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.
a) if there are only two integers, where x = 1/(-100) and z = 2/101,
then y = 3/1 is greater than x and z. I suggest limiting the
proposition to positive denominators, in which case it is truer.
b)
Even if all the denominators are positive, the proposition is false if
x = z, in which case y will equal x and z. I suggest limiting the
proposition to x < z, or proving that x <= y <= z
Edited on October 15, 2004, 6:57 pm
Counterexamples
