Divisibility property: $\gcd(a,b)=1,\, a,b\mid c\Rightarrow ab\mid c$ 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 $a|b$?
 A: If $a\mid c$ and $b\mid c$, then $ab\mid bc$ and $ab\mid ac$, so
$$ab\mid\gcd(ac,bc)=c\gcd(a,b)=c$$
A: 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 $ab|c$.
A: 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 $ab |c$.
A: 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 $\,\gcd(a,b)=1.$
