Chapter A, Problem 44E

To determine

To calculate: To show that if $0<a<b$ , then $a^2<b^2$

Answer to Problem 44E

Its proved that if $0<a<b$ , then $a^2<b^2$

Explanation of Solution

Given information: $0<a<b$

Formula Used:

If $a$ and $b$ are real numbers, where $c$ is non-zero real number, then the inequality $a<b$ is same as

  $a+c<b+c$

  $ac<bc$   if $c>0$

  $ac>bc$   if $c<0$

Calculation:

Given the following:

  $0<a<b$

On simplifying,

  $\begin{array}{l}a<b\\ a-b<0\end{array}$

Since $a>0,b>0$ , thus $a+b>0$

Multiplying both sides of equation by $\left(a+b\right)$ :

  $\begin{array}{l}\left(a-b\right)\left(a+b\right)<0\left(a+b\right)\\ \left(a-b\right)\left(a+b\right)<0\\ a^2-b^2<0\\ a^2<b^2\end{array}$

Conclusion:

Hence, proved that $a^2<b^2$