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

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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.

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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!

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

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Hint:

Your statement is

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

the negation is

the set $E$ is finite.

I am sure no point.

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Let me put it simple. In your statement you assert two things. X is odd. There infinitely many X. Your statement doesn't talk about the existence of x which are not odd. So negating it doesn't make a negation of the statement, as it statement doesn't deny that. That leaves is with the other part. Infinitely many x. Negating this denys the statement. So the negation is "there for finite number of x for (x, y).

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