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The Sum of the Parts Is Less Than And More Than the Parts (Posted on 2004-10-15) Difficulty: 2 of 5
Given a number of different fractions, create a new fraction whose numerator is the sum of all those fractions' numerators and whose denominator is the sum of the denominators. Call it y. Call the smallest of the original fractions x and the largest z.

Prove that for all cases, x < y < z.

No Solution Yet Submitted by Victor Zapana    
Rating: 2.3333 (3 votes)

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Solution Simple proof | Comment 9 of 11 |
Suppose a/b<c/d<e/f<...<s/t<u/v. We want to show that (a+c+e+...+s+u)/(b+d+f+...+t+v)<u/v. This is the same as seeing whether (a+c+e+...+s+u)v<(b+d+f+...+t+v)u.

Since a/b<u/v, av<bu, and similarly cv<du, ev<fu, ... sv<tu. Summing all these unequalities produces the result we sought.

The other proof (a/b<(a+c+e+...+s)/(b+d+f+...+t)<u/v) is similar.
Edited on October 16, 2004, 1:13 pm
  Posted by Federico Kereki on 2004-10-16 13:11:40

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