Proving the Centeredness of Fractions using the Min Function

[Miike012]

Question on proof...

Homework Statement

The theorem says ...
If: we are given a1/b1 ; a2/b2 ; an;bn
Then: the sum of the numerator divided by the sum of the denominator lies between the largest and smallest fraction.
( I understand this)
BUT the theorem only proves that the sum is bigger than the smallest or smaller than the largest... it does not prove that it is in the center.
for instance...
a/b is the smallest
a/b = k
a1/b1 > k
an/bn >k

finally... a1 + bn/b1+bn > k ; k = a/b

Homework Equations

Here is my question. first off that proved nothing... or I am obviously oblivious to something that it is trying to show me...?
2nd: is there a way to prove that
IF a/b = k = smallest
A/B > k = largest number
Sum(a)/Sum(b) < k+1
Sum(an)/Sum(bn) < k+1
(The reason I added one to k is to indicate that it is larger than a/b... idk if I would be correct though...?)

So... a/b < (Sum(a) + Sum(an)) / (Sum(b) +Sum(bn) ) < A/B

 

[grief]

Can you clarify what exactly your question is?
Are you trying to prove "min{an/bn}<=sum(an)/sum(bn)<=max{an/bn}"?

If this is what you're trying to prove, where do you get stuck?

 

[Miike012]

I don't know how to prove it.

 

[JonF]

Sorry it was hard to follow what you were saying, are you trying to prove:

Min{a₁/b₁, a₂/b₂,…,a_n/b_n} ≤ (a₁+a₂+…+a_n)/(b₁+b₂+..+b_n) ≤ Max{a₁/b₂, a₂/b₂,…,a_n/b_n}

if so do induction on n and use these facts:

Min{a₁/b₁, a₂/b₂,…,a_n/b_n, a_n+1/b_n+1}≤Min{a₁/b₁, a₂/b₂,…,a_n/b_n}
and
Max{a₁/b₁, a₂/b₂,…,a_n/b_n}≤Max{a₁/b₁, a₂/b₂,…,a_n/b_n,a_n+1/b_n+1}

 

[Miike012]

click on this link... it has the problem..
http://mathhelpp.webs.com

My issue is ... the theorem says that the sum of the numberator divided by the sum of the denominator is between the smallest and the largest ratios.

But when they prove it they only prove that the sum of the num/den is larger than the smallest ratio...

Shouldnt they prove that the sum is between the largest and smallest ratios? After all that is whawt they are claiming.

 

[JonF]

My work blocks most internet links so I can’t see your URL right now. But they should show both from what you say. The author may have just assumed since the two cases are so similar that showing you one was sufficient. If you are familiar with induction this is a fairly straight forward proof.

 

[JonF]

Since apparently you have a copy of the proof online, here is what I was thinking:

For n=1
Min{a₁/b₁} ≤ a₁/b₁ ≤ max{a₁/b₁}
is trivial

assume this relation holds for n, i.e.
Min{a₁/b₁, a₂/b₂,…,a_n/b_n} ≤ (a₁+a₂+…+a_n)/(b₁+b₂+..+b_n) ≤ Max{a₁/b₂, a₂/b₂,…,a_n/b_n}

and Min{a₁/b₁, a₂/b₂,…,a_n/b_n, a_n+1/b_n+1}≤Min{a₁/b₁, a₂/b₂,…,a_n/b_n} comes from the definition of min for any _n+1/b_n+1 term (this is easy to prove, do you see why? Consider the case where _n+1/b_n+1 is the smallest element, and where _n+1/b_n+1 is not the smallest element)

Putting these two together
Min{a₁/b₁, a₂/b₂,…,a_n/b_n, a_n+1/b_n+1}≤Min{a₁/b₁, a₂/b₂,…,a_n/b_n} ≤ (a₁+a₂+…+a_n)/(b₁+b₂+..+b_n)

now consider (a₁+a₂+…+a_n)/(b₁+b₂+..+b_n) ≤ (a₁+a₂+…+a_n)/(b₁+b₂+..+b_n) for the cases where a_n+1/b_n+1 is and isn't the smallest.

which will prove the n+1 case for Min, a similar argument can be made for Max

 

[Miike012]

Jeez, I want to be able to prove stuff like that... I havnt learned any of that yet. I am really new to math, I've only taken a sem of algebra.
What math did you go through?

 

[JonF]

Lol I’ve taken a lot of math courses; I have bachelors and I’ve taken a few graduate courses in math. But, I’m small fries on this forum. There are several PHD mathematicians who post regularly.

Do you know what the Min function does? It returns the smallest element.

So, some examples:
Min{1/2,1/4,3/5} = 1/4
Min{.1,.01,001,.0001} = .0001
Min {1,2,3,4,5,..} = 1

Why don’t you try to prove min{a₁,a₂,a₃,a₄,..,a_n,b} ≤ min{a₁,a₂,a₃,a₄,..,a_n} no matter what a₁,a₂,a₃,a₄,..,a_n, and b are. It relates to your original question. I gave you a hint how on my last post.