Question

Show that ${\left(w^R\right)}^{\mathbf{i}}\mathbf{=}{\left(w^i\right)}^{\mathbf{R}}$ whenever w is a string and i is a nonnegative integer; that is, show that the ${\mathit{i}}^{\mathbf{t}\mathbf{h}}$${\mathit{i}}^{\mathbf{th}}$ power of the reversal of a string is the reversal of the power of the string

Short answer

From the principle of mathematical induction, we have

${\left(W^R\right)}^i={\left(W^i\right)}^R$

Step-by-step solution

Introduction

Mathematical Induction is a mathematical technique which is used to prove a statement, a formula or a theorem is true for every natural number. Step 1(Base step) − It proves that a statement is true for the initial value.

Induction step

Let p (i) be the statement that ${\left(W^R\right)}^i={\left(W^i\right)}^R$

Basis step:

p (0) is true because $\begin{array}{r}{\left(w^R\right)}^0={\left(w^0\right)}^R\\ =1\end{array}$

Inductive step:

Assume that p (i+1) is true.

$i.e.{\left(w^R\right)}^i={\left(w^i\right)}^R$

We show that p(i + 1) is true.

$\begin{array}{r}\mathrm{Now}{\left({\mathrm{W}}^{\mathrm{R}}\right)}^{\mathrm{i}+1}={\left({\mathrm{W}}^{\mathrm{R}}\right)}^{\mathrm{i}}\cdot {\mathrm{W}}^{\mathrm{R}}\\ ={\left({\mathrm{W}}^{\mathrm{i}}\right)}^{\mathrm{R}}\cdot {\mathrm{W}}^{\mathrm{R}}\left(\text{From inductive step}\right)\\ ={\left({\mathrm{W}}^{\mathrm{i}}\right)}^{\mathrm{R}}\cdot {\left({\mathrm{W}}^1\right)}^{\mathrm{R}}\\ ={\mathrm{W}}^{\mathrm{iR}}\cdot {\mathrm{W}}^{1\mathrm{R}}\\ ={\mathrm{W}}^{\mathrm{iR}+1\mathrm{R}}\\ ={\mathrm{W}}^{(\mathrm{i}+1)\mathrm{R}}\\ ={\left({\mathrm{W}}^{\mathrm{i}+1}\right)}^{\mathrm{R}}\end{array}$

Therefore p (k + 1) is true.

Hence from the principle of mathematical induction, we have

${\left(W^R\right)}^i={\left(W^i\right)}^R$