I''m pretty sure that we learned this stuff in school already, though the fancy wording might''ve only officially made its way into varsity.
but yes

a/b = (a/1) * (1/b)

so
a/b*c/d
= (a/1) * (1/b) * (c/1) * (d/1)
= (a*c) * 1/(b*d)
= (a*c) / (b*d)
which looks nicer as
= (a.c)/(b.d)

since ac/bd can be misread if you feel evil enough as
a*c/b*d = (a*c*d)/b

Haeleyx76:
I think you mean commutative rather than associative. Fortunately matrix multiplication is associative, otherwise things would get really, really nasty.

Also your use of the word scalar is a little misleading, I know what you mean, you''re talking about a field, for which commuativity is given.

a/b * c/d = ac/bd holds for any field, but not necessarily for a group, which does not have to be commutative. In fact it holds for any commutative group.

I''m beginning to think this post is a little pointless, but if you try to view the problem from a pure perspective, taking it into a more general case, where it doesn''t have to be on the real numbers, you might as well go the whole way.

I''m pretty sure the most general system where it holds is an abelian group though, as the only things given are identity, closure, associativity, inverses and commutativity.