If $a,b$ are integers such that $a \mid x$ and $b \mid x$ , must it be true that $\mathrm{lcm}(a,b)\mid x$ ? I claim that it is true but I don't know how to prove it.
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that is precisely the definition of lcm – Jonathan Mar 06 '13 at 04:01
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are you allowed to use prime factorization theorem? – Maesumi Mar 06 '13 at 04:01
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1What is the purpose of $c$? – Zev Chonoles Mar 06 '13 at 04:02
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Do you know that $[a,b]=\frac{ab}{(a,b)}$? – Pedro Mar 06 '13 at 04:13
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@ZevChonoles: yeah. i know $\gcd(a,b) \times lcm(a,b) = ab$ – Idonknow Mar 06 '13 at 04:19
3 Answers
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Suppose that it is not the case that $\mathrm{lcm}(a,b)$ divides $x$. By the division algorithm, we know that there are integers $q$ and $r$, with $0\leq r<\mathrm{lcm}(a,b)$, such that $$x=\mathrm{lcm}(a,b)q+r,$$ and because $\mathrm{lcm}(a,b)$ does not divide $x$, we have that $r\neq 0$. Rewrite this as $$r=x-\mathrm{lcm}(a,b)q.$$
Because $a$ divides $x$ and $\mathrm{lcm}(a,b)$, we see that $a$ divides $r$. Similarly, $b$ must divide $r$. Therefore $r$ is a common multiple of $a$ and $b$. But $r<\mathrm{lcm}(a,b)$, so this is a contradiction.
Zev Chonoles
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Conceptually it is quite simple. The set M of all common multiples of $a,b$ is a nonempty set of integers closed under subtraction, therefore every element of M is a multiple of its least positive element $ = {\rm lcm}(a,b).\,$ In particular, $x$ is a multiple of $\,{\rm lcm}(a,b).$
Remark $\ $ Hence the lcm is not only least in absolute value, but also least in terms of divisibility. The latter is the property used to define lcm in more general rings, where no notion of absolute value need exist, i.e. generally $\ a,b\mid c\iff {\rm lcm}(a,b)\mid c,\ $ and, dually, $\ c\mid a,b\iff c\mid \gcd(a,b)$.
If you know ring theory, you can restate the first paragraph as: the set M of common multiples of $a,b$ forms a nonzero ideal in $\,\Bbb Z.\,$ Since $\,\Bbb Z\,$ is a PID, we infer that $\, {\rm M} = (m)\,$ is principal, generated by its least positive element $\,m = {\rm lcm}(a,b).\:$ Therefore $\, x\in {\rm M} = (m)\:\Rightarrow\: m\mid x$.
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In the first sentence "nonempty" should be "with a positive element" or (more cryptically) "not contained in ${0}$". – Marc van Leeuwen Oct 14 '13 at 04:34
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Using unique prime factorization:
Let $a = \prod p_i^{a_i}$ and $b = \prod p_i^{b_i}$. Then $lcm(a, b) = \prod p_i^{\max(a_i, b_i)}$.
If $a | x$ and $b | x$, write $x = \prod p_i^{x_i}$.
Since $a | x$, $\prod p_i^{a_i} | \prod p_i^{x_i}$, so $a_i \le x_i$ for each $i$.
Similarly, since $b | x$, $\prod p_i^{b_i} | \prod p_i^{x_i}$, so $b_i \le x_i$ for each $i$.
Therefore $x_i \ge \max(a_i, b_i)$ for each $i$, so $lcm(a, b) | x$.
marty cohen
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