Solution:
[tex](Integer)^{Integer}[/tex]
[tex]2^3=2 \times 2 \times 2=8\\\\ 2^{-3}=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}, \\\\ (-2)^3=-2 \times -2 \times -2= -8,\\\\ (-2)^{-3}=\frac{-1}{2}\times\frac{-1}{2}\times\frac{-1}{2}=\frac{-1}{8},[/tex]
In all cases we are getting an integer.
So, yes, Marcus is correct.
Option D : Marcus is correct. If any integer is raised to any integer exponent, the base is multiplied times the exponent. The product of two integers is always a rational number.
