Show that $\rm lcm(a,b)=ab \iff gcd(a,b)=1$

Question

Show that $$\rm lcm(a,b)=ab \iff \gcd(a,b)=1.$$

My attempt: If $\gcd(a,b)=1$ then there exist two integers $r$ and $s$ such that $$ar+bs=1.$$

and then I'm stuck... any advice?

Answer 1

Theorem: For positive integers $a$ and $b$ we have:

$$\gcd(a,b)\text{lcm}(a,b)=ab.$$

Using this, suppose $\text{lcm}(a,b)=ab$ it follows that $\gcd(a,b)=1$. Also on the other way suppose $\gcd(a,b)=1$, using the above theorem we conclude that $\text{lcm}(a,b)=ab$.

Answer 2

If we have the prime decompositions of $a$ and $b$: $$ a = 2^{a_2} \cdot 3^{a_3}\cdot 5^{a_5}\cdots \qquad b = 2^{b_2} \cdot 3^{b_3}\cdot 5^{b_5}\cdots $$ where only a finite number of $a_p, b_p$ are non-zero, then we get $$ \gcd(a, b) = 2^{\min(a_2, b_2)}\cdot 3^{\min(a_3, b_3)}\cdot 5^{\min(a_5, b_5)}\cdots \\ \operatorname{lcm}(a, b) = 2^{\max(a_2, b_2)}\cdot 3^{\max(a_3, b_3)}\cdot 5^{\max(a_5, b_5)}\cdots $$ From there it's not difficult to conclude that $\operatorname{lcm}(a, b)\cdot \gcd(a, b) = ab$ (they must be equal, since they have the same prime decomposition), and then what you want to show follows easily.

Answer 3

If you want to do this without prime factorization and without the gcd-lcm theorem, here is another approach.

By the way, I am assuming here that $a$ and $b$ are positive integers.

Let $d=\gcd(a,b)$ and let $m=\mbox{lcm}(a,b)$.

Note that $\frac{ab}{d}=\frac{a}{d}\cdot b=\frac{b}{d}\cdot a$ is a common multiple of $a$ and $b$. So if $d>1$, $\frac{ab}{d}$ is a common multiple of $a$ and $b$ that is less than $ab$. So in this case, we would have $m<ab$. Then by contrapositive, $m=ab\Rightarrow d=1$.

For the other direction, if $m<ab$, then $m=ax=by$ for some positive integers $x,y$ with $x<b$ and $y<a$. Then $\frac{a}{y}=\frac{b}{x}$, and this common value is a common factor of $a$ and $b$. However, since $y<a$, this common factor, $\frac{a}{y}$, is greater than $1$. Thus $d>1$. So again by contrapositive, $d=1\Rightarrow m=ab$.

Answer 4

Hint $\, $ Show $\,n\mapsto ab/n\,$ bijects the common divisors of $\,a,b\,$ with the common multiples $\le ab.$ Being order-$\rm\color{#c00}{reversing}$, it maps the $\rm\color{#c00}{Greatest}$ common divisor to the $\rm\color{#c00}{Least}$ common multiple, i.e. $$\begin{align} {\rm\color{#c00}{G}CD}(a,b)\,\mapsto\ &\bbox[5px,border:2px solid #c00]{ab/\color{#0a0}{{\rm GCD}(a,b)} = {\rm \color{#c00}{L }CM}(a,b)}\\[.4em] &{\rm so}\ \ \color{#0a0}{{\rm GCD}(a,b)\!=\!1}\!\iff\! {\rm LCM}(a,b)\!=\!ab \end{align}\quad$$

Remark $\ $ For more on this (involution) duality between gcd and lcm see here and here.

Answer 5

$\dfrac{a\cdot b}{gcd(a,b)}=a\cdot\left(\dfrac b{gcd(a,b)}\right)=\left(\dfrac a{gcd(a,b)}\right)\cdot b$ is a multiple of both $a$ and $b$.

As $a\cdot b$ is known to be the smallest multiple, then $gcd(a,b)=1$.