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

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!

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.

• Thank you for your response. This seemed to be the most helpful (for now) because the textbook I am using mentions the Well-Ordering Principle prior to this statement. Two questions I would like to ask: 1) Why did you define $A$ to be the set of integers greater than $t$? 2) I do not understand your reasoning with $n-s$. I am not sure how you can justify that $m-1-s > t - s - 1$. Could you please explain what you are doing?
– John
Aug 17, 2020 at 19:46
• 1) I wanted to find a set of integers which had a least element that was connected in some way to the number we were looking for. This can be done in several ways. You could for example also pick the set of integers greater than $s$. 2) The step $m - 1 - s > t - s - 1$ is justified by $m > t$ since $m \in A$. The second step $t - s - 1 > 1 - 1$ is because by your assumption, $t - s > 1$. Hope this helps, tell me if you have more questions! :) Aug 18, 2020 at 6:48

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 $$n=\lfloor s\rfloor +1$$.

• Although the textbook I am using did not mention the greatest integer function definition, this is a nice layout to frame a good response. Thank you
– John
Aug 17, 2020 at 19:49