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.

First a two fraction case a1/b1 < a2/b2

(note by cross-multiplying: a1b2 < a2b1)

The new fraction is (a1+a2)/(b1+b2)

We need to show a1(b1+b2) < (a1+a2)b1

and (a1+a2)b2 < a2(b1+b2)

in the first, distribute to get a1b1 + a1b2 < a1b1 + a2b1

subtract a1b1 from both sides to get the inequality in the note above.

similar reasoning can show the second case

Thats a QED to me.

 

For a multi-fraction case

a1/b1 < a2/b2 < a3/b3 < ... < an/bn

(note by cross multiplying: many inequalities emerge)

The new fraction is (a1+a2+a3+...+an)/(b1+b2+b3+...+bn)

Need to show this is between a1/b2 and an/bn

Cross multiply with the first and we need to show

a1(b1+b2+b3+...+bn) < (a1+a2+a3+...+an)b1

Distribute to get

a1b1+a1b2+a1b3+...+a1bn < a1b1+a2b1+a3b1+...+anb1

It can be shown that a1b2<a2b1, a1b3<a3b1,..., a1bn<anb1 by cross-multiplying.

This shows the fraction is more than the smallest.

Similar reasoning can show the fraction is less than the largest.

-Jer

Posted by Jer on 2004-10-15 17:08:32