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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.

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    $(a,b)\mid a\mid [a,b]\ \ $ – 2017-02-12

2 Answers 2

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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 : $d\mid m$.

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    Yes this makes so much sense! – 2017-02-12
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    now i need to find in what conditions (a,b)=[a,b], i do not want the answer just a way to proceed? any help? – 2017-02-12
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    We know that $(a,b)\le a\le [a,b]$. Hence $(a,b)=[a,b]$ implies ... And after that, you can do the same with $b$, so ... – 2017-02-12
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    how did you get a is between (a,b) and [a,b]? I did not get that? – 2017-02-12
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    @HananIbraheim: Given two positive integers $m,n$ such that $m\mid n$, you have $n=km$ for some positive integer $k$ and hence $n\ge m$. Here we have $(a,b)\mid a$, thus $(a,b)\le a$ and we also have $a\mid [a,b]$, thus $a\le [a,b]$. – 2017-02-12
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    I thought about it as hard as I can but I still cant get it. Thank you :) – 2017-02-12
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    According to my previous comment, if $(a ,b)=[a,b]$, then $(a,b)=a$ (which implies $a\mid b$) and $(a,b)=b$ (which implies $b\mid a$), hence $a=b$. The converse is obviously true. – 2017-02-13
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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 $lcm(a,b)$.

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    I see. Thank you so much for your input! – 2017-02-12