Question

1. reinforce when does a negative exponent not move the base to the denominator?

Explanation:

When the base is \(1\) (since \(1^{-n}=\frac{1}{1^{n}} = 1\)) or when the base is \(0\) (but \(0\) with a negative exponent is undefined in the context of moving to the denominator as \(0^{-n}=\frac{1}{0^{n}}\) which is division by zero). Also, when the entire expression with the negative - exponent is in a fraction where it can be simplified in another way. For example, in \(\frac{a^{-m}}{a^{-n}}=a^{n - m}\) (using the rule \(\frac{a^{m}}{a^{n}}=a^{m - n}\)), the negative - exponent bases \(a\) are not moved to the denominator in the traditional \(a^{-k}=\frac{1}{a^{k}}\) sense for the simplification of the fraction.

Answer:

When the base is \(1\) (because \(1^{-n} = 1\) for any non - zero real number \(n\)), when dealing with fraction simplification rules (e.g., \(\frac{a^{-m}}{a^{-n}}=a^{n - m}\)), and when the base is \(0\) (but \(0\) with a negative exponent is undefined in the simple \(a^{-n}=\frac{1}{a^{n}}\) sense due to division by zero).