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. \(\frac{a+c}{b+d} = \frac{a}{b} + \frac{c}{d}\)

Explanation: 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.