Erin Lozano

Answered

2022-06-25

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

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

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

Answer & Explanation

kpgt1z

Expert

2022-06-26Added 23 answers

HINT: Note that

$\begin{array}{rl}\frac{a+c}{b+d}-\frac{a}{b}& =\frac{a+c}{b+d}\cdot \frac{b}{b}-\frac{a}{b}\cdot \frac{b+d}{b+d}\\ & =\frac{b(a+c)-a(b+d)}{b(b+d)}\\ & =\frac{ab+bc-ab-ad}{b(b+d)}\\ & =\frac{bc-ad}{b(b+d)}\phantom{\rule{thickmathspace}{0ex}}.\end{array}$

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.

$\begin{array}{rl}\frac{a+c}{b+d}-\frac{a}{b}& =\frac{a+c}{b+d}\cdot \frac{b}{b}-\frac{a}{b}\cdot \frac{b+d}{b+d}\\ & =\frac{b(a+c)-a(b+d)}{b(b+d)}\\ & =\frac{ab+bc-ab-ad}{b(b+d)}\\ & =\frac{bc-ad}{b(b+d)}\phantom{\rule{thickmathspace}{0ex}}.\end{array}$

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.

arridsd9

Expert

2022-06-27Added 12 answers

$\frac{a}{b}=\frac{a(1+\frac{d}{b})}{b(1+\frac{d}{b})}=\frac{a+d(\frac{a}{b})}{b+d}<\frac{a+d(\frac{c}{d})}{b+d}=\frac{a+c}{b+d}$

and similarly for the other inequality.

and similarly for the other inequality.

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