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$p,q\in\mathbb{Q}$ such that $(q\neq 0)$, then $\frac { p }{ q } $ is rational.

Steps I took and my thoughts on this:

This seems awfully obvious, but yet I can't seem to organize my thoughts in any way to construct a formal mathematical proof. The most I can think of is as follows:

let $p=\frac { a }{ b } $ and $q=\frac { m }{ n }$; such that $a,b,m,n\in \mathbb{Z}$ and $(m\neq 0$, $b\neq 0$, $n\neq 0)$

$$\frac { p }{ q } =\frac { \frac { a }{ b } }{ \frac { m }{ n } } =\frac { a }{ b } \cdot \frac { n }{ m } =\frac { an }{ bm } $$

$an$ and $bm$ are both integers because $a,b,m,n$ are all integers.

I imagine that this is way off. Please put me on the right path, and forgive my amateur attempt at a proof. I have a lot to learn about writing mathematical proofs, and I am just starting.

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    $\begingroup$ This looks fine. Although, you also need the condition, $b\ne 0$. $\endgroup$ – Tim Raczkowski Oct 16 '15 at 0:09
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    $\begingroup$ A counterexample can only show that a statement is false. You must be misunderstanding something. $\endgroup$ – Tim Raczkowski Oct 16 '15 at 0:12
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    $\begingroup$ You are misreading the sentence. The phrase "using a counter example" is in the same clause as "disprove. $\endgroup$ – Tim Raczkowski Oct 16 '15 at 0:16
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    $\begingroup$ Read as: (Prove) or (disprove using a counter example). $\endgroup$ – hardmath Oct 16 '15 at 0:16
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    $\begingroup$ Yep that does it. Just one more thing: add "$\frac p q =$" to the front of the chain of equalities -- a nicety. Well done. $\endgroup$ – BrianO Oct 16 '15 at 1:10
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That's not way off. It's close. In fact you are done and you were correct.

First figure out what rational means, we did that right: p = a/b, q = m/n a,b,m,n in Z, b, n ne 0. Then you identified p/q = qm/bn and as qm, and bn are in Z and bn ne 0, p/q is rational. You did it perfectly.

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The only thing I might add would be using closure of integers for multiplication to show that the new numerator and denominator are themselves integers.

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