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If $a_n$ is Cauchy then for $q\in\mathbb{Q}$, the sequence $(q$ $a_n)_n$ is Cauchy.

Proof. Let $\varepsilon >0$. Assume $a_n$ is cauchy. Since $a_n$ is cauchy then there is a $N$ in $\mathbb{N}$ such that for all $n,m >N$ we have $\left| a_{n}-a_{n}\right| < \varepsilon$. So, xxx

Can you give a hint for proof?

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    I'm going to quibble about your notation. Does $(q a_n)n$ mean the the sequence: $\{q\cdot a_n \}$? I'm also going to quibble that $a_n$ is, presumably, a single real value. To say $a_n$ is Cauchy wouldn't make sense as $a_n$ is not a sequence but a single value. (Although if one used the notation $a_n =\{b_0, b_1, b_2.....\}$ it could be a sequence. But then the "n" bears in relevent meaning whatsoever.) – 2017-02-20
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    Hint: If $|a_i - a_j| < \epsilon$ what can you say about $|q\cdot a_i - q\cdot a_j|$? – 2017-02-20
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    Nother broader hint: If $a > 0$ then $b < c \iff ab < ac$. What if $a = |q|$? – 2017-02-20
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    @fleablood yes for second comment. – 2017-02-20

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Exploit the definition of $b_n=qa_n$: $$ |b_m-b_n|=|q||a_m-a_n|, $$ now note that $q$ does not depend on $n$, choose $\epsilon/|q|$ in the definition of Cauchy sequence (if $q$ is nonzero).

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    Except when $q=0$, but in that case it's obvious. – 2017-02-20
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    Yes, I implicitly supposed that :) correct though! – 2017-02-20
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    Nitpicking, but… `;-)` – 2017-02-20
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    Nitpicking is sometimes essential (I tend to be, usually):) appreciated. – 2017-02-20
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    So, I have a question that how can you say $b_n=qa_n$? – 2017-02-20
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    As the question is stated, seems like $q \in \mathbb{Q}$ is fixed and the sequence $b_n=(qa_n)_n$ is the sequence in which every term is $a_n$ (depending on $n$) times the rational number $q$ (fixed). – 2017-02-20