# Math Induction

• Mar 24th 2010, 06:01 PM
serious331
Math Induction
Let a be a non zero number and m and n be integers. Prove by induction that:

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

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

can anyone explain me the basis and inductive steps here?

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

$\displaystyle n=1$ in (2)?

• Mar 25th 2010, 07:43 AM
Tinyboss
Exactly what properties of exponentiation are you given to work with?
• Mar 25th 2010, 07:57 AM
serious331
Quote:

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
• Mar 25th 2010, 08:01 AM
Tinyboss
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.)
• Mar 26th 2010, 02:24 PM
novice
Quote:

Originally Posted by serious331
Let a be a non zero number and m and n be integers. Prove by induction that:

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

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

can anyone explain me the basis and inductive steps here?

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

$\displaystyle n=1$ in (2)?

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

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

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

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

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

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

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