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 Hint}\ \ d\,| \gcd(m,n)\!\!\!\overset{\ \ \rm\color{darkorange}U}\iff d\,|\, m,n\!\!\overset{\rm\color{#0a0}{\rm R}\!\!}\iff $ $ m,n\,|\, \frac{mn}d\!\!\overset{\ \ \rm\color{darkorange}U}\iff {\rm lcm}(m,n)\,|\, \frac{mn}d\!\!\overset{\rm\color{#0a0}{\rm R}\!\!}\iff d\,|\, \frac{mn}{{\rm lcm}(m,n)}$

where above we have applied $\,\rm \color{darkorange}U = $ LCM & GCD Universal Properties.

Remark $ $ Bringing to the fore the $\rm\color{#0a0}{R = cofactor\ reflection}$ symmetry we obtain a simpler proof: $ $ it is easy to show $\,d\,\color{#0a0}\mapsto\, mn/d\,$ bijects common divisors of $\,m,n\,$ with common multiples $\le mn.$ 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. $\,{\rm\color{#c00}{G}CD}(m,n)\,\color{#0a0}\mapsto\, mn/{\rm GCD}(m,n) = {\rm \color{#c00}{L }CM}(m,n).\,$

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

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

$\DeclareMathOperator{\lcm}{lcm}$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 4

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