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We argue that 2 has no rational square root.

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

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

Thm* ( p: q:  p p = 2 q q)

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:  p p = 2 q q  ( p':  p'<p & q q = 2 p' p')

(see hypothesis Hyp:4 of the proof)

We rewrite

(p p = 2 q q) to (q q = 2 p' p')

and then

(q q = 2 p' p') to (p' p' = 2 q' 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:  p p = 2 q q  ( p':  p'<p & q q = 2 p' 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:  p p = a q q) with a gloss.