- #1
Miike012
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Question on proof...
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
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
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