# How can $(\exists x\in Z)(\forall y\in Z)(x>y)$ and its negation, $(\forall x\in Z)(\exists y\in Z)(x\le y)$, both be true?

$$(\exists x\in Z)(\forall y\in Z)(x>y)$$

This statement is true since we can take any $$y\in Z$$, add $$1$$ to it which would yield $$x\in Z$$ always greater than $$y$$.

If we now negate this statement we get: $$(\forall x\in Z)(\exists y\in Z)(x\le y)$$ This statement should be false, but if we take any $$x\in Z$$, add $$1$$ to it, we get $$y\in Z$$ such that $$x\le y$$ which makes the negation of a true statement a true statement??

Now there's probably something really wrong in my reasoning so can someone clarify this a little bit?

$$(\exists x\in Z)(\forall y\in Z)(x>y)$$ is False, not True ...

There is an integer greater than all integers (including itself)?! No.

Indeed, when you say:

$$(\forall x\in Z)(\exists y\in Z)(x\le y)$$

This statement is true since we can take any $$y\in Z$$, add $$1$$ to it which would yield $$x\in Z$$ always greater than $$y$$.

what you really show is that:

$$(\forall y\in Z)(\exists x\in Z)(x>y)$$ is True (which indeed it is)

So, in case you had not yet realized this: the order of the quantifiers matters!

• Yes! I see the mistake now. I just started learning proof writing and I confused the order of quantifiers. That's a mistake I shouldn't have made. Nevertheless, thank you for your answer. I don't Know if it's Worth leaving the question online but maybe it helps someone. Oct 8 '19 at 20:34
• @ToTheSpace2 You're welcome! And yes, I think it might be helpful to someone ... your title is helpful in that regard as well. Oct 8 '19 at 20:40
• True, I could've written a more informative title. Oct 8 '19 at 21:01
• @ToTheSpace2 I think your title is just fine, actually! Oct 8 '19 at 21:11
• Now that it's corrected, It's perfect! Oct 8 '19 at 21:42