Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am going through Apostol's and wondering if I am answering this questions correctly. It is as follows.

If $x$ and $y$ are arbitrary real numbers with $x \lt y$, prove that there is at least one real $z$ satisfying $x \lt z \lt y$.

Here is my answer

Choose $n$ such that $n \gt \displaystyle \frac 1 {y-x}$. Let $\displaystyle z=x+ \frac {1}{n}$ then $y>z>x$

share|cite|improve this question
Well, you need to prove that such an $n$ exists. And that your $z$ actually satisfies the inequalities you claim. This is far from the easiest way of doing this question though. – Chris Eagle Mar 1 '13 at 22:56
This is one of those cases where drawing a picture will tell you how to do this in as easy a way as @Chris hints. – Lubin Mar 1 '13 at 22:58
There is a more direct approach. Your proof works, but you have to show that there is such an $n$. On the other hand, there is a simple formula for one $z$ between $x$ and $y$. – Thomas Andrews Mar 1 '13 at 23:03
Is there any easy way to show that such an n exists? – AlexHeuman Mar 1 '13 at 23:04
If not, what would be a hint towards a simpler method? – AlexHeuman Mar 1 '13 at 23:04

Just take the average of $x$ and $y$.

share|cite|improve this answer
-1 If the OP is seeking for hints in the comments, please do not provide a solution. Now, the OP doesn't have to think. – JavaMan Mar 1 '13 at 23:10
Sometimes the obvious needs to be pointed out. And the OP still needs to understand how to prove the answer is correct. – Jim Mar 1 '13 at 23:19
This not one of those times. – JavaMan Mar 2 '13 at 0:19
This is absolutely one of those times. – Jim Mar 2 '13 at 0:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.