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.

Consider the case of two fractions with a/b<c/d. Assuming we're dealing with all positive integers, this means that ad<bc.

ad<bc
ab+ad<ab+bc
a(b+d)<b(a+c)
a/b<(a+c)/(b+d)

ad<bc
ad+cd<bc+cd
d(a+c)<c(b+d)
(a+c)/(b+d)/c/d

So we have a/b<(a+c)/(b+d)<c/d. So the new fraction will always be between the two original ones (given that they're different).

For the case of multiple fractions, write them out in order. a1/b1<...<an/bn.

You can just simply compare them 2 at a time. We have a1/b1<(a1+a2)/(b1+b2). Then you compare the new fraction to a3/b3 so a1/b1<(a1+a2)/(b1+b2)<(a1+a2+a3)/(b1+b2+b3). And so on. Similarly, you'd do the same to show that it's less than an/bn.

Posted by np_rt on 2004-10-15 18:09:50