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Edit: The original question lacked some constraints as saying that $ a $, $b$, $c$, $d$ are positive and integers and that we know that $ \frac{a}{b} < \frac{c}{d}$. Thank you @Aqua and @LeeMosher for pointing out the flaws

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I'm trying to prove that, given $ a $, $b$, $c$, $d$ positive numbers and integers, if:

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

then:

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

For that, I simplify the inequation, removing the denominators:

\begin{align} \frac{a}{b} < \frac{a + c}{b + d} < \frac{c}{d} \\ ad(b + d) < bd(a + c) < bc(b+d) \\ abd + ad^2 < abd + bcd < b^2c + bcd \\ ad^2 - bcd < 0 < b^2c - abd \\ \end{align}

Try to clear left side of the inequation for $ a $:

\begin{align} ad^2 - bcd < 0 \\ a < \frac{bc}{d} \\ \frac{a}{b} < \frac{c}{d} \end{align}

And, same for the right side of the inequation:

\begin{align} 0 < b^2c - abd \\ a < \frac{bc}{d} \\ \frac{a}{b} < \frac{c}{d} \end{align}

Could I argue that this proves that: $ \frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}$ ?

Jon
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    Related: This question. – Axion004 Jan 18 '20 at 22:48
  • Thank you, I edited my question based on that one. – Jon Jan 18 '20 at 23:21
  • Counterexample: let $a=b=c=1$ and $d=2$, so $\frac{a}{b} = 1 > \frac{1}{2} = \frac{c}{d}$. – Lee Mosher Jan 18 '20 at 23:52
  • @LeeMosher $ \frac{a}{b} $ is less than $ \frac{c}{d} $ – Jon Jan 19 '20 at 00:09
  • Then you should add that hypothesis to your question. Nowhere is that stated. – Lee Mosher Jan 19 '20 at 04:28
  • @LeeMosher, isn't implied if what we are trying to prove is a/b < a+c/b+d < c/d? – Jon Jan 19 '20 at 11:20
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    I agree that $$\underbrace{a/b < (a+c)/(b+d) < c/d}_Q \text{ implies } \underbrace{a/b < c/d}_P$$ And I understand that your intention is to prove the converse implication, namely that $a/b < c/d$ implies $a/b < (a+c)/(b+d) < c/d$ (with additional hypotheses pointed out by @Aqua). But when you are trying to prove a statement of the form "$P$ implies $Q$", you may not use implications of $Q$ in your proof; that would make your proof circular. You must always state precisely what hypotheses you are assuming. To prove "$P$ implies $Q$", you must state $P$ as your hypothesis. – Lee Mosher Jan 19 '20 at 14:33
  • @LeeMosher Thank you. I understand now. The main problems when asking this question have been not specifying that $ a, b, c, d $ are positive and integers and that initially $ a/b < c/d $. I have edited the question. – Jon Jan 19 '20 at 14:44
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    There is an interesting "partial converse" to your question. It can be shown that if $a,b,c,d$ are positive integers such that $\frac{a}{b}<\frac{c}{d}$. Any rational number $r$ such that $\frac{a}{b}<r<\frac{c}{d}$ can be written in the form $$r=\frac{au+cv}{bu+dv}$$ where $u$ and $v$ are coprime positive integers. Your problem is a special case where $u=v=1$. $$ \phantom{a}$$ In fact, for all real numbers $a,b,c,d>0$ such that $\frac{a}{b}<\frac{c}{d}$, if $r$ is a real number such that $\frac{a}{b}<r<\frac{c}{d}$, then $r=\frac{ax+c}{bx+d}$ for some real number $x>0$. – Batominovski Jan 19 '20 at 15:11
  • So the requirement of the numbers being positive is not necessary after all. But the premise of $ a/b < c/d $ is. – Jon Jan 19 '20 at 15:44
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    You need at least that $b$ and $d$ are positive. For example, when $a=2$, $b=-1$, $c=-2$ and $d=2$, we have $$\frac{a}{b}=-2<-1=\frac{c}{d},$$ but $$\frac{a+c}{b+d}=\frac{2-2}{-1+2}=0$$ does not lie between $a/b$ and $c/d$. – Batominovski Jan 19 '20 at 18:41

2 Answers2

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No, it is not correct since you don't know if the denominators are positive (or negative). Only then you can get rid of them (easly).

Correct would be:

$$ \frac{c}{d} - \frac{a+c}{b+d} = {(b+d)c-d(a+c)\over d(b+d)} = {bc-ad\over d(b+d)}$$

But we can not say nothing explicitly for the last fraction since we know nothing about $a,b,c,d$.

nonuser
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  • Thank you for your answer. What do you mean with: "we can say nothing explicitly for the last fraction since we know nothing about a,b,c,d"? – Jon Jan 18 '20 at 23:22
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    What that means is this. First, the quantity $bc-ad$ could be positive or negative. Second, the quantity $d$ could be positive or negative. Third, the quantity $b+d$ could be positive or negative. But we have not made any hypotheses about the values of $a,b,c,d$. So we can draw no conclusions about about the first, second, or third possibilities. So we can draw no conclusions about whether $\frac{bc-ad}{d(b+d)}$ is positive or negative. So we can draw no conclusions about whether $\frac{a+c}{b+d} < \frac{c}{d}$. – Lee Mosher Jan 18 '20 at 23:53
  • Oh, I understand now. Definitely I have missed an important part when writing down the question and is that $ a $, $b $, $c$, $d$ are integers and positive – Jon Jan 19 '20 at 00:06
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Yours is correct. Here is another approach:

Let $$k=\dfrac{a}{b}<\dfrac{c}{d}=l$$ Then $$\dfrac{bk+dk}{b+d}<\dfrac{bk+dl}{b+d}<\dfrac{bl+dl}{b+d}$$ $$\iff \dfrac{a}{b}<\dfrac{a+c}{b+d}<\dfrac{c}{d}$$

Martund
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