1.2: Problem 0: Show that \([a,b] = \frac{ab}{(a,b)}\) for any integers \(a,b\).
Proof
We need the following fact. If \(m\) is any common multiple of \(a\) and \(b\). Then we know that \(a \mid m\) and \(b \mid m\). Now if \(a\) and \(b\) are relatively prime, then this implies \(ab \mid m\). So \([a,b] = ab\) whenever \((a,b) = 1\).
So now take any \(a\) and \(b \in \mathbb{Z}\). Let \(d = (a,b)\). Write \(a = dx\) and \(b = dy\) for some \(x,y \in \mathbb{Z}\). Since \(d\) is the greatest common divisor, then \((x,y) = 1\). So by the fact above, \([x,y] = xy\). Observe that
$$
\begin{align*}
[a,b] = [dx, dy] = d[x,y] = d \cdot xy.
\end{align*}
$$
while
$$
\begin{align*}
\frac{ab}{(a,b)} = \frac{dx \cdot dy}{d} = d \cdot xy.
\end{align*}
$$
so
$$
\begin{align*}
\frac{ab}{(a,b)} = [a,b].
\end{align*}
$$