For integers p, and q  0, the ratio of p to q is the square root of 2 just when pp = 2qq.

We shall consider only the non-negative integers here, so we must show that

Thm*  (p:, q:. pp = 2qq)

It is impossible to build an infinite decreasing sequence of natural numbers, as is expressed by our principle

Thm*  P:(Prop). (x:. P(x)  (x':. x'<x & P(x')))  (x:. P(x))

So, to show that 2 has no rational square root, it is enough to show that if you could find a ratio "p/q", expressed with a non-negative numerator, whose square was 2, then you could find a smaller such numerator p' of a rational root "p'/q'" i.e.,

p:, q:. pp = 2qq  (p':. p'<p & qq = 2p'p')

(see hypothesis Hyp:4 of the proof)

We rewrite

(pp = 2qq) to (qq = 2p'p')

and then

(qq = 2p'p') to (p'p' = 2q'q')

giving us a rational square root with numerator p'<p and denominator q'. EXAMPLE

These rewrites are justified by a special-purpose lemma

Thm*  p:, q:. pp = 2qq  (p':. p'<p & qq = 2p'p')

See the GLOSS of the lemma.

QED

There is also a proof generalizing from 2 to any prime. Thm*  a:. prime(a)  (p:, q:. pp = aqq) with a gloss.

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