Chapter 1.7, Problem 122E

To determine

To state: a general rule about the relationship between $a^n\text{and }{b}^n$ when $a<b$ and n is any integer.

Answer to Problem 122E

The solution $a^n<b^n$

Explanation of Solution

Given:

If $a<b,\text{ is }{a}^2<b^2$ any $a<b,\text{ is }{a}^3<b^3$ .

Calculation:

According to the question, to check all positive and negative values,

Case 1: both $a\text{ and}b$ negative. Then:

  $\begin{array}{l}a<b\\ \therefore a^2>b^2\{ifa<0,b<0\}\end{array}$

Case 2: both $a\text{negative and}b\text{positive}$ . Then:

  $\begin{array}{l}a<0,b>0\\ \therefore \text{ can't say about }{a}^2>b^2\end{array}$

Case 3: both $a\text{ and}b$ positive. Then:

  $\begin{array}{l}ifa>0,b>0\\ \therefore a^2<b^2\end{array}$

Case 4: both $a\text{ and}b$ negative. Then:

  $\begin{array}{l}a<b\\ \therefore a^3<b^3\{ifa<0,b<0\}\end{array}$

Case 5: $a\text{negative and}b\text{positive}$ . Then:

  $\begin{array}{l}a<0,b<0\\ a^3<b^3\end{array}$

Case 6: both positive. Then:

  $\begin{array}{l}a<0,b>0\\ a^3<b^3\end{array}$

Case 7: From the above observation, $a^n<b^n$ for given $a<b$ .

Case8: If n is even,

If $a<0,b<0$

Then $a^n>b^n$

  $\begin{array}{l}\text{If }a<0,b>0\\ \therefore \text{can't say about }{a}^n>b^n\end{array}$

  $ifa>0,b>0$

  $a^n<b^n$

Case9: If n is odd,

If $a<0,b<0$

Then $a^n<b^n$

  $\begin{array}{l}\text{If}a<0,b>0\\ a^n<b^n\end{array}$

  $\begin{array}{l}ifa>0,b>0\\ a^n<b^n\end{array}$

Hence, the relationship between $a^n\text{and }{b}^n$ is above solved.