Let $a$ and $b$ be positive integers. Prove that $(a,b)|[a,b]$.

Question

Hi everyone here is my question;

So, let $a$ and $b$ be positive integers. Prove that $(a,b)|[a,b]$. Here what I have so far;

$(a,b)=d$ $[a,b]=m$ so I need an equation of the form $m=d()$. I just need some hints on how to proceed.

Thank you so much in advance.

Answer 1

I suppose that $d=(a,b)$ denotes $gcd(a,b)$ and $m=[a,b]$ denotes $lcm(a,b)$.

Now, $d\mid a$ and hence $d\mid x$ for any $x$ beeing a multiple of $a$ (by transitivity of the relation $\mid$).

As a special case : $d\mid m$.

Answer 2

Another approach using prime decomposition. If $a={r_1}^{k_1}\cdots {r_j}^{k_j}\cdot{p_1}^{n_1}\cdots {p_r}^{n_r}$ and $b={r_1}^{k'_1}\cdots {r_j}^{k'_j}{q_1}^{m_1}\cdots {q_s}^{m_s}$, where $r_1, \cdots, r_j$ are the common primes, we have

$$gcd(a,b)= {r_1}^{\min{k_1, k'_1}}\cdots {r_j}^{\min{k_j, k'_j}}$$ $$lcm(a,b)= {r_1}^{\max{k_1, k'_1}}\cdots {r_j}^{\max{k_j, k'_j}}\cdot{p_1}^{n_1}\cdots {p_r}^{n_r}{q_1}^{m_1}\cdots {q_s}^{m_s}$$

so it is clear that $gcd(a,b)$ divides $lcm(a,b)$.