Proving |x-y| < b-a: Can It Be Done?

[silvermane]

1. The problem statement:
Suppose that a, b, x, and y are real numbers satisfying
a < x < b and a < y < b.
Show that |x - y| < b - a.

The Attempt at a Solution

We may start with the fact that:
i.) a < x < b
ii.) a < y < b

Subtracting a from all sides, we have that

i.) 0 = a - a < x - a < b - a
ii.) 0 = a - a < y - a < b - a

and

0 < x - a < b - a
0 < y - a < b - a

but I'm stuck here in justifying my steps. I thought about subtracting the inequalities from each other but that would just yield

0 < x - y < 0

which doesn't help my case :(
Any help and/or tips would be lovely!

 

[eibon]

(1) a < x < b
(2) a < y < b

well take -(2) which is
-b < -y < -a call that (2*)

now take (1)+(2*) which gives

a-b < x-y < b-a

 

[silvermane]

Isn't it true that when we multiply by -1, we reverse the equality?

so -(2) would be -b > -y > -a?

 

[silvermane]

Just out of curiosity, what if we wanted to prove
|x-y| < b - a

I'm thinking we break it down into two cases, one where (x-y) is positive, and one where it's negative. What do you think?

 

[eibon]

but i did reverse the equality then i mult. by-1

 

[╔(σ_σ)╝]

(1) a < x < b
(2) a < y < b

well take -(2) which is
-b < -y < -a call that (2*)

now take (1)+(2*) which gives

a-b < x-y < b-a

This is a correct solution. What more do you want ?;P

 

[silvermane]

lol I don't want anymore, I was just checking. This is just for my understanding. I guess one could call it a blonde moment ;)

I just asked about |x-y| to help deepen my understanding. I seemed to have come off a little sour and I'm sorry for that!

Either way, thank you greatly for your help. I really do appreciate it

 

[╔(σ_σ)╝]

i

lol I don't want anymore, I was just checking. This is just for my understanding. I guess one could call it a blonde moment ;)

I just asked about |x-y| to help deepen miy understanding. I seemed to have come off a little sour and I'm sorry for that!

Either way, thank you greatly for your help. I really do appreciate it

You did not come off a little sour; it was I who came off sour. Sorry about that.

You want to do the following, right ?


Case 1:

x-y < 0

a < b By definition so

This implies

b-a > 0
x-y < b-a

Thus

a-b< x-y < b-a [ Some steps are needed to get here and i didn't want to do it twice; so read case 2 . This inequality will follow from there.]

Case 2

x-y >0
x>y

y-x <0

y-x < b-a

x-y> a-b

definiton and assumption of case 2
b>x>y >a

b-y > x-y

y>a

-y < -a

b-y < b-a

b-a > x-y
( the. last of case 1 can be derive from the above 5 steps)

Thus

a-b < x-y < b-a

So case 1 and case 2 is equivalent to saying .

|x-y| < b-a



The first solution is a lot easier than the cases.

Because you have to prove an upper and lower bound of x-y when x-y < 0 and x-y >0. That is like proving 4 things instead of 1.