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suppose $a|c$ , $b|c$ and $gcd(a,b)=1$ .Then show that $ab|c$. Here $a,b,c$ are all real numbers.Can i start from the properties of divisibility as if $a|c$ and $c|b$ then $a|b$?

Bill Dubuque
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srimali
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4 Answers4

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If $a\mid c$ and $b\mid c$, then $ab\mid bc$ and $ab\mid ac$, so $$ab\mid\gcd(ac,bc)=c\gcd(a,b)=c$$

user2345215
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  • Remark that this implicitly uses uses the GCD distributive law, which can be replaced by Bezout if desired - see proofs 2 and 3 in my answer, and follow the link there for further exposition. – Bill Dubuque Mar 31 '14 at 17:48
  • Or see the previous versions of my post where I proved the required direction. I edited it out for clarity. – user2345215 Mar 31 '14 at 17:49
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You mean integers.

$a|c$ means $c = ak$ for some $k \in \mathbb{N}$.

Thus $b|c$ can be rewritten as $b|ka$, but $gcd(a,b)=1$ so $b|k$ which in turn means $k=lb$ for some $l \in \mathbb{N}$.

Finally $c = lba$, so $ab|c$.

todestrieb
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  • Remark that the proof implicitly invokes Euclid's Lemma (or unique factorization) to deduce that $,b\mid k.,$ This is the same as proof 1 in my answer, with $\ k,:=,c/b.\ \ $ – Bill Dubuque Mar 31 '14 at 17:54
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Since gcd $(a,b)=1$ hence $\exists x,y\in \mathbb{Z}$ such that $ax+by=1$. Thus, $acx+bcy=c$. Now $ab$ divides LHS (why ?) and hence $ab |c$.

wanderer
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  • Remark that the proof invokes Bezout's Identity for the gcd, which could be replaced by the GCD distributive law if so desired. See proofs 2 and 3 in my answer, where I align both versions to highlight the analogy between the two proofs (structure that will become clearer when one learns about ideals and/or divisor theory) – Bill Dubuque Mar 31 '14 at 17:56
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Below are a handful of proofs. One should suffice depending on what you already know.

If $\ \color{#0a0}{(a,b)=1}\,$ then $\smash[t]{\ a,b\mid c\,\Rightarrow\, \color{#0a0}a\mid \color{#0a0}b\,(c/b)\,\overset{\color{#c00}{\rm(E)}}\Rightarrow\, a\mid c/b\,\overset{\times\ b}\Rightarrow\,ab\mid c\ }$ by $\,\color{#c00}{\rm(E)} =$ Euclid's Lemma.

Alternatively, we can use the GCD Distributive Law $\color{#c0f}{\rm(DL)}$, or Bezout's Identity for the gcd

$\qquad a,b\mid c\,\Rightarrow\, ab\mid ac,bc\,\Rightarrow\, ab\mid \ (ac,\ \ bc)\ \, \overset{\color{#c0f}{\rm(DL)}} =\ \color{#0a0}{(a,\,b)}c = c\ \ $ by $ \ \ \color{#0a0}{(a,\ b)=1}$

$\qquad a,b\mid c\,\Rightarrow\, ab\mid ac,bc\,\Rightarrow\, ab\mid jac+kbc=\color{#0a0}{(ja\!+\!kb)}c = c\ $ by $\ \color{#0a0}{ja\!+\!kb=1}\ $ by Bezout

Or it is an straightforward consequence of the Fundamental Theorem of Arithmetic, i.e. the existence and uniqueness of prime factorizations.

Or, use the $ $ gcd * lcm Law $\ \gcd(a,b)\,{\rm lcm}(a,b) = ab,\,$ so $\,{\rm lcm}(a,b) = ab\ $ if $\,\gcd(a,b)=1.$

Bill Dubuque
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