If $x<y$ then $\exists\, M \in \mathbb{N}$ such that $\frac{1}{n}<y-x\;\forall\; n>M$

Question

I am wondering how I would

Prove for every pair of positive integers $x$ $y$ where such $x<y$ there exist a natural number $M$ such that if $n$ is a natural number and $n>M$ then $\frac{1}{n}<(y-x)$

So I did $y>x$ so then $y-x>0$

let $M=\frac{1}{y-x}+1$ and I guess you could use floor function on the $\frac{1}{y-x}$

So then if $n>M>\frac{1}{y-x}$ then $1/n<\frac{1}{y-x}$

not sure if this proof is ok.

Answer 1

You're on the right path. Just choose $M \in \mathbb{N}$ such that $M>\frac{1}{y-x}$ (existence is guaranteed by Archimedean property).

Answer 2

Let $S$ be the set of integers $n$ such that $$n>\frac{1}{y-x}$$ Then $S$ is non-empty and bounded below, and you can take $M$ to be the least element of $S$.