How do you prove$(a^m)^{-1} = (a^{-1})^m$ for a being any non-zero real and m being a positive integer from the axioms of real numbers?

Question

How do you prove $(a^m)^{-1} = (a^{-1})^m$ for $a$ being any non-zero real and $m$ being a positive integer, from the axioms of real numbers only?

Answer 1

From the existence of the inverse we have that, for $a \ne0$ there exists $b$ such that $a^mb=1$.

For $m=1$ this gives $ab=1 \Rightarrow b=a^{-1}$ .

Now suppose that $a^mb^m=1$ than: $$ a^{m+1}b^{m+1}=a^mab^mb=a^mb^mab=a^mb^m=1 $$

and this prove, by induction, the statement.

Answer 2

$$\forall a,m\in\mathbb{R}$$ Step one: I prove $(a^{m})^{-1}=(a^{-1})^{m}$ $$(x^n)^m=x^{nm}$$ Then $$(a^{m})^{-1}=(a^{-1})^{m}\Rightarrow a^{m\cdot-1}=a^{-1\cdot m}\Rightarrow a^{-m}=a^{-m}$$

Step two: $a^{-m}>0$ $$a^{-m}=a^{-1\cdot m}=(a^{-1})^{m}=(\frac{1}{a})^{m}=(\frac{1}{a^{m}}) \Longrightarrow(\frac{1}{a^{m}})>0$$

Answer 3

You should use induction. The base case is $m=1$ and the relation to be proved is $$ (a^1)^{-1}=(a^{-1})^1 $$ which is true.

Suppose the statement holds for $m$; then \begin{align} (a^{m+1})^{-1} &=(a^ma)^{-1} &&\text{definition of powers}\\ &=(a^m)^{-1}a^{-1} &&\text{because $(xy)^{-1}=x^{-1}y^{-1}$} \\ &=(a^{-1})^ma^{-1} &&\text{induction hypothesis} \\ &=(a^{-1})^{m+1} &&\text{definition of powers} \end{align}

The proof of $(xy)^{-1}=x^{-1}y^{-1}$: \begin{align} (xy)(x^{-1}y^{-1}) &=(yx)(x^{-1}y^{-1}) &&\text{commutativity} \\ &=y\bigl(x(x^{-1}y^{-1})\bigr) &&\text{associativity} \\ &=y\bigl((xx^{-1})y^{-1}\bigr) &&\text{associativity} \\ &=\dots \\ &=1 \end{align} which you can finish.