# Thread: Contradiction 1 of 3 (I'll go through these 1 by 1 )

1. ## Contradiction 1 of 3 (I'll go through these 1 by 1 )

If a and b are Integers then a^ 2 −4b 2
Proposition , then a^ 2 −4b 2
This is what I've done

Assume

a^ 2 −4b = 2
a^2 = 2 + 4b
a^2 = 2(2b+1)
a^2 is even because if we sub anything into b it has an even answer.
because a^2 is even it follows that a is even so a = 2k for some integer k.
now put 2k back into original assumption
(2k)^2-4b=2,
so 4k^2-4b=2 now lets devide by 2
we get 2k^2-2b=1
∴ 1 = 2(k^2-b)
because c^2-b ∈ Z it follows that 1 is even ..
so it is incorrect
we assumed the proposition false but the proposition is true.

2. ## Re: Contradiction 1 of 3 (I'll go through these 1 by 1 )

A little awkwardly stated but the proof is correct.

3. ## Re: Contradiction 1 of 3 (I'll go through these 1 by 1 )

Thanks Halls of Ivy , I think I have a better understanding of recent than the early days of the course where I was painfully frustrating to correct and didn't even know what a P and Q was ...cheers