Question

Let$R$ be a ring and$a,b\in R$ . Letrole="math" localid="1646829749148" $mandn$ be positive integers.

(a) Show that $a^m{a}^n=a^{m+n}$ and ${\left(a^m\right)}^n=a^{mn}$ .

(b) Under what conditions is it true that ${\left(ab\right)}^n=a^n{b}^n$?

Short answer

(a) It is proved that $a^m{a}^n=a^{m+n}$, and ${\left(a^m\right)}^n=a^{mn}$.

(b) Hence, when the ring is commutative, then ${\left(ab\right)}^n=a^n{b}^n$ is true.

Step-by-step solution

Property of Rings:

If any ringis designated as$R$,such that $a,b\in R$, then:

$a=b\iff a-b=0$

Proof:

The given exponential expression is: $a^m{a}^n$.

$\begin{array}{rcl}{a}^m{a}^n& =& \left(a·a·........a\right)·\left(a·a·........a\right)\\ & =& a·a·........a\\ & =& a^{\left(m+n\right)}\end{array}$

Hence, it is proved that $a^m{a}^n=a^{m+n}$.

Again, the given exponential expression is: ${\left(a^m\right)}^n$.

$\begin{array}{rcl}{\left(a^m\right)}^n& =& \left(a^m·a^n·.......a^m\right)\\ & =& a·a·.......a\\ & =& a^{mn}\end{array}$

Hence it is proved that ${\left(a^m\right)}^n=a^{mn}$.

Proof:

Let the given ring be commutative, such that $a,b\in R$. Then, we have:

role="math" localid="1646831755376" $$\begin{gathered} ab=ba \\ \Rightarrow {\left(ab\right)}^n=\left(ab.ab·......·ab\right) \\ =\left(ab.ab·......·ab\right)\left(b.b·......·b\right) \\ =a^n·b^n \end{gathered}$$

Hence, when the ring is commutative, then ${\left(ab\right)}^n=a^n{b}^n$.