Can the sum of two rational numbers equal the ratio of the sums of numerators to denominators?

Question

Normally this is not true:  7/8 + 2/5 is NOT equal 9/13.  The same sum can be written 35/40 + 16/40, which is NOT equal 51/80 (which is also not equal 9/13).  

Can it every be true?  Examples?  Or better, given two rationals, is there a procedure to write them in terms of (integral) numerators and denominators so that the sum of the rationals equals the ratio of the sum  of the numerators to the sum of the denominators.

This is a followup to:
http://answers.yahoo.com/question/index?qid=20100721200500AAfkma1
(Choice 3)

Will · Accepted Answer

Just do the algebra.

a/b + c/d = (a+c)/(b+d)

multiply

ad(b+d)+cb(b+d)=abd+cbd
adb+ad^2+cb^2+cbd = abd+cbd
So
ad^2 = -cb^2
Whenever this condition holds, we win. A solution would just be a=4,d=1,c=-1,b=2
4/2+(-1)/1=1=(3)/3

There are of course infinitely many solutions (pick any nonzero a,b,d and solve for c and you get one).

? · Answer

(a/b) + (c/d) = (a+c)/(b+d)

If a = -c and b = d, then you can have it be true.
------------------------------------------------------------------------
More general:

a(b+d)/b + c(b+d)/d = a+c

= a + ad/b + cb/d + c = a + c

ad/b + cb/d = 0

ad/b = -cb/d, multiply across by (bd)

ad^2 = -cb^2

Values satisfying his would be:

(a, b, c, d) = (2, 3, 4, √18i)

2/3 + 4/√(18i) - 6/(3+(√18i)). You'll have to check this.

(a, b, c, d) = (2, 3, -8, 6)

2/3 + (-8/6) = -6/9 = -(2/3). Verified.

Can the sum of two rational numbers equal the ratio of the sums of numerators to denominators?

Trending Questions