Prove $∃x ∈ N∀y ∈ N : x < y$

Question

How can I proof this formula if it is true or false.
I know that this must be false, but how can I make the proof.

$$∃x ∈ N∀y ∈ N : x < y$$

Is that the right negation of the formula:

$$∀x ∉ N ∃y ∈ N : x > y$$

Answer 1

Suppose it is true.

Then some $x\in N$ exists with $x<y$ for every $y\in N$.

But then also $x<x$ for this $x\in N$.

Preassuming that $<$ denotes an irreflexive relation on $N$ a contradiction has been found.

This allows the conclusion that the statement is not true.

Answer 2

If you have $0 \in \Bbb N$ you can just say that the statement fails for $y=0$ because there is no $x \lt 0$. If not, do the same with $1$.

Answer 3

To prove that the first statement isn't true, I would provide a counterexample.

It is stipulated that there exists some $x\in \mathbb{N}$, so that for all $y\in \mathbb{N}$, $x < y$ holds. So let $x_0$ be that $x$. It is now stipulated that for all $y\in \mathbb{N}$, $x_0<y$. Now put $y = 0$ (if $0\in \mathbb{N}$ for you), since it is one of those "all $y \in \mathbb{N}$". Now it is stipulated that $x_0 < 0$. Which is false for any $x_0$. So there cannot exist such an $x_0$. The $y = 0$ provides a counterexample to the statement that some property holds for all $y$, given some $x$.

But the negation of the formula is certainly true: $$ \forall x\in \mathbb{N} \ \exists y\in \mathbb{N} : x \geq y$$