$(∀x)(∃y)(x>y)$ is false. But then why is $ (∀x)(∃y)(x\geq y)$ true? Given the Universe is the set of natural numbers, then $(∀x)(∃y)(x>y)$ is false. But then why is $(∀x)(∃y)(x\geq y)$ true?
The first equation and the second equation is the same except for "=" in the second equation.
so i don't get how that affects the statement to this degree.
Thank you.
 A: Look at the smallest number in the natural numbers $\mathbb N$ (since you mention that the universe is the set of all natural numbers): the smallest number in $\mathbb N$ would be $0$ or $1$, depending on your definition.  Let's just go with $0$. 
Since $0 \in \mathbb N$, the universally quantified "x" means the inequality must hold for every natural number $x$, including $x = 0.\;$ Now, does there exist any $y \in \mathbb N$ such that that $0 > y\;$? 
On the other hand, if we allow equality too, then we have that it is true that there exists a $\,y \in \mathbb N\,$ such that $\;0 \geq y,\,$ namely, $\,y = 0:$ That is, it is certainly true that $\;0 \geq 0.$
The same logic applies if $x = 1$ instead of $x = 0$, if you are working with a definition of the natural numbers $\{1, 2, 3, \ldots\}$.
A: Any ordered set with a minimum element provides an example where the first is false and the second is true, because there can be nothing below the minimum element. 
A: Another take on this question is to try and write the latter in terms of the former:
\begin{align}
& \langle \forall x :: \langle \exists y :: x \geq y \rangle \rangle \\
\equiv & \;\;\;\;\;\text{"write $\;\geq\;$ in terms of $\;>\;$ -- since know something about that already"} \\
& \langle \forall x :: \langle \exists y :: x > y \:\lor\: x = y \rangle \rangle \\
\equiv & \;\;\;\;\;\text{"logic: $\;\exists\;$ distributes over $\;\lor\;$"} \\
& \langle \forall x :: \langle \exists y :: x > y \rangle \:\lor\: \langle \exists y :: x = y \rangle \rangle \\
\equiv & \;\;\;\;\;\text{"logic: one-point rule"} \\
& \langle \forall x :: \langle \exists y :: x > y \rangle \:\lor\: \textrm{true} \rangle \\
\equiv & \;\;\;\;\;\text{"logic: simplify using $\;P \lor \text{true} \equiv \text{true}\;$"} \\
& \text{true} \\
\end{align}
Now we see that we didn't even use the fact that $\;\langle \forall x :: \langle \exists y :: x > y \rangle \rangle\;$ is false on $\;\mathbb N\;$...
