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The fact that this is true seems clear to me. But I'm not quite sure how to write a formal proof. Obviously if $a$ is the least common multiple of $a$ and $b$, then a will be divisible by $b$. How would I go about expressing this idea in the form of a proof?

I know you begin by proving:

$a=\text{lcm}(a,b) \rightarrow b|a$

Then prove:

$b|a \rightarrow a=\text{lcm}(a,b)$

I imagine there is a simple way to express the $\text{lcm}$ that may help.

Thank you in advance for your answers.

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If $a = \text{lcm}(a,b)$, then in particular $a$ is a common multiple of $a$ and $b$, and so $b \mid a$.

The other way: Suppose $b \mid a$. Since we also have $a \mid a$, we get that $a$ is a common multiple of $a$ and $b$. Hence, $ \text{lcm}(a,b) \leq a$. Also, since $a \mid \text{lcm}(a,b)$, we get $a \leq \text{lcm}(a,b)$, so $a = \text{lcm}(a,b)$.

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Using the definitions of $\operatorname{lcm}$ and $\gcd$ this is a simple task: $$ \begin{align} \operatorname{lcm}(a,b) = \frac{ab}{\gcd(a,b)} = a \iff \frac{b}{\gcd(a,b)} = 1 \iff b = \gcd(a,b) \iff b|a \end{align} $$

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  • $\begingroup$ What is $lcd(a,b)$? $\endgroup$ – L. F. Nov 27 '18 at 9:31
  • $\begingroup$ @L.F. A typo :) I suppose LCD would be the least common divisor, which would be 1. $\endgroup$ – Frank Vel Dec 6 '18 at 15:58
  • $\begingroup$ Well $\frac{1}{n}$ is also a common divisor where $n \in \mathbb{N}^+$ :P $\endgroup$ – L. F. Dec 7 '18 at 8:45
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If $b\mid a \to a = kb \to \text{lcm}(a,b) = \text{lcm}(kb,b) = b\text{lcm}(k,1) = bk = a$, and the other direction is quite clear.

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