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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.

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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?

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  • $\begingroup$ If there is a positive rational solution $2\left(\frac{a}{b}\right)^2 = \left(\frac{c}{d}\right)^2$ then there is a positive integer solution since $2(ad)^2=(bc)^2$. And if there is a positive integer solution then there is a smallest positive integer solution $\endgroup$
    – Henry
    May 18 at 13:58
  • $\begingroup$ I didn't understand what you meant at all. Sorry I'm almost school-level math. Does positive rational solution for that expression means value for a,b,c,d? $\endgroup$
    – J...S
    May 18 at 14:03
  • $\begingroup$ It means positive integer values for $a,b,c,d$ so $\frac ab,\frac cd$ are positive rationals $\endgroup$
    – Henry
    May 18 at 14:08
  • $\begingroup$ Okay, if there are positive rational value for a,b,c,d such that 2(a/b)² = (c/d)², then there are integer values for a,b,c,d as well? Is that how it is? $\endgroup$
    – J...S
    May 18 at 14:13

2 Answers 2

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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.

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  • $\begingroup$ An integer isosceles triangle means a triangle with all sides integers, right? $\endgroup$
    – J...S
    May 18 at 14:25
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    $\begingroup$ and the explanation of 'no largest prime' was lucid and could understand easily. I hadn't known it before. A separate thanks for that. :) $\endgroup$
    – J...S
    May 18 at 14:25
  • $\begingroup$ Okay I googled and see that it is indeed a trianlge with all sides integers. Sorry should've searched before. $\endgroup$
    – J...S
    May 18 at 14:26
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    $\begingroup$ Oh wait. p is an integer. It's p/q that's not an integer. $\endgroup$
    – J...S
    May 18 at 14:30
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    $\begingroup$ Thanks! With your and Tony's help, I think I get it now. $\endgroup$
    – J...S
    May 18 at 14:54
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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.

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  • $\begingroup$ 'infinite sequence of positive integers' part makes me see it more clearly. We can make an infinite series of triangles but once we start from an integer we can't count down infinitely because at some point in time it got to go below 0. Right? $\endgroup$
    – J...S
    May 18 at 14:22
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    $\begingroup$ Yes, that's right. You can turn this into a purely algebraic proof if you like, by noting that if $p^2=2q^2$, then (i) $(2q-p)^2=2(p-q)^2$ and (ii) $p-q<q$. So $\frac{2q-p}{p-q}$ is a "smaller" rational representation of $\sqrt 2$. $\endgroup$
    – TonyK
    May 18 at 14:26

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