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I'm posting this doubt since this user is not active for many years

Let and be positive integers, show that $\gcd(, ) \cdot \mathrm{lcm}(, ) = b$

My attempt : I found the answer here but I'm not getting this answer in my mind.

User answer given :

Let $\gcd(a,b)=d$. Then for some $a_0,b_0$ such that $a_0$ and $b_0$ are relatively prime, we have $a=da_0$ and $b=d b_0$. If we can show that the lcm of $a$ and $b$ is $da_0b_0$, we will be finished.

Certainly $da_0b_0$ is a common multiple of $a$ and $b$. We must show that it is the least common multiple.

Let $m$ be a common multiple of $a$ and $b$. We will show that $da_0b_0$ divides $m$.

Since $m$ is a multiple of $a$, we have $m=ka=ka_0d$ for some $k$. But $b$ divides $m$, so $db_0$ divides $ka_0d$, and therefore $b_0$ divides $ka_0$. Since $a_0$ and $b_0$ are relatively prime, it follows that $b_0$ divides $k$, and we are finished.

My doubt : user is saying if we can show that the lcm of $a$ and $b$ is $da_0b_0$, we will be finished.

My question is that how it can be finished? Suppose if I take $\operatorname{lcm}(a,b) = da_0b_0 = \gcd(a,b) a_0b_0$

then this will not match what the question is exactly asking?

Question is about $\operatorname{lcm}(a,b) = \frac{(a,b)}{\gcd(a,b)}$

Bill Dubuque
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jasmine
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1 Answers1

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This is basically this:$\DeclareMathOperator{\gcd}{\text{gcd}}$ $\DeclareMathOperator{\lcm}{\text{lcm}}$

Let $a, b~$ be as described in question. Consider $\gcd(a, b) = d$. Then by the definition of $\gcd$ we have that, $$ d|a ~\text{ and }~ d|b $$ Thus, we have that $a = dk_1$ and $b = dk_2$. Again, from the definition of $\lcm$ we have $$ \lcm(a, b) = d \cdot k_1 \cdot k_2 $$ Hence: $$ a \cdot b = d k_1 \cdot d k_2 = d \cdot (d \cdot k_1 \cdot k_2) = \gcd(a, b) \cdot \lcm(a, b) $$

Ralph Clausen
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