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$\newcommand{\lcm}{\operatorname{lcm}}$ Just a quick one.

I have to prove that the lcm of two consecutive numbers is its product.

Using the identity $\gcd(a,b) \cdot \lcm(a,b) = a \cdot b$, you can find $\lcm(a,b) = \frac{a \cdot b}{\gcd(a,b)}$

I can prove separately that the gcd of consecutive numbers is 1 (coprime) -- which leaves $a \cdot b$ and thus proving the statement.

I'm pretty sure this is complete but for the sake of clarity is there something I'm missing?

PrincessEev
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Adam Mansfield
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1 Answers1

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Nothing. That is correct.

On the other hand, when two coprime numbers $a$ and $b$ divide a number $c$, then $ab\mid c$ too. So, $ab$ divides every common multiple of $a$ and $b$ and this also proves that $\operatorname{lcm}(a,b)=ab$, without using the fact that $\gcd(a,b)\operatorname{lcm}(a,b)=ab$.

José Carlos Santos
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