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Here goes.

Theorem. √3 is irrational, i.e., there is no pair of integers p,q (q ≠ 0) satisfying (p/q)

^{2}= 3.

Proof (by contradiction). Assume, for contradiction, that there are two integers p,q (q≠ 0) such that

(p/q)

^{2}= 3

Then p/q is odd (it is easily proved by contradiction that if n

^{2}is odd, then n must be odd for all integers n), i.e., there is an integer n such that

p/q = 2n + 1

and therefore

p = q(2n + 1).

This can be substituted into the first equation, giving

(2n + 1)

^{2}= 3

which expands to

2n(2n + 1) + 2n + 1 = 3 → 4n

^{2}+ 4n = 2 → 2(n

^{2}+ n) = 1

Since n

^{2}+ n is an integer (closure under addition and multiplication), this last statement implies that 1 is an even integer, which it is not. Therefore there is no pair of integers p,q (q ≠ 0) satisfying (p/q)

^{2}= 3, i.e., √3 is irrational, which is what was required to be shown.

That's the proof I came up with. My specific question is this: Most proofs of this theorem are similar to the proof of the irrationality of √2, i.e. they start by assuming p and q have no common factor and end by deriving a contradiction of this assumption. My proof does not derive a contradiction of a hypothesis but instead derives a contradiction of a known fact, i.e. that 1 is an odd integer. Is this a 'legal' move, or not?

Thanks for taking time to help.