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I am looking for proof that, if you take any two different fractions and add the numerators together then the denominators together, the answer will always be a fraction that lies between the two original fractions.

Would be grateful for any suggestions!

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    $\begingroup$ Not to be picky, but if we consider the fractions $a/b,c/d$ with $a=b=c=1$ and $d=-2$, then $(a+c)/(b+d)=-2$, which does not lie between $a/b=1$ and $c/d=-1/2$.... $\endgroup$ – Barry Cipra Apr 27 '16 at 18:55
  • $\begingroup$ Good point Barry. Let's say that a, b, c and d are all positive integers. $\endgroup$ – Joeywald Apr 27 '16 at 19:24
  • $\begingroup$ mathsguy, not a lot as "proof" is new to me. I know that adding the same number to the top and bottom of a fraction moves the derived fraction closer to 1 and that adding two fractions as above gives a resultant fraction between the two, but I can't prove why this happens. $\endgroup$ – Joeywald Apr 27 '16 at 19:33
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Suppose we have positive $a,b,c,d$ with $\frac{a}{b}\ge \frac{c}{d}$. Then multiplying through by $bd$ we get $ad\ge bc$. Adding $ab$ to both sides we get $a(b+d)\ge b(a+c)$. Dividing by $b(b+d)$, we get $\frac{a}{b}\ge\frac{a+c}{b+d}$.

Similarly, add $cd$ to both sides of $ad\ge bc$ to get $d(a+c)\ge c(b+d)$. Dividing by $d(b+d)$ we get $\frac{a+c}{b+d}\ge \frac{c}{d}$.

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Assuming $a,b,c,d\gt0$, we have

$$\begin{align} {a\over b}\lt{a+c\over b+d}\lt{c\over d}&\iff a(b+d)\lt b(a+c)\quad\text{and}\quad(a+c)d\lt(b+d)c\\ &\iff ad\lt bc\quad\text{and}\quad ad\lt bc\\ &\iff ad\lt bc\\ &\iff {a\over b}\lt{c\over d} \end{align}$$

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  • $\begingroup$ One usually starts with the premise and ends with the conclusion... No? $\endgroup$ – Najib Idrissi Jun 7 '16 at 17:38
  • $\begingroup$ @NajibIdrissi If it's $“\Leftrightarrow”$ signs the whole way through, it shouldn't matter. $\endgroup$ – Akiva Weinberger Jun 7 '16 at 17:47
  • $\begingroup$ @AkivaWeinberger Sure, but I think it's easier to read and looks more sound if you put it in the other order. But of course from a logical point of view, equivalence of propositions is symmetrical... $\endgroup$ – Najib Idrissi Jun 7 '16 at 18:09
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Here is another way that uses calculus:

Let $\phi(t) = {(1-t)a+t c \over (1-t)b + t d}$ and note that $\phi(0) = {a \over b}, \phi(1) = {c \over d}$. Furthermore, $\phi'(t) = {bc-ad \over ((1-t)b + t d )^2}$. Since $ad < bc$, we see that $\phi$ is increasing, and so $\phi(0) \le \phi({1 \over 2}) \le \phi(1)$. Since $\phi({1 \over 2}) = {a+c \over b+d}$, we have the desired result.

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Assume that $b,d >0$. Note that $$\frac{a+c}{b+d} = \frac{b}{b+d}\frac{a}{b} +\frac{d}{b+d}\frac{c}{d}.$$ Remark that $0<\frac{b}{b+d}<1$ and the same for $\frac{d}{b+d}$. Hence you have written $\frac{a+b}{c+d}$ as a convex combination of $\frac{a}{b}$ and $\frac{c}{d}$ so you get $$\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}.$$

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Here's one way to look at it:

You're taking a class. Suppose you get $a$ points out of $b$ possible on Quiz 1, and $c$ points out of $d$ possible on Quiz 2.

Your overall points are $a+c$ out of $b+d$ possible. And your overall percentage should be between your lower quiz score and your higher quiz score.

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  • $\begingroup$ The question asks for "proof". $\endgroup$ – AlohaSine Apr 30 '16 at 23:54
  • $\begingroup$ @MathematicsStudent1122 Actually the presenter asked for suggestions. One could use this idea as the basis for a proof. $\endgroup$ – paw88789 May 1 '16 at 2:49
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Hint $\ $ The mediant $(a\!+\!b)/(c\!+\!d)\,$ is the slope of the diagonal of the parallelogram with vector sides $(b,a),\ (d,c).\:$ But the slope of the diagonal lies between the slopes of the sides.

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