For each $x,y∈\mathbb N$, if $x≤y$ and $y≤x$, then $x=y$

Question

Proof: For each $x,y∈\mathbb{N}$, if $x≤y$ and $y≤x$, then $x=y$

By definition of order, $x≤y$ if and only if $x<y$ or $x=y$

By definition of order, $x<y$ if there exist a $K∈\mathbb{N}$ such that $y=x+K$

and similarly the same for $y≤x$ is $y<x$ or $y=x$

$y<x$ is $x=y+M$ for some $M∈\mathbb{N}$.

now from here I'm having trouble seeing exactly where I need to go to actually prove $x=y$

I would assume I could do this inductively but I'm having trouble setting it up.

Answer 1

You should say $x \le y$ if $x < y$ or $x = y$.

Assume $x \le y$ and $y \le x$. Suppose to the contrary that $x \not= y$. Then

1. $x \le y \implies x < y \implies y = x + k$ for some $k \in \mathbb N$, and
2. $y \le x \implies y < x \implies x = y + n$ for some $n \in \mathbb N$.

You get $x = x + n + k$. Can you get a contradiction?

Answer 2

$x\le y\iff y=x+A$ for some $A\in\Bbb Z_{\ge 0}$.

$y\le x\iff x=y+B$ for some $B\in\Bbb Z_{\ge 0}$.

Therefore $y=y+A+B$, i.e. $A+B=0$, i.e. $A=B=0$.