Question

Show that the set of all non-zero complex numbers under the
operation of multiplication

Answer 1

Answer:

Step-by-step explanation:

Well, it trivially follows from the set of rationals being a field, but let’s assume we do not know this.

We have associativity under operation

We know that  1  is in the set of nonzero rationals

For any  x  in nonzero rationals we have  y=1x  such that  xy=1

For any  a,b  in nonzero rationals we see that  ab  is in nonzero rationals

We have associativity, identity element, inverses in set, and closure under operation.

We have a group!