# Prove there is an integer larger than a given real number

A homework problem guides me to prove that there is a rational number $$\frac{m}{n}$$ between every two real numbers $$x$$ and $$y$$. The first step requires me to prove that there exists an integer $$n$$ such that $$\frac{1}{y-x} < n$$. That you can always find a natural number larger than any given real number seems trivial obvious to me, but I can't figure out how to prove it. I thought of trying to use the fact that there is no largest natural number, but that doesn't prove that there is always one larger than a given real number. The assignment says that I can use the fact that $$\frac{1}{y-x}$$ has a decimal expansion to prove this first step, but I don't see how that helps.

Can someone please help me figure out how to prove that there exists an integer $$n$$ such that $$\frac{1}{y-x} < n$$ for all real $$x$$ and $$y$$?

• Have you heard of the Archimedean Property? – AnotherJohnDoe Oct 1 '18 at 12:43
• When needing to prove some "trivial" result you need to have a clear picture of what axioms you are beginning with. Your question appears to be lacking any statement about that. – user334732 Oct 1 '18 at 12:44
• But between $\frac1{y-x}$ and $(\frac1{y-x})+1$ might be a logical place to look for such a number. – user334732 Oct 1 '18 at 12:47

Since $$y-x>0$$, by Archimedean property, there exits $$n \in \Bbb{N}$$ such that $$n(y-x)>1$$
[For, suppose it is false. Then we have $$n \leq \frac{1}{y-x}, \forall n.$$ which means $$\frac{1}{y-x}$$ is an upper bound for $$\Bbb{N}$$, a contradiction! ]
Assuming the existence of such an $$\frac{m}{n}$$ with $$n > 0$$. So, we have $$x < \frac{m}{n}. That is, $$nx < m < ny$$. Thus we are claiming that the interval $$(nx, ny)$$ contains an integer. It is geometrically obvious that a sufficient condition for an interval $$J = (nx, ny)$$ to have an integer in it is that its length $$ny-nx$$ is greater than $$1$$.That is , $$n(y-x)>1$$. Archimedean property assures of such $$n$$'s.