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If $a,b\in \mathbf{Q}^+$, how to prove $\{p\ |\ p\in \mathbf{Q}, 0<p<ab\}\subset \{rs\ |\ r,s\in \mathbf{Q},\ 0<r<a,\ 0<s<b\} $?

Here is my try:

Suppose $q\in \{p\ |\ p\in \mathbf{Q}, 0<p<ab\}$. Then $q\in\mathbf{Q}$ and $0<q<ab$. So $1<\frac{ab}{q}$ and $1<\sqrt{\frac{ab}{q}}$. So, we have $q=\frac{a}{\sqrt{\frac{ab}{q}}}\frac{b}{\sqrt{\frac{ab}{q}}}$. Here, $0<\frac{a}{\sqrt{\frac{ab}{q}}}<a$ and $0<\frac{b}{\sqrt{\frac{ab}{q}}}<b$. The problem is that $\frac{a}{\sqrt{\frac{ab}{q}}}$ and $\frac{b}{\sqrt{\frac{ab}{q}}}$ may not in $\mathbf{Q}$. As a result, I fail to prove it. Could you help me? Thank you.

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1 Answer 1

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The main point is that if $a,b$ are positive rationals and $0<p<ab$ then there exists a number $b’$ which is less than $b$, such that $0<p<ab’$.

Proof: Pick a small enough positive rational $\delta$ less than $b$, such that $p+a\delta<ab$ then we have $p<a(b-\delta)$ and set $b’=b-\delta$.

Now to prove the statement let $r=p/b’$ and $s=b’$.

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  • $\begingroup$ Thank you. Your answer is amazing!!!!! It seems that there is another method -- just use the proved lemma twice. But your method is more elegant. $\endgroup$
    – studyhard
    Commented Mar 7, 2022 at 16:01

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