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Proofs homework question, here's what I've figured out thus far.

Suppose m doesn't divide k. We need to then prove that m|l. If m doesn't divide k and m is a prime then we know m and k are co-prime - hcf (m,k) = 1. Which means 1 = ks + mt (for some integers; s,t).

I'm not really sure where to go from here to prove that m|l based on these facts, help?

Martin Sleziak
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user122661
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3 Answers3

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From your work, you can conclude that

$$l = (kl) s + m (lt)$$

Now $m | kl$ by assumption, so $m$ divides the right hand side of this equation. Hence it divides the left.

  • Alright, I think I understand..

    The right hand side of the equation is made up of the sum of two terms that are both divisible by m ( (kl)*s and m(lt) ) correct?

     – user122661 May 02 '14 at 00:21
  • Yes, exactly. At some point, you've probably shown that if $a | b$ and $a | c$, then $a | b + c$. –  May 02 '14 at 00:22
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The Bezout-based proof is a bit more intuitive interpreted modulo $\,m.\,$ Notice that the Bezout identity $\,\color{#c00}{1 = \gcd(k,m)} = a k\! +\! b m\,$ becomes $ $ mod $\,m\!:\ 1\equiv ak,\,$ hence $\,\color{#c00}{k^{-1}} \equiv a$ exists mod $\,m.\,$ Hence $\, m\mid k\,\ell\,\Rightarrow\,k\,\ell\equiv 0\!\!\!\!\! \overset{\large\quad \times\,\color{#c00}{k^{-1}}}\Rightarrow\!\!\ell\equiv 0\,\Rightarrow\,m\mid \ell.\ $ Therefore, mod $\,m,\,$ the proof amounts to the simple fact that elements $\,k\,$ coprime to the modulus are, by Bezout, invertible, so cancellable.

Remark $\ $ If you know the gcd Distributive Law $\rm\color{#0a0}{(DL)}$ you can write the proof very concisely

$$m\mid k\ell,m\ell\ \Rightarrow\ m\mid (k\ell,m\ell)\overset{\rm\color{#0a0}{(DL)}} = \color{#c00}{(k,m)}\ell = \ell\qquad\qquad$$

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Bill Dubuque
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$$\frac\ell m=\frac{\ell\cdot1}m=\frac{\ell(ks+mt)}m=s\cdot\frac{k\ell}m+\ell t.$$Since $\ell,s,t$ are integers, and since $\frac{k\ell}m$ is an integer, it follows that $\frac\ell m$ is an integer, Q.E.D.

bof
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