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?