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The question was like this. "Prove that there exists an irrational number $\\a$ such that $x< au< y$ if $x< y$ and $u> 0$" This is my proof.

Let $a=\sqrt{2}$ which is irrational. Since $x<y$, we have $\frac{x}{\sqrt{2}}<\frac{y}{\sqrt{2}}$. By density of rational number, there exists a rational number $u>0$ such that $\frac{x}{\sqrt{2}}<u<\frac{y}{\sqrt{2}}$. Thus we have $x<\sqrt{2}u<y$ as desired.

Is my proof legit?

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  • $\begingroup$ they only want to find $a$ which is rational. In fact, $\sqrt{2} u$ may be rational that's why i didn't mention this. $\endgroup$ – Ling Min Hao Feb 28 '18 at 8:45
  • $\begingroup$ Sry I misunderstood something :-) So $x$, $y$, and $u$ are given and you have to find an irrational $a$ such that the inequality holds. Then you can choose $u$ like you did. But the main idea isn't bad. $\endgroup$ – Mundron Schmidt Feb 28 '18 at 8:48
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You're asked to find $a$ in terms of $x$, $y$, and $u$, so you can't just pick a value for $u$. Here's my version.


Choose any $x, y, u \in \mathbb R$ such that $x < y$ and $u > 0$. Then we know that $\frac{x}{u\sqrt 2}, \frac{y}{u\sqrt 2} \in \mathbb R$ and $\frac{x}{u\sqrt 2} < \frac{y}{u\sqrt 2}$. By the density of $\mathbb Q$ in $\mathbb R$, we know that there exists some $q \in \mathbb Q$ such that: $$ \frac{x}{u\sqrt 2} < q < \frac{y}{u\sqrt 2} \iff x < (q\sqrt 2)u < y $$ But then since $q \in \mathbb Q$ and $\sqrt 2 \in \mathbb R - \mathbb Q$, we may take $a = q\sqrt 2 \in \mathbb R - \mathbb Q$, as desired. $~~\blacksquare$

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  • $\begingroup$ You should take care of the case when $x<0$ and $y>0$, where $q=0$ would not be a good choice. But it's easy, use a rational $q$ between $0$ and $y/(u\sqrt{2})$. $\endgroup$ – egreg Feb 28 '18 at 10:39

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