# Thread: proving rational number inequalities

1. ## proving rational number inequalities

Suppose a,b,x, and y are positive real numbers with a/b < x/y.
Explain why a/b < (a+x)/(b+y) < x/y.
I guess I know the fact is true however how can I prove this?

2. Originally Posted by mathsohard
Suppose a,b,x, and y are positive real numbers with a/b < x/y.
Explain why a/b < (a+x)/(b+y) < x/y.
I guess I know the fact is true however how can I prove this?
From the given if follows at once that

3. Originally Posted by mathsohard
Suppose a,b,x, and y are positive real numbers with a/b < x/y.
Explain why a/b < (a+x)/(b+y) < x/y.
I guess I know the fact is true however how can I prove this?
a/b < x/y
so ay < bx (since y and b are positive)
so ab+ay < ab + bx
so a(b+y) < b(a+x)
so a/b < (a+x)/(b+y)

EDIT: sorry didn't see Plato's post.

4. I would have been inclined to do an indirect proof:
Suppose a/b < (a+x)/(b+y) < x/y is NOT true. Then either a/b>= (a+ x)/(b+y) or (a+ x)/(b+y)>= x/y.

If a/b>= (a+ x)/(b+ y) then a(b+y)>= b(a+ x) which is the same as ab+ ay>= ab+ bx and so ay>= bx which (assuming b and y are non-zero which is implied by the hypothesis) a/b>= x/y, contradicting the hypothesis.

Similarly, if (a+ x)/(b+ y)>= x/y, then y(a+ x)>= x(b+y) which is the same as ay+ xy>= bx+ xy and so ay>= bx again which gives the same contradiction.