# Why does x^0 = 1 and x^(-n) = 1/x^n? (Part 1)

I distinctly remember when, in my second year as a college professor, a really good college student — with an SAT Math score over 650 — asked me why $x^0 = 1$ and $x^{-n} = \displaystyle \frac{1}{x^n}$. Of course, he knew that these rules were true and he could apply them in complex problems, but he didn’t know why they were true. And he wanted to have this deeper knowledge of mathematics beyond the ability to solve routine algebra problems.

He also related that he had asked his math teachers in high school why these rules worked, but he never got a satisfactory response. So he asked his college professor.

Looking back on it, I see that this was one of the incidents that sparked my interest in teacher education. As always, I never hold a grudge against a student for asking a question. Indeed, I respected my student for posing a really good question, and I was upset for him that he had not received a satisfactory answer to his question.

This is the first of two posts where I give two answers to this question from two different points of view.

Answer #1. Let’s recall the definition of $x^n$ for positive integers $x$: $x^4 = x \cdot x \cdot x \cdot x \cdot x$ $x^3 = x \cdot x \cdot x$ $x^2 = x \cdot x$ $x^1 = x$

Starting from the bottom, the exponents increase by $1$ with each step up, while an extra $x$ is multiplied with each step up.

Of course, there’s no reason why we can’t proceed downward instead of upward. With each step down, the exponents decrease by $1$, while an extra $x$ is divided from the right-hand side. So it makes sense to define $x^0 = \displaystyle \frac{x}{x} = 1$.

We can continue decreasing the exponent — into the negative numbers — by continuing to divide by $x$: $x^{-1} = \displaystyle \frac{1}{x}$ $x^{-2} = \displaystyle \frac{1/x}{x} = \displaystyle \frac{1}{x^2}$ $x^{-3} = \displaystyle \frac{1/x^2}{x} = \displaystyle \frac{1}{x^3}$ $x^{-4} = \displaystyle \frac{1/x^3}{x} = \displaystyle \frac{1}{x^4}$

We see that, if $n$ is positive and $-n$ is negative, that $x^{-n} = \displaystyle \frac{1}{x^n}$.

Technically, this only provides an explanation for this rule for negative integers. However, I haven’t met a student that didn’t believe this rule held for negative rational exponents (or negative irrational exponents) after seeing the above explanation for negative integers.