Why is $\sqrt 2 \sqrt 3 = \sqrt{2\cdot 3}$?

Question

It seems very obvious that $\sqrt 2 \sqrt 3 = \sqrt 6$. But if we think that $\sqrt 2$ is irrational and that $\sqrt 3$ is also irrational, how can we prove that product of two irrational numbers is equal to another irrational number? Please correct me.

Answer 1

Product of two irrational numbers can be both rational (for example, $\sqrt 2 \cdot \sqrt {0.5} = \sqrt 1 = 1 $) or irrational (your example), so the fact that $\sqrt 2$ and $\sqrt 3$ are irrational is not a proof that $\sqrt 6$ is irrational.

However, proof that $\sqrt 6$ is irrational is not hard. Assume the opposite (i.e. $\sqrt 6$ is rational). Therefore $\sqrt 6$ can be expressed as $\frac{p}{q}$, where $p$ and $q$ are coprime intergers. $$ \sqrt 6 = \frac{p}{q}$$ $$ 6 = \frac{p^2}{q^2}$$ $$ p^2 = 6q^2$$ The latter implies that $p$ is even, so $p^2$ is divisible by $4$. So $q^2$ is also even and $q$ is even, however that means that $p$ and $q$ are not coprime since they are both divisible by $2$. Contradiction.

Answer 2

As already stated in the comments, the product of two irrationals is usually an irrational, but there are infinitely many counterexamples. For instance, pick $\sqrt2 \cdot \sqrt2 = 2$, or any irrational number with its reciprocal (which must be also irrational), resulting in $1$.

Your question is not really clear because the title says a thing and the body says another. Tackling the title: let $x = \sqrt2 \sqrt 3$. So $x^2 = (\sqrt2 \sqrt 3 )(\sqrt2 \sqrt 3 )$. Since the product of real numbers is commutative, this is equal to $ (\sqrt2 \sqrt2 )( \sqrt3 \sqrt3) = 2\cdot 3 \implies x = \sqrt 6$. The rules for real numbers aren’t arbitrary - they were built based on real life necessities and follow naturally, once the axioms are defined.

Now tackling the body: as I’ve said before, the product of two irrational numbers usually is not an irrational number. For this specific case, suppose that $\frac pq = \sqrt 6$ for coprime integers $p, q$. So $p = q \sqrt6 \implies p^2 = 6q^2$. Since $6 = 2\times3, 2$ and $3$ divide $p^2$ so both of them divide $p$; i.e., $p = 6p’$. Then $36p’^2 = 6q^2 \implies 6p’^2 = q^2$. By the same argument, $6 | q$. But then $6 | \operatorname{gcd}(p,q) = 1$, contradiction. So there are no coprime integers $p,q$ such that their quotient squared equals $6$; i.e., $\sqrt{6}$ is not rational. Try to prove this for any integer which is not a perfect square.