Prove by Contradiction: Let $a,b,k$ be an element of $\Bbb Z$. If $a|k$ or $b|k$ then $(ab)|k$.
How should I proceed? I have $a=\frac{k+s_1}{l}$, $b=\frac{k}{r}$ and $k=abp+s_2$ where $s_1,s_2,l,r,p$ are elements of an interger.
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This isn't true. If $a=b=k=3$, then $ab\not\mid k$. – Clayton Oct 29 '15 at 22:04
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The assertion is not true, so you can't prove it. The simplest counterexample is $a=1$, $b=2$, $k=1$. Note, however, that the statement $a\mid k$ translates into $k=as$, for some integer $s$. – egreg Oct 29 '15 at 22:07
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Well the statement says that a|k OR b|k so the statement could be true. – Natalie VelaCruz Oct 30 '15 at 00:02
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let a =2, b=4, k =12.
2|12 and 4|12, but 8 doesn't divide 12. That's a counter example
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Well the statement says that a|k OR b|k so the statement could be true – Natalie VelaCruz Oct 30 '15 at 00:03