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


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


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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?)

  • $\begingroup$ (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
  • 1
    $\begingroup$ @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

let $a$ be rational and $b$ be irrational $\frac{a+b}{2}$ is between them and irrational.


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