(ab)^-1=a^-1b^-1

We will prove an identity from Micheal Spivak's Calculus
Problem: Ch1 #3.3

[@hollowshiningami3080]

cant you use the property that (a^n)*(b^n)=(ab)^n ?

[@shavojohn4241]

You need this first to prove exponent laws for integer exponent

[@willnewman9783]

You say that ab is not zero because a and b are not zero, but I don't think this is clear until you do some of the later steps in the proof.

[@debikk4204]

No, it is. A product of multiplication is only zero if one of the factors sis zero. Otherwise you would get a = 0/b = 0 which is not true