Inspired by Sandeep Bhardwaj

Not always sir .See this link .

It's better to give a proof of it which leaves no ambiguity for the conclusion.

The notion of LCM that we're using here is that if $k=\textrm{LCM}(\sqrt2,\sqrt3)$ , there exists positive integers $a,b$ such that,

$a\sqrt2=b\sqrt3=k$

We prove that this cannot happen by drawing out a contradiction using the fundamental theorem of arithmetic. Square the equation above. We get,

$2a^2=3b^2$

Since $2,3$ are primes, it follows that for equality to happen, we should have $2\mid b^2$ and $3\mid a^2$ . Since $a,b$ are integers, we subsequently have $2\mid b$ and $3\mid a$ which gives $2^2\mid b^2$ and $3^2\mid a^2$ .

Furthermore, note that if we supposedly had equality, both sides should have the same prime power (say, power of $2$ ). But, note that we always have an odd power of $2$ in LHS while an even power of $2$ in RHS whatever the values of $a,b$ be. But, this is absurd by the fundamental theorem of arithmetic.

Hence, no such positive integers $a,b$ exist which further implies that $k$ doesn't exist.