Why is the following expression is false

Question

Can anybody explain following expression and why is it false?

$$\exists x\in \mathbb N: \forall y\in \mathbb Z: y = x^2 $$

$\mathbb N$ : Natural numbers,
$\mathbb Z$ : Integers
When there several quantifiers, what is the best way to convert that expression into plain English? I'm having troubles with this so much. Thanks!

Answer 1

Read it in order from left to right:

there is a natural number $x$ such that every integer $y$ is equal to $x^2$.

This is saying that there is one particular natural number, which we’re calling $x$, that is simultaneously the square of every integer. This is clearly impossible: the integers $1$ and $2$ have different squares, so there is no natural number that is both $1^2$ and $2^2$, let alone the square of every integer.

Answer 2

"There is a natural number $x$ such that every integer $y$ is the square of $x$."

or, slightly more formal:

"There is an $x$ in $\mathbb{N}$ such that every $y$ in $\mathbb{Z}$ satisfies $y=x^2$."

Now, why is it false? Well, you cannot find any single natural number that is the square of all integers at the same time...