Answer
No
Work Step by Step
We know that $a^{-n}=\frac{1}{a^{n}}$ for some nonzero real number a and some positive integer n.
Therefore, we know that $a^{-2}=\frac{1}{a^{2}}$. We know that the square of any real number will be positive, which would make the denominator of $\frac{1}{a^{2}}$ positive. Since, both the numerator and denominator are positive, we know that $\frac{1}{a^{2}}$ will also be positive.
So, there is no number a such that $a^{-2}$ will be a negative number.
