if $a, b, c$ and $d$ are positive integers and $\frac{a}{b}

Question

if $a, b, c$ and $d$ are positive integers and $\frac{a}{b} < \frac{c}{d}$, which of the following must be true? i. $\frac{a+c}{b+d} < \frac{c}{d}$ II. $\frac{a+c}{b+d} < \frac{a}{b}$ III. $\frac{a+c}{b+d} = \frac{a}{b} + \frac{c}{d}$

Accepted Answer

সঠিক উত্তর: । ব্যাখ্যা: Given: $a, b, c, d$ are positive integers, and $\frac{a}{b} < \frac{c}{d}$. **Statement i:** $\frac{a+c}{b+d} < \frac{c}{d}$. Since $c$ and $d$ are positive, cross-multiply: $d(a+c) < c(b+d)$. $ad + cd < cb + cd$. Subtract $cd$ from both sides: $ad < cb$. Divide by $bd$ (which is positive): $\frac{ad}{bd} < \frac{cb}{bd}$. $\frac{a}{b} < \frac{c}{d}$. This is the original given inequality. Since the steps are reversible and the original inequality is given as true, **Statement i must be true.**. **Statement ii:** $\frac{a+c}{b+d} < \frac{a}{b}$. Since $a$ and $b$ are positive, cross-multiply: $b(a+c) < a(b+d)$. $ab + bc < ab + ad$. Subtract $ab$ from both sides: $bc < ad$. Divide by $bd$: $\frac{bc}{bd} < \frac{ad}{bd}$. $\frac{c}{d} < \frac{a}{b}$. This contradicts the given inequality $\frac{a}{b} < \frac{c}{d}$. So, **Statement ii must be false.**. **Statement iii:** $\frac{a+c}{b+d} = \frac{a}{b} + \frac{c}{d}$. Let $a=1, b=2, c=3, d=4$. $\frac{a}{b} = \frac{1}{2}$, $\frac{c}{d} = \frac{3}{4}$. $\frac{1}{2} < \frac{3}{4}$ is true. LHS: $\frac{a+c}{b+d} = \frac{1+3}{2+4} = \frac{4}{6} = \frac{2}{3}$. RHS: $\frac{a}{b} + \frac{c}{d} = \frac{1}{2} + \frac{3}{4} = \frac{2}{4} + \frac{3}{4} = \frac{5}{4}$. Since $\frac{2}{3} \neq \frac{5}{4}$, **Statement iii is false.**. Only Statement i must be true.

if \(a, b, c\) and \(d\) are positive integers and \(\frac{a}{b} < \frac{c}{d}\), which of the following must be true? i. \(\frac{a+c}{b+d} < \frac{c}{d}\) II. \(\frac{a+c}{b+d} < \frac{a}{b}\) III. \(\frac{a+c}{b+d} = \frac{a}{b} + \frac{c}{d}\)