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Prove by contradiction that if b is an integer such that b does not divide k for every natural number k, then b=0.

I understand to begin by assuming the false statement: There exists an integer b such that b does not divide k (for every natural number k) and b is not equal to 0. But I am unsure how to proceed from here?

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    $\begingroup$ In particular, $b$ would not divide itself. $\endgroup$
    – Bernard
    Mar 15, 2016 at 19:47

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Suppose $b\neq 0$. Then $|b|>0$. Thus $|b|\in\mathbb{N}$. But every integer divides his absolute value. That is $b||b|$. However, by hypothesis, $b$ doesn't divide $|b|$. Contradiction!! Therefore $b=0$

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