Let $s$ and $t$ be real numbers such that $t -s > 1$. Prove that there exists a positive integer $n$ such that $s < n < t$.

Question

I hope you are doing well!

I am doing a proof in elementary Real Analysis and wanted to explain my thought thus far:

Proof: By contradiction, suppose not. Then $\forall p \in \mathbb{N}$, we have $s \geq p$ or $p \leq t$.

I tried to go through the first case ($s \leq p$), but am having trouble at arriving to a contradiction. I believe I have to use the Archimedean Property of Real Numbers somehow. Could anyone please let me know if I am on the right track? If I am completely off, would someone please explain, in a basic way, how you would go about this proof?

Thank you and cheers!

Answer 1

I would use the well-ordering principle. It can be done as follows:

Let $A$ be the set of integers greater than $t$. By the well-ordering principle $A$ has a least element $m$. We will show that $n = m - 1$ satisfies your property. Indeed, $n < t$ as otherwise $n \in A$ means $m$ is not the least element. Further

$$n - s = m - 1 - s > t - s - 1 > 1 - 1 = 0$$

Hence, $s < n$, as required.

Answer 2

As $t-s>1$, we can be sure that $t > s+1 \geq \lfloor s+1\rfloor=\lfloor s\rfloor +1$. As $\lfloor s\rfloor +1$ is an integer, it is enough to choose $n=\lfloor s\rfloor +1$.