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I have gotten this far, but I'm not sure how to make it apply to all rational and irrational numbers....

http://i.imgur.com/6KeniwJ.png">

BTW, I'm quite newbish so please explain your reasoning to me like I'm 5. Thanks!

UPDATE:

enter image description here

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Let $p/q$ be a rational number and $r$ be an irrational number.

Consider the number $w = \dfrac{p/q+r}2$ and prove the following statements.

$1$. If $p/q < r$, then $w \in ]p/q,r[$. (Why?)

$2$. Similarly, if $r < p/q$, then $w \in ]r,p/q[$. (Why?)

$3$. $w$ is irrational. (Why?)

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  • $\begingroup$ (p/q,r) means all the numbers between p/q and r not including r, right? thanks! $\endgroup$ – papercuts Jan 25 '13 at 21:18
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    $\begingroup$ @papercuts Yes. $(a,b)$ denotes $\{x \in \mathbb{R}: a < x < b\}$. I have changed $(a,b)$ to $]a,b[$, which also denotes the same thing. Sometimes the notation $(a,b)$ might be confused with the notation for coordinates in a $2$D coordinate system. Hence, I have used $]a,b[$ to denote the open interval i.e. $\{x \in \mathbb{R}: a < x <b\}$. $\endgroup$ – user17762 Jan 25 '13 at 21:18
  • $\begingroup$ Could you have a look at my update? I'm just struggling with the last bit now. Can I say something like "this works for any rational and any irrational number, therefore QED" or something? There was something about "Without a loss of generality" in our textbook, that I feel might be something we should use here, but I'm not sure how to word it. THANKS! $\endgroup$ – papercuts Jan 25 '13 at 21:48
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let $a$ be rational and $b$ be irrational $\frac{a+b}{2}$ is between them and irrational.

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