Intermediate Algebra (12th Edition)

As shown on page 445, we know that $a^{\frac{m}{n}}=\sqrt[n] a^{m}=(\sqrt[n] a)^{m}$. Therefore, $(-36)^{\frac{1}{2}}=\sqrt[2] (-36)^{1}=\sqrt -36=$not a real number. We know this because there is no real number $a$, such that $a^{2}=-36$.