# Proof by contradiction using divisibility

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?

• In particular, $b$ would not divide itself. Mar 15, 2016 at 19:47

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 $b=0$