Setup: We're in a Euclidean domain $R$ and have elements $a,b\beta \x88\x88R$.

We've shown that a least common multiple of $a$ and $b$ is any element $e\beta \x88\x88R$ such that $\beta \x9f\xa8e\beta \x9f\copyright =\beta \x9f\xa8a\beta \x9f\copyright \beta \x88\copyright \beta \x9f\xa8b\beta \x9f\copyright $, and we've shown such an element is unique up to multiplication by a unit.

We're now showing that the element $e=\frac{ab}{d}$ is a least common multiple of $a$ and $b$, where $d$ is a greatest common divisor of $a$ and $b$.

We've established that $e$ is a common multiple of $a$ and $b$. Now we'll show it's the least common multiple. First recall that a property of a greatest common divisor is that it can be written as an $R$-linear combination of $a$ and $b$. In other words, there are elements $r,s\beta \x88\x88R$ such that

$$d=ra+sb.$$

Note that since $d$ divides $a$ and $b$, we can actually "divide" this equation by $d$ to get

(What we're really doing is writing $a=kd$ and $b=ld$ for some $k,l\beta \x88\x88R$, and then using cancelation on both sides to get $1=rk+sl$.) We'll use this equality in a little bit.

Suppose ${e}^{\beta \x80\xb2}$ is a least common multiple of $a$ and $b$. Since $e$ is a common multiple and ${e}^{\beta \x80\xb2}$ is a least common multiple, this means $e=t{e}^{\beta \x80\xb2}$ for some $t\beta \x88\x88R$. We'll show that $t$ must be a unit, which will prove $e$ is also a least common multiple of $a$ and $b$. We also know ${e}^{\beta \x80\xb2}=xa=yb$ for some $x,y\beta \x88\x88R$. Combining this with the equality $e=t{e}^{\beta \x80\xb2}$ yields

This proves $t$ is a unit in $R$, with inverse given by $ry+sx$. Since ${e}^{\beta \x80\xb2}$ is a least common multiple of $a$ and $b$, and since $e=t{e}^{\beta \x80\xb2}$ with $t$ a unit, we've thus proven $e$ is also a least common multiple of $a$ and $b$.