1. Define multiplication by integers- that's the part you accept, right?
2. Define multiplication by numbers of the form 1/n, n a non-zero integer. m*1/n is the number x, such that nx= m. It's not to difficult to show that such a thing exists and is unique (i.e. that "multiplication by 1/n is well-defined).
3. Define multiplication by numbers of the form m/n, m an integer, n a non-zero integer, that is, all rational numbers, by x*(m/n)= (x*m)*(1/n).
4. Finally, define multiplication by irrational numbers in the following way: If x is irrational number, then there exist a sequence {an} or rational numbers that converge to x. xb= lim (anb). Again, it is possible to show that sequence always converges.
That last is the one that will give you [itex]\sqrt{3}*\sqrt{2}[/itex]. There exist a sequence of rational numbers {an} that converges to [itex]\sqrt{3}[/itex] and a sequence of rational numbers {bn} that converges to [itex]\sqrt{2}[/itex]. [itex]\sqrt{3}*\sqrt{2}[/itex] is defined as lim (anbn).
The fact that [itex]\sqrt{3}[/itex] can be represented by the infinite decimal number 1.73205... means that {1, 1.7, 1.73, 1.732, 1.7320, 1.73205, ...} is an infinite sequence of rational numbers (each is a terminating decimal) which converges to [itex]\sqrt{3}[/itex]. Similarly, [itex]\sqrt{2}[/itex] can be represented by the infinite decimal number 1.414213... so that {1, 1.4, 1.41, 1.414, 1.4142, 1.41421, 1.414213, ...} is a sequence of rational numbers converging to [itex]\sqrt{2}[/itex]. That gives a proof that we can multiply irrational numbers "digit by digit".