Prove $[a, b]$, where $a$ and $b$ are positive integers, is unique.
We use the uniqueness of prime decomposition.
The prime decompositions of $a$ and $b$ are
$a = p_1^{e_1} \times p_2^{e_2} \times \ldots \times p_n^{e_m}$
$b = p_1^{f_1} \times p_2^{f_2} \times \ldots \times p_n^{f_n}$
where $p_i, e_i, f_i, n$ are non-negative integers. Note that $e_i, f_i$ can be zero.
Since the prime decomposition is unique, the LCM
$$[a,b] = p_1^{\max(e_1, f_1)} \times p_2^{\max(e_2, f_2)} \times \ldots \ p_n^{\max(e_n, f_n)}$$
is unique.
Let's also try a different argument.
Let $q = [a,b]$ and also $r=[a,b]$ are both the LCM of $a$ and $b$.
For the purpose of contradiction, let's assume $r>q$. If $q$ is the lowest common multiple, then $r$ is not the lowest common multiple because it is strictly greater than $q$. So $r$ cannot be greater than $q$.
Again, for the purpose of contradiction, let's assume $r<q$. If $r$ is the lowest common multiple, then $q$ is not the lowest common multiple, because it is strictly greater than $r$. So $q$ cannot be greater than $r$.
We have shown both $q>r$ and $q<r$ lead to contradictions. Therefore $q=$, that is, the LCM of $a$ and $b$ is unique.