Math Induction

**serious331** · March 24th 2010, 06:01 PM

Let a be a non zero number and m and n be integers. Prove by induction that:

(1) $a^{m+n} = a^{m} a^{n}$

(2). ${(ab)}^{n} = {a^n}{b^n}$

can anyone explain me the basis and inductive steps here?

Do I have to start by supposing $n=0$ in (1) and

$n=1$ in (2)?

what about the inductive step?

**Tinyboss** · March 25th 2010, 07:43 AM

Exactly what properties of exponentiation are you given to work with?

**serious331** · March 25th 2010, 07:57 AM

Originally Posted by Tinyboss

Exactly what properties of exponentiation are you given to work with?

the question doesnt state anything about properties of exponentiation. However, induction has to be used here

**Tinyboss** · March 25th 2010, 08:01 AM

If you don't know what exponentiation is, you have no hope of proving anything about it. (Not saying you don't know how to exponentiate--just that your book/prof is not expecting you to prove this without giving you something to start from. Look in your chapter or ask your prof what you're allowed to assume about exponentiation to begin with.)

**novice** · March 26th 2010, 02:24 PM

Originally Posted by serious331

Let a be a non zero number and m and n be integers. Prove by induction that:

(1) $a^{m+n} = a^{m} a^{n}$

(2). ${(ab)}^{n} = {a^n}{b^n}$

can anyone explain me the basis and inductive steps here?

Do I have to start by supposing $n=0$ in (1) and

$n=1$ in (2)?

what about the inductive step?

When $n=0 \text{ and } m=0$ , $a^{m+n} = a^{m} a^{n}$ holds since $a^{0+0}=1=a^{0 \cdot 0}$

For induction hypothesis, assume $a^{p+q} = a^{p} a^{q}$ , where $p$ and $q$ are integers.

We begin by $a^{p+q} = a^{p} a^{q}$ .

Multiplying through by a, we obtain $a \cdot (a^{p+q}) = a \cdot a^{p} a^{q}$ .

Plug in the assumption for the left handside in the parethesis, we obtain

$a \cdot (a^{p} a^{q})= a \cdot a^{p} a^{q}$ . The result is as desired.

Consequently, by induction hypothesis $a^{m+n} = a^{m} a^{n}$ for all integers $m$ and $n$ .

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