Negative exponents and positive numbers. Why is it that when we raise a positive number to a negative power, we don't get a negative number?
 A: When you define negative exponents, you define them in a way that the power laws still hold. In particular,
$$a^{-n}a^n = a^{-n+n} = a^0 = 1$$
From this you immediately see that $a^{-1}$ is positive, since $a^n$ and $1$ are both positive, but the product of a negative and a positive number is negative.
Indeed, the formula above just says that
$$a^{-n} = \frac{1}{a^n}$$
A: A negative exponent is just another way of writing $x^y$ where $x,y > 0$ as $\dfrac{1}{x^y}$.
Since a positive number raised to a positive exponent is positive and the reciprocal of a positive number is positive. We have that $x^{-y}$ is positive. 
Using the familiar power laws we have that $$a^{b}a^{-b} = a^{b - b} = a^0 = 1 \implies a^{-b} = \frac{1}{a^{b}}$$
So we have shown why we can write $x^{-y}$ as $\dfrac{1}{x^y}$.
A: You could also approach it graphically. If you look at the graph of $e^x$ when $x > 0$, this function is always positive. Likewise when $x < 0$, it is also positive. Hence, by definition, it will always be positive whether you raise it to a positive power or otherwise.  
A: Because negative exponents are defined such that $a^{-b} = \dfrac{1}{a^b}$. Its just a convention.
Think about it this way: because of the exponent law $x^{a} x^{b} = x^{a+b}$, we must have $a^b a^{-b} = a^{b+(-b)} = a^0 = 1$. Therefore, because $a^b a^{-b} = 1$, we have $a^{-b} = \dfrac{1}{a^b}$.
