Do the positive irrationals form a group with respect to multiplication

Question

I think it's not form a group because multiple of two irrational number is rational number Please give your answer

Answer 1

It is not even a grupoid, since the multiplication of two irrational numbers is sometimes a rational number.

Answer 2

Irrational numbers form a semigroupoid under multiplication; the only axiom this imposes is that $(xy)z=x(yz)$ if all the products in that equation exist within the algebra of interest. Well, OK, it's something a bit more specific than that: it's a commutative semigroupoid (when the products exist), with one other nice property we'll get to in a moment. We don't have totality, because $\sqrt{2}\frac{1}{\sqrt{2}}\in\Bbb Q$; we don't have an identity, because $1\in\Bbb Q$; and without an identity we can't have invertibility in the sense $xx^{-1}=1$, although we do have the weaker $xyx^{-1}=y$, again if all products exist.