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where the conditions are: $a \neq 0$, $b \neq 0$ and $a$ and $b$ are integers.

maybe i'm missing something very basic about the properties of an absolute values.

My approach was to supposed, on the contrary, that |b| >= |a|, but I'm always getting that |b| is indeed >= |a|

could someone help me?

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Are you missing a condition on $b/a$? – ADF Oct 19 '12 at 13:06
@ADF: Perhaps Draconar means $b\mid a$ (and in the title $|b|\le|a|$), but it certainly has to be cleared up before the question can be answered. – Brian M. Scott Oct 19 '12 at 13:09
I re-wrote the conditions. Is it clearer? – Draconar Oct 19 '12 at 13:13
Hint: IF $x$ is an integer and $x\neq 0$ then $|x|\geq 1$. – Thomas Andrews Oct 19 '12 at 13:14
up vote 1 down vote accepted

HINT: If $a\mid b$, then there is an integer $n$ such that $b=na$, and therefore $|b|=|n||a|$. Since $b\ne 0$, we know that $n\ne 0$, and therefore $|n|$ is a positive integer. Therefore $|n|\ge 1$. If you now multiply this inequality by the right thing, you’ll get the desired result.

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by 'this' inequality you mean $|n| \ge 1$, right? – Draconar Oct 19 '12 at 15:46
@Draconar: I do indeed. – Brian M. Scott Oct 19 '12 at 16:00
thank! got it :D hope it is right. Are we allowed to post solutions in here? – Draconar Oct 19 '12 at 17:03
@Draconar: Absolutely. In this case, though, I can just finish the argument, and you can compare that with what you have. Multiplying $|n|\ge 1$ by $|a|$, we get $|b|=|n||a|\ge 1\cdot|a|=|a|$, which is exactly what we wanted to prove. – Brian M. Scott Oct 19 '12 at 17:13

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