Simple fun algebraic proof, help please?
- Thread starter markallen
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Without loss of generality, since we can just do it the other way, let's assume one is greater than the other.
\(\displaystyle \frac{a}{c}<\frac{b}{d}\)
Since everything is positive, we have:
\(\displaystyle \frac{a}{c}<\frac{b}{d} \iff a\cdot d < b\cdot c\)
Do that two more times
\(\displaystyle \frac{a+b}{c+d}<\frac{b}{d}\implies\)
\(\displaystyle \frac{a}{c}<\frac{b}{d}\)
Since everything is positive, we have:
\(\displaystyle \frac{a}{c}<\frac{b}{d} \iff a\cdot d < b\cdot c\)
Do that two more times
\(\displaystyle \frac{a+b}{c+d}<\frac{b}{d}\implies\)
D
Deleted member 4993
Guest
markallen said:
TK showed that
(a+b)/(c+d) < b/d
Now following his logic - twist it a bit - and yo'll get to
(a+b)/(c+d) > a/c
J
JeffM
Guest
Well here is a different way if you are still puzzled by TK's direct method.markallen said:
I admit that my way is longer, but I have an affection for proofs by contradiction.
Without loss of generaility, it can be assumed that (a/c) < (b/d). Same as TK's first step.
Assume, FOR PURPOSES OF CONTRADICTION, that (a + b)/(c + d) >= (b/d).
But c and d positive so d(c + d) positive.
So, (a + b)d >= b(c + d).
So ad + bd >= bc + bd.
So ad >= bc.
But c and d positive so cd positive.
So, (a/c)>= (b/d), which contradicts hypothesis that (a/c) < (b/d).
So (a + b)/(c + d) < (b/d).
So there is half of a proof by reductio ad absurdam.
Can you do the other half?
J
JeffM
Guest
I absolutely agree, but it is sometimes valuable to demonstrate that there is usually more than one way to skin a cat. Math is a way of thinking, and the more ways you know how to think the better. I suspect we agree on that too.tkhunny said:
Hello, markallen!
This is a slight variation of JeffM's proof . . .
\(\displaystyle \text{WLOG, assume }\,\frac{a}{c} \,<\,\frac{b}{d} \quad\Rightarrow\quad ad \:<\:bc\;\;[1]\,\)
[Edit: corrected typo]
\(\displaystyle \text{Add }ac\text{ to both sides of [1]: }\:ac + ad \:<\:ac + bc \quad\Rightarrow\quad a(c+d) \:<\:c(a+b)\)
. . \(\displaystyle \text{Divide by }c(c+d)\!:\:\;\frac{a}{c}\:<\:\frac{a+b}{c+d}\;\;[2]\;\)
\(\displaystyle \text{Add }bd\text{ to both sides of [1]: }\:ad + bd \:<\:bc + bd \quad\Rightarrow\qiad d(a+b) \:<\:b(c+d)\)
. . \(\displaystyle \text{Divide by }d(c+d)\!:\;\;\frac{a+b}{c+d} \:<\:\frac{b}{d}\;\;[3]\;\)
\(\displaystyle \text{Combine [2] and [3]: }\;\frac{a}{c}\:<\:\frac{a+b}{c+d}\:<\:\frac{b}{d}\)
This is a slight variation of JeffM's proof . . .
\(\displaystyle \text{WLOG, assume }\,\frac{a}{c} \,<\,\frac{b}{d} \quad\Rightarrow\quad ad \:<\:bc\;\;[1]\,\)
[Edit: corrected typo]
\(\displaystyle \text{Add }ac\text{ to both sides of [1]: }\:ac + ad \:<\:ac + bc \quad\Rightarrow\quad a(c+d) \:<\:c(a+b)\)
. . \(\displaystyle \text{Divide by }c(c+d)\!:\:\;\frac{a}{c}\:<\:\frac{a+b}{c+d}\;\;[2]\;\)
\(\displaystyle \text{Add }bd\text{ to both sides of [1]: }\:ad + bd \:<\:bc + bd \quad\Rightarrow\qiad d(a+b) \:<\:b(c+d)\)
. . \(\displaystyle \text{Divide by }d(c+d)\!:\;\;\frac{a+b}{c+d} \:<\:\frac{b}{d}\;\;[3]\;\)
\(\displaystyle \text{Combine [2] and [3]: }\;\frac{a}{c}\:<\:\frac{a+b}{c+d}\:<\:\frac{b}{d}\)
soroban,
it should have been "ad < bc" below.
Or this method:
\(\displaystyle Continuing \ with \ the \ information \ in \ the \ quote \ box:\)
\(\displaystyle * \ (ad - bc) \ < \ 0,\ \ and \ similarly,\)
\(\displaystyle ** \ \ (bc - ad) \ > \ 0.\)
\(\displaystyle Long \ division \ of \ \ \frac{a + b}{c + d} \ \ yields \ \ \frac{a}{c} + \frac{bc - ad}{c(c + d)}\)
\(\displaystyle By \ using \ *, \ \ \frac{a + b}{c + d} \ > \ \frac{a}{c}.\)
----------------------------------------------------------------------------
\(\displaystyle Rewrite \ \ \frac{a + b}{c + d} \ \ as \ \ \frac{b + a}{d + c}.\)
\(\displaystyle Long \ division \ of \ \ \frac{b + a}{d + c} \ \ yields \ \ \frac{b}{d} + \frac{ad - bc}{d(d + c)}.\)
\(\displaystyle By \ using \ **, \ \ \frac{a + b}{c + d} \ < \ \frac{b}{d}.\)
-----------------------------------------------------------------------------
\(\displaystyle Therefore,\)
\(\displaystyle \ \frac{a}{c} \ < \ \frac{a + b}{c + d} \ < \ \frac{b}{d}.\)
it should have been "ad < bc" below.
soroban said:
Or this method:
\(\displaystyle Continuing \ with \ the \ information \ in \ the \ quote \ box:\)
\(\displaystyle * \ (ad - bc) \ < \ 0,\ \ and \ similarly,\)
\(\displaystyle ** \ \ (bc - ad) \ > \ 0.\)
\(\displaystyle Long \ division \ of \ \ \frac{a + b}{c + d} \ \ yields \ \ \frac{a}{c} + \frac{bc - ad}{c(c + d)}\)
\(\displaystyle By \ using \ *, \ \ \frac{a + b}{c + d} \ > \ \frac{a}{c}.\)
----------------------------------------------------------------------------
\(\displaystyle Rewrite \ \ \frac{a + b}{c + d} \ \ as \ \ \frac{b + a}{d + c}.\)
\(\displaystyle Long \ division \ of \ \ \frac{b + a}{d + c} \ \ yields \ \ \frac{b}{d} + \frac{ad - bc}{d(d + c)}.\)
\(\displaystyle By \ using \ **, \ \ \frac{a + b}{c + d} \ < \ \frac{b}{d}.\)
-----------------------------------------------------------------------------
\(\displaystyle Therefore,\)
\(\displaystyle \ \frac{a}{c} \ < \ \frac{a + b}{c + d} \ < \ \frac{b}{d}.\)
J
JeffM
Guest
Soroban
Your proof is undoubtedly more elegant and concise than mine. But then I do not know much math, and I have this odd fondness for proofs by contradiction.
Jeff
Your proof is undoubtedly more elegant and concise than mine. But then I do not know much math, and I have this odd fondness for proofs by contradiction.
Jeff
D
Deleted member 4993
Guest
markallen said:
Ofcourse you do - you got your work done without doing any work!!
In your other post you said (without showing any work):
But when it comes to actually answering the question, i.e. when Person B picks 3 boxes... I can't quite get my head round the calculation.
and all your work was done..... good system....
I'll give the point.
Whenever you post, *show* as much work as you can,
not claim you did work that you did not show.
Go to and the "Read Before Posting" thread above the problems for your
part of the responsibility.
I was wrong to show as much work to you as I did (without you showing any work),
and I got caught up in the moment. No one else here should have laid out any solutions
to you, either.
Whenever you post, *show* as much work as you can,
not claim you did work that you did not show.
Go to and the "Read Before Posting" thread above the problems for your
part of the responsibility.
I was wrong to show as much work to you as I did (without you showing any work),
and I got caught up in the moment. No one else here should have laid out any solutions
to you, either.
- Joined
- Apr 12, 2005
- Messages
- 11,339
Occasionally, we're relaxed enough to give you a chance to get through your first post.
Please show all work you have done. It is so much easier to provide what is needed, rather than rewriting text books. Please also provide your thoughts about your planned solution and some indication of why you think you are stuck.
Please show all work you have done. It is so much easier to provide what is needed, rather than rewriting text books. Please also provide your thoughts about your planned solution and some indication of why you think you are stuck.