# What is the negation of “there are infinitely many integer solutions $(x,y)$ where $x$ is odd”

Is the negation "there are infinitely many integer solutions $(x,y)$ where $x$ is even"?

no its there are only finite integer solututions (x,y) where x is odd, otherwise the statement and its negation could both be true i.e consider all integer pairs (x,y) then the first and second statements i.e. infinite even and odd x would still occur.

No, it's "There are only finitely many integer solutions $(x, y)$ where $x$ is odd."

Your proposed negation doesn't work because the original statement might still be true!

• ha beat you by 18 seconds :P – shai horowitz Nov 1 '16 at 19:05
• Darnit, lol. Was so close D= – Asker Nov 1 '16 at 19:05

If your equation is $x = y$, then there are infinitely many even solutions and infinitely many odd solutions, so these clearly aren't negations of each other.

The negation is "there are only finitely many integer solutions $(x,y)$ where $x$ is odd".

Hint:

the set $E=\{(x,y)\in \mathbb N^2 : (x,y)$is a solution and $x$ is odd$\}$ is infinite.
the set $E$ is finite.