8. When evaluated, which of the following is not an integer?
A $1^{-1}$
B $4^{-\frac{1}{2}}$
C $6^{0}$
D $8^{\frac{2}{3}}$
E $16^{\frac{3}{4}}$
Solution
B
We have
$$
\begin{array}{l}
1^{-1}=\frac{1}{1}=1, \\
4^{-\frac{1}{2}}=\frac{1}{4^{\frac{1}{2}}}=\frac{1}{\sqrt{4}}=\frac{1}{2}, \\
6^{0}=1, \\
8^{\frac{2}{3}}=\left(8^{\frac{1}{3}}\right)^{2}=(\sqrt[3]{8})^{2}=2^{2}=4,
\end{array}
$$
and
$$
16^{\frac{3}{4}}=\left(16^{\frac{1}{4}}\right)^{3}=(\sqrt[4]{16})^{3}=2^{3}=8 .
$$

We therefore see that only the expression of option B gives a number which is not an integer when it is evaluated.