negative powers $(x^{-2} = 1/x^2)$

Question

I need clarification for negative power of a number.

I understand $x$ to the power of $2$ is equal to $x\cdot x$

But how $x$ to the power of $-2$ is equal to $\dfrac{1}{x^2}$ ?

Answer 1

If you agree that for $x\neq 0$ you have $x^0=1$ then $x^{m-m}=1$ for any positive integer $m$

If you agree that $x^a\cdot x^b=x^{a+b}$ then you have $x^{m-m}=x^m\cdot x^{-m}=1$

If you agree that $pq=1\Rightarrow q=\frac{1}{p}$ then you have $x^{-m}=\frac{1}{x^m}$

Answer 2

This is just a rule of exponents:

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

Answer 3

For $n\in\mathbb N$, the equation $$x^{-n} = \frac{1}{x^n}$$ is a matter of definition. Mathematitians decided to extend the notation of $x^n=x\cdot x\cdots x$ (with $n$ repetitions) to negative values of $n$. Since, for positive numbers of $m,n$, you have $$x^{m+n} = x^mx^n,$$ you want to define $x^{-n}$ in such a way that it will still be compatible. This means that, for $m,n\in\mathbb N$, you want $$x^{m-n}$$ to equal $$x^m\cdot x^{-n},$$ and this can only be achieved by setting $x^{-n}$ as $\frac1{x^n}$.