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I suggest the following simplification of the proof:

Assume (p/q)^2 = 2, AND p/q is written in lowest terms, i.e. p,q have no common factor. If they do, just cancel it to MAKE p/q be in lowest terms.

(It is not hard to prove: If X is rational, there exists a lowest terms representation X = p/q)

Then: p^2/q^2 = 2

and p^2 = 2q^2.

So p is even and can be written p = 2n.

Then p^2 = 4n^2, so 4n^2 = 2q^2 and 2n^2 = q^2

So q is even. Therefore, p/q is NOT in lowest terms, which is your contradiction.

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