Mathematical Induction?

Question

How do you prove using induction that

b^(x+y) = b^x * b^y

I tried solving it

P(1) holds good where both x+y = 1;

I assume P(n) holds good where n = x+y;

Now to prove if P(n+1) i.e P(x+y+1) I don't know how to go about doing this.

Answer 1

Actually, it's better to do the induction on y, not on the sum of x and y.

Note that b^(x+0) = b^x = b^x * 1 = b^x * b^0, so the case y=0 holds.

Suppose that it holds for some y. Then b^(x+(y+1)) = b^((x+y)+1) = b^(x+y) * b. But by the inductive hypothesis, b^(x+y) = b^x * b^y, so we have b^(x+y) * b = (b^x * b^y) * b = b^x * (b^y * b) = b^x * b^(y+1), so by transitivity b^(x+(y+1)) = b^x * b^(y+1), and it holds for y+1, and by induction, for all nonnegative integers y.