Chapter 1.7, Problem 104E

To determine

To show:that $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$ where $a,b,c,\text{and }d$ are positive numbers.

Explanation of Solution

Given:

The given condition is $a,b,c,\text{and }d$ are positive numbers such that $\frac{a}{b}<\frac{c}{d}$ .

Calculation:

According to the question, the given inequality is

  $\frac{a}{b}<\frac{c}{d}$ .

The inequality is expressed as

  $\begin{array}{l}\frac{a}{b}<\frac{c}{d}\\ ad<bc\{cross-multiplication\}\\ ad+ab<bc+ab\left\{addabonboththesides\right\}\\ a\left(d+b\right)<b\left(c+a\right)\\ \therefore \frac{a}{b}<\frac{a+c}{b+d}\mathrm{....}.\mathrm{....}(1)\end{array}$

Also, the expression is expressed as

  $\begin{array}{l}\frac{a}{b}<\frac{c}{d}\\ ad<bc\{cross-multiplication\}\\ ad+cd<bc+cd\left\{addcdonboththesides\right\}\\ d\left(a+c\right)<c\left(b+d\right)\\ \therefore \frac{a+c}{b+d}<\frac{c}{d}\mathrm{....}.\mathrm{....}(2)\end{array}$

From (1) and (2), it is found that

  $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$

Hence, the equation is proved.