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I'm reading a introductory book on mathematical proofs and I am stuck on a question.

Let $a, b, c, d$ be positive real numbers, prove that if $\frac{a}{b} < \frac{c}{d}$, then $\frac{a}{b} < \frac{a+c}{b+d} <\frac{ c}{d}$.

miracle173
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Math
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  • Not a proof, but worth noting: If you interpret the fraction $a/b$ as the score on a quiz, with $a$ correct answers on $b$ questions, and likewise for $c/d$, then $(a+c)/(b+d)$ is the combined average score for two quizzes -- i.e., a total of $a+c$ correct answers on $b+d$ questions -- so it makes sense that it would lie between the higher and the lower of the two individual scores. – Barry Cipra Aug 17 '16 at 00:39
  • Very true. Thank you. – Math Aug 18 '16 at 15:04

3 Answers3

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HINT: Note that

$$\begin{align*} \frac{a+c}{b+d}-\frac{a}b&=\frac{a+c}{b+d}\cdot\frac{b}b-\frac{a}b\cdot\frac{b+d}{b+d}\\ &=\frac{b(a+c)-a(b+d)}{b(b+d)}\\ &=\frac{ab+bc-ab-ad}{b(b+d)}\\ &=\frac{bc-ad}{b(b+d)}\;. \end{align*}$$

You’d like to show that this difference is positive, so you’d like to know that $bc>ad$. Use the fact that $\frac{a}b<\frac{c}d$ to show this.

The other inequality can be handled similarly.

Brian M. Scott
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  • Hi Mr. Scott, could I ask for some guidance? I really want to be a Mathematics Professor, but am very undecided in which field to pursue. How did you decide? – Math Aug 13 '16 at 21:26
  • @Alan: In my case it was easy: I fell in love with topology when I first encountered it. I’ve always liked other areas of mathematics as well, but general and set-theoretic topology always came first for me. – Brian M. Scott Aug 13 '16 at 21:28
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$$\frac{a}{b}=\frac{a(1+\frac{d}{b})}{b(1+\frac{d}{b})}=\frac{a+d(\frac{a}{b})}{b+d}<\frac{a+d(\frac{c}{d})}{b+d}=\frac{a+c}{b+d}$$

and similarly for the other inequality.

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Another hint: For positive numbers, $\; \dfrac ab<\dfrac cd\iff ad<bc$.

Bernard
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