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Can the sum of two rational numbers equal the ratio of the sums of numerators to denominators?
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=201007...
(Choice 3)
2 Answers
- WillLv 51 decade agoFavorite 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).
- JimLv 61 decade ago
(a/b) + (c/d) = (a+c)/(b+d)
If a = -c and b = d, then you can have it be true.
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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.
