A proof that I came up with

by Dr4gon39, Apr 5, 2016, 1:09 PM

Given: $\frac{a}{b}<\frac{c}{b}$ , where $a,b,c,d$ are positive integers, and $\frac{a}{b}$ and $\frac{c}{d}$ are simplified fractions, prove that $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$ .

In order to prove that $\frac{a}{b}<\frac{a+c}{b+d}$ , we attempt to prove that their difference is positive.

We have:
$\frac{a+c}{b+d}-\frac{a}{b}$
$=\frac{b(a+c)-a(b+d)}{b(b+d)}$
$=\frac{ab+cb-ab-ad}{b(b+d)}$
$=\frac{bc-ad}{b(b+d)}$

Due to the fact that $b$ and $d$ are both positive, $b(b+d)$ must be positive, so now we are attempting to prove that:
$bc-ad>0$
Or in other words:
$bc>ad$

Suppose we have two positive fractions $\frac{x}{y}=\frac{w}{z}$
Then suppose we have another positive fraction, $\frac{e}{f}$
Then, $\frac{x}{y}<\frac{w}{z}+\frac{e}{f}$
We add the two fractions and we get,
$\frac{x}{y}<\frac{wf+ez}{zf}$
We now attempt to examine the relation between the product of the numerator of the first fraction and the denominator of the second fraction and the product of the denominator of the first fraction and the numerator of the second fraction.

We subtract the product of the numerator of the first fraction and the denominator of the second fraction from the product of the denominator of the first fraction and the numerator of the second fraction.

We have:
$y(wf+ez)-xzf=wyf+yez-xzf$
Since $\frac{x}{y}=\frac{w}{z}$
then $xz=wy$
Therefore, $wyf=xzf$
So:
$wyf+yez-xzf=yez$
Since $y,e,z$ are all positive numbers, then $yez$ must be postivie too.
Hence, the product of the numerator of the first fraction and the denominator of the second fraction is less than the product of the denominator of the first fraction and the numerator of the second fraction.

From this we can conclude:
if we have two fractions where $\frac{a}{c}<\frac{b}{d}$ then $ad<bc$

Back to the original proof:

We now know that the difference between $\frac{a+c}{b+d}$ and $\frac{a}{b}$ is positive, so $\frac{a+c}{b+d}>\frac{a}{b}$

Now, we must prove that $\frac{c}{d}>\frac{a+c}{b+d}$ .
We do this by subtracting these two terms.
We get:
$\frac{c}{d}-\frac{a+c}{b+d}$
$=\frac{c(b+d)-d(a+c)}{d(b+d)}$
$=\frac{cb+cd-da-dc}{d(b+d)}$
$=\frac{bc-ad}{d(b+d)}$
Since $b$ and $d$ are positive numbers, then the denominator is positive.
Since $bc>ad$ the numerator is positive.
Therefore the difference between $\frac{c}{d}$ and $\frac{a+c}{b+d}$ is positive.
Therefore $\frac{c}{d}>\frac{a+c}{b+d}$

From above we now have the inequality $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$ .

Thus, our proof is complete.

Application of the above statement:

MagMaR team round 2016 problem 15.

Find a fraction $\frac{m}{n}$ between $\frac{59}{80}$ and $\frac{45}{61}$ for which $n<200$ .

By the above statement, $\frac{59+45}{80+61}=\frac{104}{141}$ must be between those two fractions. Ergo, problem solved.

Credit:

Credit should be given to Mr. Li, one of my old math teachers for telling me about this statement.
I, myself first wrote this proof a year ago in Chinese.
A few days ago, after using this statement on MagMaR, I felt the urge to write a proof in English.

Disclaimer:

If this is an actual theorem/lemma, please tell me.
No copyright infringement or infringement upon intellectual property was intended.

If anyone has found more errors in my proof, please PM me, leave a comment, or give me a shout.

An easier proof written by RedSoxFan:

Part 1:

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

$ad < bc$

$ad+ab < bc+ab$

$a(b+d) < b(a+c)$

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

Part 2:

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

$ad < bc$

$ad + dc < bc + dc$

$d(a + c) < c(b + d)$

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

And we are done.

No copyright infringement or infringement upon intellectual property was intended

I'd just like to point out to all of the haters that I was the one who solved the MagMaR problem in 3 seconds, not the rest of y'all with these super simple proofs so.... Peace out guys *winks* *drops mic*

I'd also like to submit that RedSoxFan's proof is kind of just mine worked backwards and not as technical and thorough so.... *cues Taylor Swift voice* "haters gonna hate hate hate hate hate"

DISCLAIMER:

I in no way own the lyrics to Taylor Swift's song entitled "Shake it off", all rights to that song belong to her (or the company she's signed to or whatever)

Basically, I just don't want to get my pants sued off...

This post has been edited 13 times. Last edited by Dr4gon39, Apr 5, 2016, 11:24 PM

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Let's steal the show.

We wish to prove $\frac{a+c}{b+d}>\frac{c}{d}$

Thus, we are trying to prove that $\frac{c}{d}-\frac{a+c}{b+d}$ is positive. Finding a common denominator, we get
$\frac{c(b+d)-d(a+c)}{d(b+d)}$

Simplifying, we have
$\frac{cb-da}{db+d^2}$

And since $d$ and $b$ are positive, $db+d^2$ is too, and therefore we simply need to prove $cb-da$ is, or $cb>da$ .

Given: $a/b<c/d$ , so by cross multiplication, $cb>da$ .

And we are done.

by willwin4sure, Apr 5, 2016, 5:57 PM

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Whoops I mean $\frac{a+c}{b+d}<c/d$

by willwin4sure, Apr 5, 2016, 5:59 PM

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XD That pro proofreading skill

by stephy2003, Apr 5, 2016, 8:14 PM

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Wait, shouldn't I be required to prove that cross multiplying thing though??

by Dr4gon39, Apr 5, 2016, 8:55 PM

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They are all positive, so I think it is fine

by willwin4sure, Apr 5, 2016, 9:22 PM

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No copyright infringement or infringement upon intellectual property was intended

by RedSoxFan, Apr 5, 2016, 10:21 PM

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Let's steal the stolen show with a much easier proof:

Part 1:

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

$ad < bc$

$ad+ab < bc+ab$

$a(b+d) < b(a+c)$

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

Part 2:

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

$ad < bc$

$ad + dc < bc + dc$

$d(a + c) < c(b + d)$

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

And we are done.

No copyright infringement or infringement upon intellectual property was intended

by RedSoxFan, Apr 5, 2016, 10:43 PM

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@RedSoxFan
Thank you for the much less convoluted proof.

@Dr4gon39
See!?! I told you there would be a simpler proof!

by dckx15, Apr 5, 2016, 11:18 PM

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^for once, @dr4gon39 did not offer the most elegant solution this is amusing

This post has been edited 2 times. Last edited by Dr4gon39, Apr 5, 2016, 11:37 PM

by cedric-the-reindeer, Apr 5, 2016, 11:32 PM

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@above For the record, I still believe that my proof and RedSoxFan's are the same, besides the fact that he skipped steps, skipped details, worked backwards from mine etc.
Unfortunately, step skipping isn't tolerated in rigorous mathematical proofs.
Ergo, *cues Death Metal Voice* "BEHOLD THE KINGS! THE KING OF KINGS!"

This post has been edited 1 time. Last edited by Dr4gon39, Apr 5, 2016, 11:40 PM

by Dr4gon39, Apr 5, 2016, 11:39 PM

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I will threaten you with cats if you continue to say that

by cedric-the-reindeer, Apr 5, 2016, 11:41 PM

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I think that RedSoxFan's Solution was much better

by Supermath7676, Apr 6, 2016, 10:54 PM

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Unforunately Dr4gon39 I don't even understand your proof so please turn your blog over to RedSoxFan who will post some actually meaningful content

by ghghghghghghghgh, Nov 28, 2016, 1:27 AM

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Submit

ooh hello
by jj_ca888, Nov 18, 2019, 10:36 PM
it's been a good ride.
by ghghghghghghghgh, Jul 30, 2019, 2:17 AM
Vincent's Fan
by Supermath7676, Feb 14, 2017, 11:48 PM
Clearly just a function convolution.
by willwin4sure, Nov 28, 2016, 5:11 PM
yvincent
by ghghghghghghghgh, Nov 28, 2016, 1:22 AM
oh dearr..
by stephy2003, Oct 12, 2016, 4:55 PM
Lol @ghghghghghghgh when I delete your comment and the admins refer to it as taking out the "trash"
Touche
by Dr4gon39, Oct 11, 2016, 9:18 PM
Vincent's Fan
by Supermath7676, Oct 11, 2016, 8:13 PM
China......
by willwin4sure, Aug 4, 2016, 11:34 PM
all the many facets of me
by RedSoxFan, Jul 20, 2016, 10:21 PM
We need a chocolate bar
by stephy2003, Jun 23, 2016, 7:17 PM
Post on LMT?
by willwin4sure, Apr 20, 2016, 8:53 PM
King of kings. XD I thought you were the doctor of sides Dr. Gon.

@dckx because /Wilma/ didn't elaborate with instructions. I is creary veri clueless
by stephy2003, Apr 17, 2016, 3:27 AM
You lost your chance...
by dckx15, Apr 16, 2016, 9:24 PM
AHEM.
BEHOLD THE KING!
THE KING OF KINGS!
by Dr4gon39, Apr 16, 2016, 1:45 PM
~~He's coming back soon, do we want to kill his blog or not?~~
by stephy2003, Apr 15, 2016, 12:13 AM
You're very welcome
by willwin4sure, Apr 13, 2016, 9:22 PM
~~Thanks for knowing my name~~
by stephy2003, Apr 13, 2016, 2:07 PM
Stephen you can post right?
by willwin4sure, Apr 13, 2016, 12:23 PM
~~Guys his in D.C. this week, let's kill his blog~~
by stephy2003, Apr 12, 2016, 11:44 PM
Are you not going to talk about LMT?
by dckx15, Apr 11, 2016, 4:13 PM
Lolololol RedSoxFan just rekt your proof
by willwin4sure, Apr 5, 2016, 10:50 PM
Yay you fixed the typos
by RedSoxFan, Apr 5, 2016, 10:15 PM
@willwin4sure, i dunno, is that allowed?
by Dr4gon39, Apr 5, 2016, 8:44 PM
No Mr. Gon, willwin4sure is
by stephy2003, Apr 5, 2016, 8:14 PM
Can't you just cross multiply to determine if a/b > c/d then ad>bc?
by willwin4sure, Apr 5, 2016, 5:22 PM
You have too many typos
by RedSoxFan, Apr 5, 2016, 5:13 PM
#**********CLUBTAKEOVER.
Harvard is next.
by willwin4sure, Apr 5, 2016, 11:58 AM
I'm obnoxious?
by Dr4gon39, Apr 5, 2016, 11:03 AM
yes he was
and to me
and redsoxfan
by ilikepie2003, Apr 5, 2016, 12:46 AM
Were you obnoxious to Emc?
by stephy2003, Apr 5, 2016, 12:41 AM
Post about your amazing MagMar experience.
by willwin4sure, Apr 4, 2016, 10:10 PM
~~a phat lizard~~ someone who can post, but usually doesn't. I kind of just use it to bookmark blogs. amd this is why summitwei never gets any views from me
by stephy2003, Mar 16, 2016, 2:58 AM
Dunno. HWAT is a contrib?
by Dr4gon39, Mar 16, 2016, 2:31 AM
Can I be a contrib? Thx!
by ilikepie2003, Mar 4, 2016, 3:59 AM
Good luck on the AIME today!
by stephy2003, Mar 3, 2016, 11:05 PM
Well I guess the only thing different are the amount of h's in UGHHHHHH
by ilikepie2003, Feb 29, 2016, 12:12 AM
Agreed @stephy
by dckx15, Feb 28, 2016, 3:54 PM
Barely refrains from saying real name. *@Dr4gon39*, you have the most creative post names, I mean they totally aren't annoyingly repetitive
by stephy2003, Feb 24, 2016, 12:51 AM
ALLO
C'EST MOI
by ilikepie2003, Feb 20, 2016, 10:30 PM
Second Shout.
by stephy2003, Feb 19, 2016, 3:41 PM
first shout >
by doitsudoitsu, Feb 17, 2016, 12:17 AM

42 shouts

My blog. Duh.

A proof that I came up with

by Dr4gon39, Apr 5, 2016, 1:09 PM

13 Comments