How is this a contradiction? If [itex]p^2 = 2q^2[/itex] just says that [itex]p^2[/itex] is even, it doesn't say that 'p' is even. It is a very obvious mistake. So.. is this a stupidity on their part or something that i may have missed [or a stupidity] on my part..

I generally wouldn't have posted this here.. but since cut-the-not.org has quality articles posted.. i would like to rethink this problem before concluding that they made a mistake.

Slider142 never said p/q was an integer if p and q were integers, but did say p/q was rational if p and q were integers. 1 and 2 are nice integers =]

We have that p is an integer, and that p^2 is divisible by 2. In fact, p^2, if divisible by 2, must also be divisble by 4. This is because if we express p^2 as a product of its factors, [tex]p^2 = 2^2 * k^2 * p^2 ...[/tex] where k and p are other integers/factors. All the powers must be even, in particular for 2. This is because if the power were odd, then in the factorization for p, there would be a non-integer power - ie not allowed.

If p is odd, then p= 2n+1 for some integer n. Then p^{2}= (2n+1)^{2}= 4n^{2}+ 4n+ 1= 2(2n^{2}+ 2n)+ 1, "2 times an integer plus 1", and so odd. That is: If p is an odd integer, then so is p^{2}". It follows that if p^{2} is even then p must be even.

I would suggest that you not throw around words like "obviously wrong" and "stupidity" so easily. They tend to come back at you.

They were talking oranges but you are talking bananas. The contradiction is that p and q were originally considered coprime!! Since if the square root of 2 = p/q where p and q are integers, we can always reduce p and q to coprime integers by dividing by the common factor.

Yes 2p is always even but that has nothing to do with what he said. He, and your original post, was talking about squares of numbers. As I showed above, the square of any even number is even, the square of any odd number is odd.

If n^{2} is even, then n itself must be even. If it were odd, then n^{2} would have to be odd, not even.

That is fine but Can you follow HallsofIvy's Logic and show that If x is square then y can not be odd? Thus the contradictiopn since y was assumed to be odd?

Hi Ramsey, I am not sure what you want to know about my post...

given x = 2y, if y is odd then x cannot be a square since x is even.

proof: if x is a square it can be written as 4a^2, for an any a, and 4a^2 = 2y ==> 2a^2 = y ==> y is even, and this is a contradiction because we assumed that y is odd

I am not sure if this is what you want, so I'll reproduce a proof of irrationality of 2

supose [tex]\sqrt{2} = p/q[/tex] in the lowest terms ==> gcd(p,q) = 1

[tex]2 = \frac{p^2}{q^2}\ ==> \ 2q^2 = p^2\ ==>[/tex] p is even (this is the conclusion about the elementary principle: two numbers cannot be equal and one be even and the other be odd, and I am not talking about q, I am talking about 2q^2, this number is even)

as p is even, p = 2k ==> p^2 = 4k^2 ==> [tex]q^2 = 2k^2[/tex] which means that q is even also, so p/q is not in the lowest terms, and this is the contradiction, and then [tex]\sqrt{2}[/tex] cannot be expressed as a rational ==> [tex]\sqrt{2}[/tex] is irrational

[itex]p^2 = 2q^2[/itex]
so [itex]p^2[/itex] is even
so [itex]p[/itex] is even
so [itex]p[/itex]=0 mod 2
so [itex]p^2[/itex]=0 mod 4
so [itex]p^2=2q^2[/itex]=0 mod 4
so [itex]q^2[/itex]=0 mod 2
so [itex]q[/itex]=0 mod 2
so q is even
so 2|gcd(p,q)
so gcd(p,q)>1