)$, $d((x, y), (x + z, y + w)) = d((0, 0), (z, w))$
- For any point $(x, y)$, $d((0, 0), (x, y))$ is nonnegative
- For any nonnegative real number $x$, $d((0, 0), (x, 0)) = x$
- For any point $(x, y)$, $d((0, 0), (x, -y)) = d((0, 0), (x, y))$
- For any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$
Suppose a function $d$ from $(mathbb{R}^2)^2$ to $mathbb{R}$ satisfies those conditions, then for any point $(x, y)$, $d((0, 0), (x, y))^2 = d((0, 0), (x, y))d((0, 0), (x, y)) = d((0, 0), (x, y))d((0, 0), (x, -y)) = d((0, 0), (x^2 + y^2, 0)) = x^2 + y^2$ so $d((0, 0), (x, y)) = sqrt{x^2 + y^2}$ so for any points $(x, y)$ and $(z, w)$, $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ Now I will show that $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ actually satisfies those properties. It's trivial to show that it satisfies the first 4 conditions. It also satisfies the fifth condition because for any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = sqrt{(xz - yw)^2 + (xw + yz)^2} = sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = sqrt{(x^2 + y^2)(z^2 + w^2)} = sqrt{x^2 + y^2}sqrt{z^2 + w^2} = d((0, 0), (x, y))d((0, 0), (z, w))$
As a result of this, from now on, I will define the distance from any point $(x, y)$ to any point $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$ and denote it as $d((x, y), (z, w))$. I will also use $d(x, y)$ as shorthand for $d((0, 0), (x, y))$ For any square, for any displacement along one of its edges, let $x$ be the $x$ component of its displacement and $y$ be the $y$ component of its displacement.
This image shows that the area of that square is $(x - y)^2 + 2xy = (x^2 - 2xy + y^2) + 2xy = x^2 + y^2 = (d(x, y))^2$
Source:
- https://www.maa.org/press/periodicals/convergence/proportionality-in-similar-triangles-a-cross-cultural-comparison-the-student-module
answered Jan 22 at 5:29
TimothyTimothy
326214
$endgroup$
add a comment |
$begingroup$
None of the other answers give a rigourous proof that the Pythagorean theorem holds for all right angle triangles in $mathbb{R}^2$, even ones whose legs are not parallel to the axes so I will give my own proof of that in this answer. It is technically not a valid proof of the Pythagorean theorem. However, once you give a precise definition of distance, you can deduce from the distance formula the Pythagorean theorem and the statement that for any square, its area is the square of the length of its edge.
We seek a definition of distance from any point in $mathbb{R}^2$ to $mathbb{R}^2$, a function from $(mathbb{R}^2)^2$ to $mathbb{R}$ that satisfies the following properties.
- For any points $(x, y)$ and $(z, w)$, $d((x, y), (x + z, y + w)) = d((0, 0), (z, w))$
- For any point $(x, y)$, $d((0, 0), (x, y))$ is nonnegative
- For any nonnegative real number $x$, $d((0, 0), (x, 0)) = x$
- For any point $(x, y)$, $d((0, 0), (x, -y)) = d((0, 0), (x, y))$
- For any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$
Suppose a function $d$ from $(mathbb{R}^2)^2$ to $mathbb{R}$ satisfies those conditions, then for any point $(x, y)$, $d((0, 0), (x, y))^2 = d((0, 0), (x, y))d((0, 0), (x, y)) = d((0, 0), (x, y))d((0, 0), (x, -y)) = d((0, 0), (x^2 + y^2, 0)) = x^2 + y^2$ so $d((0, 0), (x, y)) = sqrt{x^2 + y^2}$ so for any points $(x, y)$ and $(z, w)$, $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ Now I will show that $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ actually satisfies those properties. It's trivial to show that it satisfies the first 4 conditions. It also satisfies the fifth condition because for any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = sqrt{(xz - yw)^2 + (xw + yz)^2} = sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = sqrt{(x^2 + y^2)(z^2 + w^2)} = sqrt{x^2 + y^2}sqrt{z^2 + w^2} = d((0, 0), (x, y))d((0, 0), (z, w))$
As a result of this, from now on, I will define the distance from any point $(x, y)$ to any point $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$ and denote it as $d((x, y), (z, w))$. I will also use $d(x, y)$ as shorthand for $d((0, 0), (x, y))$ For any square, for any displacement along one of its edges, let $x$ be the $x$ component of its displacement and $y$ be the $y$ component of its displacement.
This image shows that the area of that square is $(x - y)^2 + 2xy = (x^2 - 2xy + y^2) + 2xy = x^2 + y^2 = (d(x, y))^2$
Source:
- https://www.maa.org/press/periodicals/convergence/proportionality-in-similar-triangles-a-cross-cultural-comparison-the-student-module
answered Jan 22 at 5:29
TimothyTimothy
326214
$endgroup$
add a comment |
$begingroup$
None of the other answers give a rigourous proof that the Pythagorean theorem holds for all right angle triangles in $mathbb{R}^2$, even ones whose legs are not parallel to the axes so I will give my own proof of that in this answer. It is technically not a valid proof of the Pythagorean theorem. However, once you give a precise definition of distance, you can deduce from the distance formula the Pythagorean theorem and the statement that for any square, its area is the square of the length of its edge.
We seek a definition of distance from any point in $mathbb{R}^2$ to $mathbb{R}^2$, a function from $(mathbb{R}^2)^2$ to $mathbb{R}$ that satisfies the following properties.
- For any points $(x, y)$ and $(z, w)$, $d((x, y), (x + z, y + w)) = d((0, 0), (z, w))$
- For any point $(x, y)$, $d((0, 0), (x, y))$ is nonnegative
- For any nonnegative real number $x$, $d((0, 0), (x, 0)) = x$
- For any point $(x, y)$, $d((0, 0), (x, -y)) = d((0, 0), (x, y))$
- For any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$
Suppose a function $d$ from $(mathbb{R}^2)^2$ to $mathbb{R}$ satisfies those conditions, then for any point $(x, y)$, $d((0, 0), (x, y))^2 = d((0, 0), (x, y))d((0, 0), (x, y)) = d((0, 0), (x, y))d((0, 0), (x, -y)) = d((0, 0), (x^2 + y^2, 0)) = x^2 + y^2$ so $d((0, 0), (x, y)) = sqrt{x^2 + y^2}$ so for any points $(x, y)$ and $(z, w)$, $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ Now I will show that $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ actually satisfies those properties. It's trivial to show that it satisfies the first 4 conditions. It also satisfies the fifth condition because for any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = sqrt{(xz - yw)^2 + (xw + yz)^2} = sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = sqrt{(x^2 + y^2)(z^2 + w^2)} = sqrt{x^2 + y^2}sqrt{z^2 + w^2} = d((0, 0), (x, y))d((0, 0), (z, w))$
As a result of this, from now on, I will define the distance from any point $(x, y)$ to any point $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$ and denote it as $d((x, y), (z, w))$. I will also use $d(x, y)$ as shorthand for $d((0, 0), (x, y))$ For any square, for any displacement along one of its edges, let $x$ be the $x$ component of its displacement and $y$ be the $y$ component of its displacement.
This image shows that the area of that square is $(x - y)^2 + 2xy = (x^2 - 2xy + y^2) + 2xy = x^2 + y^2 = (d(x, y))^2$
Source:
- https://www.maa.org/press/periodicals/convergence/proportionality-in-similar-triangles-a-cross-cultural-comparison-the-student-module
answered Jan 22 at 5:29
TimothyTimothy
326214
$endgroup$
None of the other answers give a rigourous proof that the Pythagorean theorem holds for all right angle triangles in $mathbb{R}^2$, even ones whose legs are not parallel to the axes so I will give my own proof of that in this answer. It is technically not a valid proof of the Pythagorean theorem. However, once you give a precise definition of distance, you can deduce from the distance formula the Pythagorean theorem and the statement that for any square, its area is the square of the length of its edge.
We seek a definition of distance from any point in $mathbb{R}^2$ to $mathbb{R}^2$, a function from $(mathbb{R}^2)^2$ to $mathbb{R}$ that satisfies the following properties.
- For any points $(x, y)$ and $(z, w)$, $d((x, y), (x + z, y + w)) = d((0, 0), (z, w))$
- For any point $(x, y)$, $d((0, 0), (x, y))$ is nonnegative
- For any nonnegative real number $x$, $d((0, 0), (x, 0)) = x$
- For any point $(x, y)$, $d((0, 0), (x, -y)) = d((0, 0), (x, y))$
- For any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$
Suppose a function $d$ from $(mathbb{R}^2)^2$ to $mathbb{R}$ satisfies those conditions, then for any point $(x, y)$, $d((0, 0), (x, y))^2 = d((0, 0), (x, y))d((0, 0), (x, y)) = d((0, 0), (x, y))d((0, 0), (x, -y)) = d((0, 0), (x^2 + y^2, 0)) = x^2 + y^2$ so $d((0, 0), (x, y)) = sqrt{x^2 + y^2}$ so for any points $(x, y)$ and $(z, w)$, $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ Now I will show that $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ actually satisfies those properties. It's trivial to show that it satisfies the first 4 conditions. It also satisfies the fifth condition because for any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = sqrt{(xz - yw)^2 + (xw + yz)^2} = sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = sqrt{(x^2 + y^2)(z^2 + w^2)} = sqrt{x^2 + y^2}sqrt{z^2 + w^2} = d((0, 0), (x, y))d((0, 0), (z, w))$
As a result of this, from now on, I will define the distance from any point $(x, y)$ to any point $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$ and denote it as $d((x, y), (z, w))$. I will also use $d(x, y)$ as shorthand for $d((0, 0), (x, y))$ For any square, for any displacement along one of its edges, let $x$ be the $x$ component of its displacement and $y$ be the $y$ component of its displacement.
This image shows that the area of that square is $(x - y)^2 + 2xy = (x^2 - 2xy + y^2) + 2xy = x^2 + y^2 = (d(x, y))^2$
Source:
- https://www.maa.org/press/periodicals/convergence/proportionality-in-similar-triangles-a-cross-cultural-comparison-the-student-module
answered Jan 22 at 5:29
TimothyTimothy
326214
edited Feb 19 at 6:09
answered Jan 22 at 5:29
TimothyTimothy
326214
answered Jan 22 at 5:29
TimothyTimothy
326214
answered Jan 22 at 5:29
TimothyTimothy
326214
326214
add a comment |
add a comment |
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$begingroup$
Related: See my answer to this question math.stackexchange.com/questions/675522/…. This also provides an answer to your question: Most proofs of the PT use existence of a certain function called "area" satisfying certain properties. One can prove that there is no function satisfying all but one of these axioms and also, as you suggested, with $A(b)=b$.
$endgroup$
– Moishe Cohen
Jun 20 '17 at 3:25
$begingroup$
proofwiki.org/wiki/Area_of_Square
$endgroup$
– Ryan A
Jun 20 '17 at 3:27