Prove $\,\gcd(m,n)\operatorname{lcm}(m,n) = mn$

Question

I'm trying to prove that $\operatorname{lcm}(n,m) = nm/\gcd(n,m)$ I showed that both $n,m$ divides $nm/\gcd(n,m)$ but I can't prove that it is the smallest number. Any help will be appreciated.

Answer 1

${\bf\small Proof}\ \, d\,| \gcd(m,n)\!\!\!\overset{\ \ \rm\color{darkorange}U}\iff d\,|\, m,n\!\!\overset{\rm\color{#0a0}{\rm R}\!\!}\iff $ $ m,n\,\raise -2pt{\Large |}\, \color{#c00}{\frac{mn}d}\!\!\overset{\ \ \rm\color{darkorange}U}\iff {\rm lcm}(m,n)\,\raise -2pt{\Large |}\, \frac{mn}d\!\!\overset{\rm\color{#0a0}{\rm R}\!\!}\iff d\,\raise -2pt{\Large|}\,\frac{mn}{{\rm lcm}(m,n)}$ where above we used: $\,\rm \color{darkorange}U = $ LCM & GCD Definition / Universal Property. $\,\ \small\bf QED$

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More conceptually: elevating the key $\rm\color{#0a0}{R =}$ cofactor reflection symmetry yields a simpler proof: note $\,d\,\color{#0a0}\mapsto\, \color{#c00}{mn/d}\:$ bijects common divisors $\,d\,$ of $\,m,n\,$ with common multiples (that divide $\,mn).$ 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. $\,{\rm\color{#c00}{G}CD}(m,n)\,\color{#0a0}\mapsto\, mn/{\rm GCD}(m,n) = {\rm \color{#c00}{L }CM}(m,n).\,\ \small\bf QED$

See these posts for more on this $\:\!\rm\color{#0a0}{involution\ (reflection)}$ symmetry at the heart of gcd, lcm duality.

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Note $ $ For completeness, in $\,\Bbb N\!:\ (d\mid g\!\iff\! d\mid \ell)\,\Rightarrow\, g = l\,$ follows as here: $ $ both $\,g\,$ and $\,\ell\,$ have the same set of divisors $\,d\,$ so they have the same greatest divisor (themselves). Or, putting $\,d\in\{g,\,\ell\}\,$ in $\,d\mid g\iff d\mid \ell\,$ shows $\,g,\ell$ divide each other $\,g\mid\ell\mid g\,$ so $\,g\le \ell\le g\,$ so $\,\ell = g\,$ (in a general domain we deduce they are associates $\,g\approx \ell,\,$ hence equal up to a unit multiple).

Answer 2

Hint: For any $a,b$ real numbers: $\min(a,b)+\max(a,b)=a+b$.

Now, if we have $a=a_1^{p_1} a_2^{p_2}\ldots$ and similarly with $b$, if you use the equation I just mentioned for all $p_i$, you will get, that $\gcd(a,b)\cdot\operatorname{lcm}(a,b)=ab$.

Answer 3

Let's just do this directly. Let $g = \gcd(m,n)$. We need to prove that $\operatorname{lcm}(m,n) = \dfrac{mn}{g}$.

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STEP $0$. (Preliminary stuff.)

DEFINITION $1$. $L = \operatorname{lcm}(m,n)$ if and only if

 1. L is a multiple of m and of n.
 2. If C is a multiple of m and of n, then C is a multiple of L.

LEMMA $2$. If $\gcd(a,b) = 1$ and $a \mid bc$, then $a \mid c$.

PROOF. If $\gcd(a,b) = 1$, then there exists integers $A$ and $B$ such that $aA + bB = 1$. It follows that $acA + bcB = c$. Since $a | acA$ and $a \mid bcB$, then $a \mid c$.

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STEP $1$. $\dfrac{mn}{g}$ is a common multiple of $m$ and of $n$.

This is true because $\dfrac m g$ and $\dfrac n g$ are integers and $\dfrac{mn}{g} = \dfrac{m}{g}n = m \dfrac{n}{g}$.

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STEP $2$. If $G$ is a common multiple of $m$ and of $n$, then $G$ is a multiple of $\dfrac{mn}{g}$.

Suppose $G = mM = nN$ for some integers $M$ and $N$. Then $\dfrac G g = \dfrac m g M = \dfrac n g N$.

Since $\gcd\left( \dfrac m g, \dfrac n g \right) = 1$, and $\dfrac m g M = \dfrac n g N$, then, by LEMMA $2$, $\dfrac m g \mid N$, say $N = \dfrac m g N'$ for some integer $N'$.

So $\dfrac{G}{g} = \dfrac{n}{g} N = \dfrac{m}{g} \dfrac{n}{g} N'$. It follows that $G = \dfrac{mn}{g} N'$ and so $G$ is a multiple of $\dfrac{mn}{g}$.

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From STEP $1$, STEP $2$, and DEFINITION $0$, we can conclude that $\operatorname{lcm}(m,n) = \dfrac{mn}{\gcd(m,n)}$.

Answer 4

$\DeclareMathOperator{\lcm}{lcm}$Here is one way without using the Fundmental theorem of arithmetic just using the definitions

The definition of $lcm(a,b)$ is as follows:

$t$ is the lowest common multiple of $a$ and $b$ if it satisfies the following:

i) $a \mid t$ and $b \mid t$

ii) If $a \mid c$ and $b \mid c$, then $t \mid c$.

Similiarly for the $\gcd(a,b)$.

Here is my proof:

Case I: $\gcd(a,b) \neq 1$

Suppose $\gcd(a,b) = d$.

Then $ab = dq_1b = dbq_1 = d\cdot (dq_1q_2)$

Claim: $\lcm(a,b) = dq_1q_2$

$a = dq_1 \mid dq_1q_2$

$b = dq_2 \mid dq_2q_1$.

Supppose $\lcm(a,b) = c$. Hence $c \leq dq_1q_2$ .

To get the other inequality we have $dq_1|a$ and $dq_2|b$. Hence $dq_1 \leq a \leq c \leq dq_1q_2$. Similiarly for $dq_2$.

Suppose that $c$ is strictly less than $dq_1q_2$, so we have $dq_1q_2 < cq_2$ and $dq_1q_2 < cq_1$.

So $dq_1q_2 < c < cq_2 < dq_2^2q_1$ and $dq_1q_2 < c < cq_2 < dq_1^2q_2$, but $dq_1^2q_2 > dq_1q_2$ so $c < dq_1q_2$ and $c > dq_1q_2$ contradiction. Hence $c = dq_1q_2$.

Notice that the case where $\gcd(a,b) = 1$ we can just set $q_1 = a$ and $q_2 = b$, and the proof will be the same.

Answer 5

Here is a proof that works only in $\def\Z{\Bbb Z}\Z$, while the result is valid much more generally (basically in any rings where $\gcd$ and $\def\lcm{\operatorname{lcm}}\lcm$ are well defined). So it certainly will seem inferior to other approaches to certain tastes, but it has a hands-on quality that may please others.

Let $n,m$ be two positive integers. Consider the list of remainders modulo$~n$ of the multiples $0,m,2m,\ldots$ of$~m$ until some multiple $km$ has a remainder that is already present in the list (this remainder is not added). The term in the list to which it was equal can only be the initial$~0$, since otherwise the the two terms in the list at the positions just before the two that were found to be equal would also have to be equal to each other, which would contradict that this was the first occurrence of such an equality. Having a remainder$~0$, this $km$ is a multiple of$~n$, so by definition the least common multiple of $m$ and $n$.

Now the set of classes represented by the numbers of the list forms a subgroup of the additive group $\Z/n\Z$ (the class of $0$ is among them, and the set is closed under taking differences in $\Z/n\Z$). If the initial$~0$ is the only number on the list, namely if $n$ divides $m$, then the result to prove is obvious (since $\gcd(n,m)=n$ and $\lcm(n,m)=m$), Having dealt with this case, we can assume there is a nonzero remainder on the list; let $d>0$ be the smallest such remainder. By the mentioned subgroup property, $d$ divides $n$, and the numbers on the list are all multiples of $d$. In particular $d$ divides the remainder modulo$~n$ of $m$ (the second number on the list) and therefore$~m$ itself: $d$ is a common divisor of $n$ and $m$. Also $d$ was obtained as remainder of some multiple $sm$ of$~m$, so $d=sm-tn$ for some integers $s,t$ which shows that all common divisors of $n,m$ divide $d$ (since they divide both $sm$ and $tm$), which makes $d$ the greatest common divisor of $n$ and $m$.

Now it remains to count the numbers on the list in two ways. On one hand we produced $k$ terms, where $km=\lcm(n,m)$, so the number is $k=\lcm(n,m)/m$. On the other hand the list consists (in some permutation) of all multiples $ld$ of$~d$ with $0\leq{ld}<n$, so the number is $n/d$. Using $d=\gcd(n,m)$, we get $$ \frac{\lcm(n,m)}m = \frac{n}{\gcd(n,m)} \qquad\hbox{so}\qquad \lcm(n,m) = \frac{nm}{\gcd(n,m)}. $$ An example can clarify the construction: for $n=24$ and $m=110\equiv14\pmod{24}$, the list of remainders is $$ 0, 14, 4, 18, 8, 22, 12, 2, 16, 6, 20, 10, $$ with $12m=1320=\lcm(n,m)$ producing a remainder$~0$ at what would be the next position in the list, and of which $2=7m-32n=\gcd(n,m)$ is the smallest positive remainder. The equality comes from equating $12=24/2$.