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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 $y>z>x$

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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
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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

1 Answer 1

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

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-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

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