Understanding a 'geometrical proof' of irrationality of √2 I had been having trouble understanding a proof of the irrational nature of √2.
I found this proof in the first page of the foreward to 17 theorem provers of the world where a 'geometrical proof' (is that a right  term?) of the irrationality of √2 is mentioned (page 2 of the pdf).
It goes like this:

Call the original triangle ABC, with the right angle at C. Let the hypothenuse AB = p, and let the legs AC = BC = q. As remarked, p²  = 2q².
Reflect ABC around AC obtaining the congruent copy ADC. On AB position E so that BE = q. Thus AE = p − q. On CD position F so that BF = p. Thus DF = 2q − p. The triangle BF E is congruent to the original triangle ABC. EF is perpendicular to AB, the lines EF and AD are  parallel.
Now, position G on AD so that AG = EF = q. Since AEFG is a rectangle, we find AG = q. Thus, DG = F G = AE = p − q. So, the triangle   DFG is an isosceles right triangle with a leg = p − q and hypothenuse = 2q − p.
If there were commensurability of p and q, we could find an example with integer lengths of sides and with the perimeter p + 2q a     minimum. But we just constructed another example with a smaller perimeter p, where the sides are also obviously integers. Thus, assuming commensurability leads to a contradiction.


What I understood was, we assume that we have got the smallest possible isosceles right triangle: ABC.
It has hypotenuse p and leg q, where p,q ∈ ℤ.
ie, p and q are smallest pair integers for which ABC would be a right isosceles triangle.
That means p² = 2q².
Then we derive a smaller isosceles right triangle (DFG) with hypotenuse 2q-p and leg p-q.
Thus, the assumption that ABC was the smallest right isosceles triangle was wrong.
But how can that be used to say that p and q are not commensurable?
(p and q being commensurable means p/q is rational, right?)
My level of math knowledge is very basic.
Could you help me understand this?
 A: This is an example of argument by contradiction. The key idea is to take a fact that you hope to eventually show is false, and instead assume that it is true, but that the assumption of truth leads you to a contradiction.
The classic example of this style of argument is the proof that there's no largest prime; if there was, then that would mean I could make a finite list of all primes, such as $\{2, 3, \dots, p\}$, where $p$ was the largest prime. But if that were true, I would notice that the expression $2 \cdot 3 \cdots p + 1$ would not be divisible by any of my primes -- hence, it must be prime itself, and it's necessarily larger than $p$, which leads me to a contradiction. This implies that my assumption ("there exists a largest prime") must have been incorrect.
In this case, we start by assuming that there exists an rational isosceles right triangle (meaning one whose sides are all fractions) -- that is, $\left( \frac a b \right)^2 + \left( \frac a b \right)^2 = \left(\frac c d \right)^2$. If so, multiplying both sides by $(bd)^2$ would give us an integer isosceles right triangle, as we'd have $(ad)^2 + (ad)^2 = (cb)^2$.
If there are integer isosceles right triangles, then by looking at the length of the shorter sides we see that there must be a smallest integer isosceles right triangle (this is an application of the well-ordering principle). But, the argument above shows that we could find a smaller one than the one we had, which contradicts that we had the smallest one.
Since there can't be a smallest integer isosceles right triangle, there can't be any integer isosceles right triangles at all, which also means there couldn't have been a rational isosceles right triangle to begin with.
A: If $\sqrt 2$ is rational, it can be expressed as $p/q$ where $p,q$ are integers. (Is this what you were missing?) The proof then goes on to construct a rational representation of $\sqrt 2$ with smaller integers in the numerator and denominator. But if this were possible, you could repeat this process indefinitely, which would lead to an infinite sequence of decreasing positive integers -- a contradiction.
