There are many elementary proofs for the Pythagorean Theorem, but no matter they use areas, similarities, even algebraic proofs, it is not straightforward to tell why it is true tracing back to the (Euclidean geometry) axioms. Are all these proofs equivalent? Do they all track back to the same axioms?

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    $\begingroup$ Studious gives a nice answer here, for proofs that rely on the notion of area, math.stackexchange.com/questions/675522/… . $\endgroup$
    – Sawarnik
    Mar 31 '14 at 14:05
  • $\begingroup$ @Sawarnik, thanks. Studious answer fills the gap between the area proofs and the underlying theorem. $\endgroup$
    – ahala
    Mar 31 '14 at 14:16

Sure, the Pythagorean theorem is an item in the theory of Euclidean geometry, and it can be derived from the modern axioms of Euclidean geometry.

A full set of Euclidean geometry axioms contains the information about similarity and area that are sufficient to prove the Pythagorean theorem "synthetically," that is, directly from the axioms. The algebraic proofs are a little different, though!

It turns out that after defining the real numbers and basic algebra, you can create a model of Euclidean geometry in $\Bbb R\times \Bbb R$ which obeys all the Euclidean axioms. The algebraic operations in an algebraic proof reflect the synthetic axioms being used, but the direct connections are not obvious. You are still indirectly using the synthetic axioms, but they are all hidden assumptions about $\Bbb R\times\Bbb R$ and coordinate geometry.


Your Question is much more complicated than it looks.

First some philosophy:

Are all proof equivalent?

What do you mean by equivalent in relation to proofs? (not just in relation to this proof but to any proof at all)

Euclidean geometry:

Euclid was a bit lacks with his Postulates and Common Notions, The axioms of his geometry were only found in late 19 , beginning 20 century so it is reasonably fresh.

see for example: (there are many more and even within these examples there are different options)

PS: Make sure you use the axioms for Euclidean geometry, you need to add the parallel axiom or an axiom that (together with the other axioms) can proof it.

Theories (and Euclidean geometry is a theory) are defined by their theorems (everything that follows from the axioms and rules of inference) not by their axioms, so many different axiomatisations can give the same Theory. But for being the same theory they have to proof the same theorems.

Then the Pythagorean Theorem.

Yes you can proof the Pythagorean Theorem in any axiomatisation, if anything if you could not prove it it would not be Euclidean geometry.

Do they all track back to the same axioms?

No, there are different axiomatisations possible so they are not even able to track all back to on and the same set.

Hope it all helps

  • $\begingroup$ Equivalent = Relying the same set of axioms, usually. Therefore a proof that uses less axioms is "better", and a proof that uses different non-equivalent axioms is simply "different". $\endgroup$
    – yo'
    Mar 31 '14 at 19:45
  • $\begingroup$ @tohecz but then you get in a spin what are non-equivalent axioms, for example Tarski's axioms don't use angles does that make them 'better'? and you can string all axioms to one big axiom, (not very readable) again better? $\endgroup$
    – Willemien
    Mar 31 '14 at 20:19
  • $\begingroup$ @William if $A_1\wedge A_2\wedge\dots\wedge A_n \Leftrightarrow B$, then $B$ is equivalent to the system of axioms $A_1,\dots,A_n$, so I'm no quite sure what you speak to in the second part. And if $A\wedge B\Rightarrow T$ and $A\wedge C\Rightarrow T$ and $A\wedge B\not\Rightarrow C$ and $A\wedge C\not\Rightarrow T$, then you got two non-equivalent proofs of your theorem $T$. I would never "guess" which axioms are "better" in any other way than possibility to derive ones from the others. (But maybe it's just too late and I overlook some stupidity in my arguments.) $\endgroup$
    – yo'
    Mar 31 '14 at 20:31

If you skipped 10th grade plane geometry because you just loved algebra, your path would take you right to the complex numbers. You don't know any geometry but you like the idea of representing complex numbers

z = x + yi

as points/vectors in a two coordinate Real Number 'Plane', Of course you naturally put the pure imaginary numbers (y axis) 'orthogonal' to the pure real numbers (x axis).

As you play with your new toy, you begin to recognize geometric type things going on. You decide that no matter what, the 'vectors'

1 + 0i, 0 + 1i, -1 + 0i, 0 - 1i

must all be at a distance of 1 from 0, and you want to create an absolute value (length) for any complex number. You also want the length to be an extension of the absolute value function defined for (pure) real numbers.

If $z = 5i$ and $w = 3$, $zw = 15i$, so you think that, tentatively, you want to keep

$\lvert z w\rvert$ = $\lvert z\rvert \lvert w\rvert$

One day while fooling around you start to examine the conjugates of complex numbers. It then hits you! The distance traveled by going x steps horizontally and then y steps vertically is invariant under the sign of y.

So your 'killer' axiom is:

The absolute value of a complex number is equal to the absolute value of its conjugate:

$\lvert\bar z\rvert = \lvert z\rvert$

But now that is all you need, since

$z \bar z = x^2 + y^2 = \lvert z \bar z\rvert = \lvert z\rvert \lvert\bar z\rvert = \lvert z\rvert ^2$

You define the absolute value of z to be the square root $x^2 + y^2$, and wonder where that will take you.

Ans: The Euclidean 2-Dimensional Plane.

Exercise: Show that with this definition, the absolute value of a product is the product of the absolute values.


Pythagoras is equivalent to the parallel postulate.

  • $\begingroup$ To improve your answer I suggest you explain what the parallel postulate is, as not everyone knows it. $\endgroup$
    – Jori
    Mar 31 '14 at 14:17
  • $\begingroup$ Ok not a bad idea - however I would not want to merely recapitulate the discussion here en.m.wikipedia.org/wiki/Parallel_postulate so I will merely link to it. $\endgroup$ Mar 31 '14 at 14:55
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    $\begingroup$ I think this statement is misleading. To prove the equivalence, it is required to first state the pythagorean theorem in neutral geometry, and this is not obvious (see math.stackexchange.com/questions/1292453/…). $\endgroup$ Jun 28 '17 at 11:51

All proofs I have seen so far trackback to the parallel axiom, or the axioms of similarity(which I think are equivalent). I'd be interested to see these "algebraic" proof though. Anything proved by pure algebra alone must be universally true, which is not so in the case of the Pythagorean theorem. It must use (euclidean) geometry in some form or the other.

  • $\begingroup$ Hmm, "universally true": this is a dangerous and misleading concept :) $\endgroup$
    – rschwieb
    Mar 31 '14 at 13:48
  • $\begingroup$ @rschwieb I don't fully get what you mean. Please explain? $\endgroup$
    – Guy
    Mar 31 '14 at 13:48
  • $\begingroup$ Actually, I am the one who needs to as you to explain what "universally true" means. Otherwise I don't know how I would respond. $\endgroup$
    – rschwieb
    Mar 31 '14 at 13:49
  • $\begingroup$ @rschwieb okay. so I meant "universally true" to mean that algebra doesn't use a lot of axioms(apart from things like $a+b=b+a$, which can very well be seen as part of definition of +) so it doesn't have any implicit "A is true, but only so long as B true". What has been shown to be true, is always true. I might be wrong about algebra's (non)axiomatic status though. $\endgroup$
    – Guy
    Mar 31 '14 at 13:53
  • $\begingroup$ OK, that helps me get a better picture of what you're thinking :) It looks like you have a lot of ideas about truth which I probably am not very good at addressing. Here's my best shot: in mathematics, only concern yourself about truth within a set of axioms. Synthetic geometry has its own axioms, and algebra (usually via set theory) has its own axioms. Within the axioms of algebra, you can make a model of the axioms of Euclidean geometry, so you can do geometry in algebra and prove things there that the synthetic axioms prove. $\endgroup$
    – rschwieb
    Mar 31 '14 at 14:03

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