Proving Index Law for negative power

Question

I came across a proof in this book which shows that $(x^m)^n=x^{mn}$ where $n=-k$ and $m$ is a positive integer but there's a step that I am having trouble justifying. The proof is as follows: \begin{align} (x^m)^n&=(x^m)^{-k}\\ &=((x^m)^{-1})^k \\ &=((x^{-1})^m)^k \\ &=(x^{-1})^{mk} \\ &=x^{-mk}\\ &=x^{mn} \end{align} I don't get how they justified going from line 2 to 3. The author said by "the rule for the inverse of a product" which is shown on page 209 but I don't see the connection. Any thoughts?

Answer 1

The justification is as follows-

Consider some $x\gt 0$.

Then, ${1\over x^{m}}= {1^{m}\over x^{m}}, m\in \mathbb{N}$ (Since, $1^{m}=1$ for all $m\in \mathbb{N}$)

$\Rightarrow {1\over x^{m}}=( {1\over x})^{m}$ (Since, ${a^{m}\over b^{m}}=({a\over b})^{m}$ for all positive integral $m$)

It should be obvious that the above is equivalent to -

$(x^{m})^{-1}=(x^{-1})^{m}$ as required.

The steps in the justification follow from the power laws for positive integral indices, with which you must undoubtedly be familiar.

Answer 2

$(x^{-1})^m\cdot x^m=$

$(x^{-1}(x^{-1}(....... (x^{-1}\cdot x).....)x)x) $ by associativity.

And by induction that is equal to the identity.

So $(x^m)^{-1}=(x^{-1})^m$

......

Or. By inverse product $(ab)^{-1}=b^{-1}a^{-1} $ so $(x^m)^{-1}=(xxxxxxx.....xxx)^{-1}=x^{-1} x^{-1} x^{-1}... x^{-1} x^{-1} x^{-1}= (x^{-1})^m $.