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!

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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 Kohan Jun 20 '17 at 3:25
  • $\begingroup$ proofwiki.org/wiki/Area_of_Square $\endgroup$ – Ryan A Jun 20 '17 at 3:27

Any reasonable definition of areas will make the area of a square of side $s$ proportional to $s^2$. I'll give the proof and indicate [in brackets] what "reasonableness" is required to make the proof work.

Suppose $A$ is a reasonable definition of areas; $A(s)$ is the area of a square of side $s$. [This already presupposes one aspect of "reasonableness", namely that all squares with the same side length have the same area.] For any positive integer $q$, we can partition a square of side $1$ into $q^2$ smaller squares of side $1/q$, just by drawing perpendicular grid lines, at $1/q$ spacing, parallel to the sides of the original square. So each of these smaller squares has area $A(1)/(q^2)$. [Here I'm assuming another aspect of reasonableness: If several squares overlap only along edges, then the area of their union is the sum of their areas.] So $A(1/q)=A(1)(1/q^2)$. Now let $p$ be another positive integer and consider a square of side $p/q$. It can be split by grid lines into $p^2$ squares of side $1/q$. So, by the same reasonableness properties already used above, we get $$ A(p/q)=p^2A(1/q)=p^2A(1)(1/q^2)=A(1)(p/q)^2. $$ So we have $A(x)=A(1)x^2$ for all positive rational $x$.

As a final reasonableness property, I'll assume that $A$ is a monotone function: if $x<y$ then $A(x)\leq A(y)$. Then, for any positive real number $r$, we have that $A(r)$ is sandwiched between $A(x)$ and $A(y)$ for all positive rational $x$ and $y$ with $x<r<y$. That is, for all such positive rational $x$ and $y$, we have $A(1)x^2\leq A(r)\leq A(1)y^2$. The only number $A(r)$ that satisfies these inequalities, for all such $x$ and $y$, is $A(r)=A(1)r^2$.

So, up to a constant factor $A(1)$ which amounts to a choice of units, the usual notion of area is the only reasonable one, provided you accept that a reasonable notion of area must have the properties I stated during the proof.

(The last of the reasonableness assumptions could be replaced by assuming that $A$ is continuous. The required conclusion would still follow. I prefer to use monotonicity, because it's easier to state and to understand. In particular, I think my whole argument would have made good sense to Euclid; indeed, for all I know, it might be in Euclid's "Elements".)

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    $\begingroup$ I don't think you'll find this in Euclid because Euclid doesn't use multiplication of real numbers, and uses areas (and ratios between areas) everywhere that we would now use multiplication of lengths. If we (anachronistically) turn this into a definition of multiplication of real numbers, we would get that Euclid defines the product $xy$ of real numbers as the ratio between the area of an $x\times y$ rectangle and a $1\times 1$ square. $\endgroup$ – Eric Wofsey Jun 25 '19 at 2:49
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    $\begingroup$ This argument certainly does have the flavor of many exhaustion-type arguments in Euclid though and I suspect if he saw it he would consider it to be something he basically already "knew", even if he had never expressed it quite in those words. $\endgroup$ – Eric Wofsey Jun 25 '19 at 2:53
  • $\begingroup$ Actually, I take that back: Euclid does have a notion of multiplication separate from areas, given by repeated ratios (given ratios $a:b$ and $b:c$, you can take the ratio $a:c$). Book VI Proposition 23 then says that the two notions of multiplication (via area which is essentially taken axiomatically as a "reasonable" notion as in your answer, and via ratios which is essentially multiplication of Dedekind cuts of rationals) coincide. $\endgroup$ – Eric Wofsey Jun 25 '19 at 3:55

Fix an angle $\theta$, i.e. a number between 0 and $2\pi$. 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)

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  • $\begingroup$ 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 '19 at 5:37

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.

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  • $\begingroup$ 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 '19 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 '19 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 '19 at 5:59
  • $\begingroup$ I think you mean different definition for distance, not different definition for area. Area can be defined as an integral and be computed first and then you can pick a definition of distance based on area. $\endgroup$ – Timothy Jun 25 '19 at 0:27
  • $\begingroup$ My answer must be seen in the framework of euclidean geometry, I had in mind the same kind of argument that Andreas Blass exposed in his recent answer. $\endgroup$ – Aretino Jun 25 '19 at 13:35

Area can be defined using calculus. Suppose you have a square in $\mathbb{R}^2$. enter image description here Let $x$ be the change in the first coordinate going from E to D. Let $y$ be the change in the second coordinate going from E to D. We can see that the area of the square is $(x - y)^2 + 2xy = x^2 - 2xy + y^2 + 2xy = x^2 + y^2$. With no assumptions about what properties the distance formula follows, that just proves that the area is $x^2 + y^2$ and proves nothing about what the distance formula is. We could seek a function $d$ that's a binary function from $\mathbb{R}^2$ to $\mathbb{R}$, in other words, a function from $(\mathbb{R}^2)^2$ to $\mathbb{R}$ satisfying certain properties where we say that $\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((x, y), (z, w))$ is the distance from $(x, y)$ to $(z, w)$. We seek a function $d$ satisfying the following properties:

  1. The distance from any point to itself is 0
  2. For any square, the distance from any vertex to the one adjacent to it in the counterclockwise direction is the square root of the area

It can be easily shown that $d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$ is the unique function satisfying those properties. That just shows that using this definition of distance, the Pythagorean theorem holds for all right angle triangles whose legs are parallel to the axes. To show that the Pythagorean theorem holds for all right angle triangles, we also have to show that that function satisfies the following property

  • $\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$

That can be done as follows. $d((0, 0), (xz - yw, xw + yz)) = \sqrt{(xz - yw)^2 + (xw + yz)^2} = \sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = \sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = \sqrt{(x^2 + y^2)(z^2 + w^2)} = \sqrt{x^2 + y^2}\sqrt{z^2 + w^2} = d((0, 0), (x, y))d((0, 0), (z, w))$

Some people find other properties of distance so intuitive. How do we know that there exists a way of defining distance that satisfies all of them? Because it has been proven in this answer that $d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$ is the unique function satisfying the following properties

  1. $\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((x, y), (x + z, y + w)) = d((0, 0), (z, w))$
  2. $\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((x, y), (z, w))$ is nonnegative
  3. $\forall \text{ nonnegative } x \in \mathbb{R}d((0, 0), (x, 0)) = x$
  4. $\forall x \in \mathbb{R}\forall y \in \mathbb{R}d((0, 0), (x, -y)) = d((0, 0), (x, y))$
  5. $\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$

and it also satisfies the additional properties

  1. The area of any square is the square of the length of its edges
  2. $\forall x \in \mathbb{R}d((0, 0), (\cos(x), \sin(x))) = 1$

Image source: https://www.maa.org/press/periodicals/convergence/proportionality-in-similar-triangles-a-cross-cultural-comparison-the-student-module

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