- #1
captainquarks
- 10
- 0
Not sure if this is the correct place to put this question, but here it goes (sorry about the vague title, but not sure how to describe it):
We are asked to consider two rational fractoins, for example:
a/b & x/y
We are now asked to do the following:
(a + x)/(b + y)
Now, we are told that this new, rational expression lies in the range between our initial rational fractionss..
We are asked to prove that this is generally true, with our constants as integers of course:
e.g. 7/3, 9/2 => (7+9)/(3+2) = 16/5, which is between the originals.
I started by doing basic inequalities
x/y < (a + x)/(b + y) < a/b
I split these up:
x/y < (a + x)/(b + y)
x(b + y) < y(a + x)
bx + xy < ay + yx
bx < ay
#
(a + x)/(b + y) < a/b
b(a + x) < a(b + y)
ba + bx < ab + ay
bx < ay
I have no idea (or havn't found that spark yet) as to how, or where to solve this problem, which is very frustrating indeed. Does anyone have any insight into what I'm doing? Input would be greatly appreciated.
We are asked to consider two rational fractoins, for example:
a/b & x/y
We are now asked to do the following:
(a + x)/(b + y)
Now, we are told that this new, rational expression lies in the range between our initial rational fractionss..
We are asked to prove that this is generally true, with our constants as integers of course:
e.g. 7/3, 9/2 => (7+9)/(3+2) = 16/5, which is between the originals.
I started by doing basic inequalities
x/y < (a + x)/(b + y) < a/b
I split these up:
x/y < (a + x)/(b + y)
x(b + y) < y(a + x)
bx + xy < ay + yx
bx < ay
#
(a + x)/(b + y) < a/b
b(a + x) < a(b + y)
ba + bx < ab + ay
bx < ay
I have no idea (or havn't found that spark yet) as to how, or where to solve this problem, which is very frustrating indeed. Does anyone have any insight into what I'm doing? Input would be greatly appreciated.