$\forall x,y \in \mathbb R ,x \lt y \implies \exists z,$s.t. $x \lt z \lt y$.

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 $x \lt z \lt y$.

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

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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 $y$.