# Proof regarding fractions

by captainquarks
Tags: fractions, proof
 P: 10 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.
P: 606
 Quote by captainquarks 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.

Shouldn't the condition "positive integers" be added? For example, we have $$\frac{1}{-4}<\frac{3}{2}\,\,\,but\,\,\, \frac{3+1}{-4+2}=-2<\frac{1}{-4}...$$.
Now, if all the numbers are natural ones (no zero included), then we have $$\frac{x}{y}<\frac{a}{b}\Longrightarrow \frac{x+a}{y+b}<\frac{a}{b}\Longleftrightarrow bx+ab<ay+ab\Longleftrightarrow bx<ay$$ and this last inequality is the very same we started with.

Something similar can be done to show the other side's inequality

DonAntonio
 Sci Advisor P: 6,112 bx < ay implies x/y < a/b (assuming all integers involved are positive). So start with assuming x/y < a/b and work your derivations in reverse order.
 P: 10 Proof regarding fractions Thanks DonAntonio and mathman for your replies and help. I think I get where you're coming from, but I'm not the best Mathematician in the world (Below 1st year university level Id say)... A little more explanation/algebra would be great... I was thinking, that, if I got my inequalities as i did, then if i work back, i obviously get the same inequality, does the answer lie in the crux on this? I just can't seem to express myself mathematically as well as others
 P: 10 Does anyone else have any ideas? I just cant seem to get my head around it =(
 Emeritus Sci Advisor PF Gold P: 4,500 You wrote x/y < (a + x)/(b + y) x(b + y) < y(a + x) bx + xy < ay + yx bx < ay Now the key to these calculations is that they are all reversible - i.e. not only can you go from bx + xy < ay + yx to bx < ay, but from bx < ay you can get bx + xy < ay + yx Now we have a,b,x,y with x/y < a/b. So from here we can derive that bx
P: 606
 Quote by captainquarks Does anyone else have any ideas? I just cant seem to get my head around it =(

More ideas?! You've already been shown by at least two different people the very solution to your problem. If you still

don't get it then either you don't know the basic properties and operations with fractions yet, and then you need to