ciphergoth | Irrationality of sqrt(2) (Reply)

From:

Hopefully I wrote these without errors:

even/odd:

sqrt(2) = p/q for some integers p and q that are relatively prime (i.e. smallest reduced form)
2 = p²/q²
p² = 2q²
note: p is clearly even, so q must be odd (by our premise)
note that the square of an even integer must be of the form 4k for some integer k . So, we can substitute: 4k = 2q²
2k = q² , clearly q is even as well (contradiction)

well-ordering:

sqrt(2)*q = p for some integers p and q . w.l.o.g. let q by the smallest such integer
note that 1 < sqrt(2) < 2
so: sqrt(2) - 1 < 1
multiply both sides by q: (sqrt(2) - 1)*q < q
distribute: sqrt(2)*q - q
let: sqrt(2)*q - q = s where s is an integer
s*sqrt(2) is also an integer:
1. s*sqrt(2) = (sqrt(2)*q - q)*sqrt(2)
2. distribute: s*sqrt(2) = sqrt(2)*q*sqrt(2) - q*sqrt(2)
3. simplify: s*sqrt(2) = 2*q - q*sqrt(2)
4. substitute: s*sqrt(2) = 2*q - p
So s is both smaller than q and s*sqrt(2) is an integer. This violates our assumption that q was the smallest such integer.