Motivate why $a^{-n}$ equals to $\frac{1}{a^n}$

Question

I have to prove that $a^{-n} = \frac{1}{a^n}$ with $\frac{3^4}{3^7}$, but before I can do that I have to understand the background.

The background says: we know that $\frac{3^4}{3^7} = \frac{1}{3^3}$ and that $3^{4-7} = 3^{-3} = \frac{1}{3^3}$

I don't understand why $\frac{3^4}{3^7} = \frac{1}{3^3}$, before I can move on and solve why $a^{-n}$ equals to $\frac{1}{a^n}$.

What is $a$? What set is it from? What axioms are you using? Is $a \in \mathbb{R}$? etc. etc. — anak, Jul 31 '15 at 17:43
Well, if $n$ is positive, $a^{-n}$ is defined to be equal to $1/a^n$. At most, then, what you can do is explain what motivates this definition! — Mariano Suárez-Álvarez, Jul 31 '15 at 17:43

score 5 · Answer 1 · answered Jul 31 '15 at 17:44

If $a^b \cdot a^c = a^{b + c}$ generalizes from positive $b, c$ to all integer values of $b, c$, then we have

$$1 = a^0 = a^{1 - 1} = a^1 \cdot a^{-1} = a \cdot a^{-1}.$$

That is, $$a \cdot a^{-1} = 1$$ Hence we are motivated to now define

$$a^{-1} = \frac{1}{a}$$

Similarly,

$$a^n \cdot a^{-n} = 1$$

and hence we also define

$$a^{-n} = \frac{1}{a^n}$$

1 Answers1