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 $y$?

  • $\begingroup$ Have you heard of the Archimedean Property? $\endgroup$
    – user418131
    Oct 1, 2018 at 12:43
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    $\begingroup$ 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. $\endgroup$ Oct 1, 2018 at 12:44
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    $\begingroup$ But between $\frac1{y-x}$ and $(\frac1{y-x})+1$ might be a logical place to look for such a number. $\endgroup$ Oct 1, 2018 at 12:47

1 Answer 1


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! ]

Question : Why we need that ?


Assuming the existence of such an $\frac{m}{n}$ with $n > 0$. So, we have $x < \frac{m}{n}<y$. 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.

  • $\begingroup$ Thank you. Is there a way to use decimal expansion? $\endgroup$
    – The Ledge
    Oct 1, 2018 at 15:58
  • $\begingroup$ Wow, I forgot to accept an answer. Sorry that took me so long. $\endgroup$
    – The Ledge
    Sep 30, 2020 at 17:07

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