What can we know from this inequality?

Algebra Level 3

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

If $a, b, c, d$ (are real variables that) satisfy the above inequality, what can we say about

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

2 solutions

Calvin Lin Staff
Jul 29, 2015

If some of these terms are negative, then it need not be true that $\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}$ . For example,
$\frac{2}{-1} < \frac{1}{2}, \text{ but } \frac{2}{-1} < \frac{1}{2} < \frac{ 3 } { 1 }$
As pointed out by Julian, if $b+d =0$ , then the fraction is undefined.

-2/3 < -1/3 then -2/3 < -3/6 < -1/3 ... If -2/3 < 1/2 then -2/3 < -1/5 < 1/2. Wait I see -2/3 < 1/-3 then -1/0 has no meaning so then Calvin I now will buy what you are selling

Julian Fuller - 5 years, 10 months ago

That's a good point, that if $b+d = 0$ , then the fraction is undefined!

Calvin Lin Staff - 5 years, 10 months ago

Maybe it could have been explicitly stated that a,b,c,d belong to the set of integers (and not necessarily to the set of whole numbers)

Pavan Kumar Vaitheeswaran - 5 years, 10 months ago

Well, they are real numbers, not necessarily integers.

Calvin Lin Staff - 5 years, 10 months ago

@Calvin Lin sir, the options given were little ambiguous. Now u asked "what can we say about $\frac{a+c}{b+d}$ ". U should've mentioned " $\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}$ for all a,b,c,d" Then your question would've been clear.

Aditya Kumar - 5 years, 10 months ago

Isn't that a given? These are variables, and not specific constant that you have chosen.

Calvin Lin Staff - 5 years, 10 months ago

ooh i got it

Aditya Kumar - 5 years, 10 months ago

Completely unclear.answer from Calvin and what is more, not valid. For all real values except b and d equal zero, the solution a/b less than (a+c)/(b+d) less than c/d is one and only one correct answer.

Oleg Yovanovich

Oleg Yovanovich - 7 months, 2 weeks ago

Hi Oleg, I gave an explicit to the claim that $\frac{ a}{b} < \frac{ a+c} { b+d} < \frac{ c}{d}$ . In particular, we have $bd < 0$ , which was necessary to create the counterexample.

I believe that you're making the assumption that $bd > 0$ . I'm guessing that this happened implicitly when you multiplied by the denominator without checking the signs, to claim that $ad < bc$ .

Calvin Lin Staff - 7 months, 2 weeks ago

Figel Ilham
Aug 2, 2015

Since there are no boundaries for $a,b,c,d$ , we may free to assume either the numerators are non-positive (either positives) or denumerators are non-positive (either positives). If the statement will be true on $\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}$ , consider the statement before. $b+d$ may also result to zero and even negatives, so it contradicts the statement.

What can we know from this inequality?

2 solutions

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