Please could I get some feedback on the correctness and overall quality of the following proof, which is meant to show that all Pythagorean triples contain unique integers.
To Prove: all Pythagorean triples contain unique integers.
First we note that in a right-angled triangle, if $c$ is the hypotenuse and $a,b$ are the remaining sides, $c > a$ and $c > b$ by triangle inequality. Now it remains to prove that
$\forall (a,b,c) \in \mathbb{N}$, where $a^2+b^2=c^2$, $a \ne b$
We use contradiction.
Assume that $a = b$. Then we have $a^2 + a^2 = c^2$
Thus, $c = \sqrt{2a^2} = a\sqrt{2}$
Since an integer multiple of an irrational number remains irrational, $c \notin \mathbb{N}$, which is a contradiction of our original assumption.
Therefore $a \ne b$
Since $a \ne b$ and $c > a$ and $c > b$,
$a,b,c$ are distinct integers and theorem is proven.
$\square$
Best Answer
I think your proof seems to work fine. However, maybe it’s worth mentioning when you state that ‘an integer multiple of an irrational number remains irrational’, you are excluding the case where that integer is zero. Obvious, but worth mentioning for rigour. Perhaps state at the beginning that $a,b,c$ are positive integers.
The only other thing I think you could make clearer is how you know that $c>a$ and $c>b$. The triangle inequality actually says that $a + b \geq c$ (or $a+b>c$ if you exclude degenerate triangles), so perhaps you should give a few more steps to show how you arrive at your conclusion.