which axiom(s) are behind the Pythagorean Theorem
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)
- Hilbert's axioms http://en.wikipedia.org/wiki/Hilbert_axioms
- Tarski's axioms http://en.wikipedia.org/wiki/Tarski_axioms
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
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.