You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**Jake****Guest**

a function f: R -> R has the property that for any four real numbers a, b, c, d such that a - b > c - d, we have f(a) - f(b) > f(c) - f(d). prove that f is a linear function, ie f(x) = mx + n for all x belonging to R, where m, n belong to R and m > 0

does anyone know the name of this theorom or have any ideas how to prove it?!

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Are you supposed to prove that f *can* be a linear function, or f *has to* be a linear function?

If it's the first:

We know a - b > c - d. We also know that f(x) = mx + n. So let's apply m to both sides of our original equation. m(a-b) > m(c-d). Note that m is positive, and thus, this is legal. So ma - mb > mc - md. Now let's add and subtract n. ma - mb + n - n > mc - md + n - n. After a bit of rearrangement, we get ma + n - mb - n > mc + n - md - n. Grouping that negative, we get ma + n - (mb + n) > mc + n - (md + n). Now we can write this as f(a) - f(b) > f(c) - f(d). So the linear function holds.

Edit:

To do this problem, you must work it backwards first.

Start out assuming f(a) - f(b) > f(c) - f(d). Then you can replace that with ma + n - (mb + n) > mc + n - (md + n). After this, you can get ride of the n's, divide the whole thing through by m, and you should end up with a - b > c - d. Then when writing the proof, you start at the bottom (a - b > c - d) and work your way back up to end at f(a) - f(b) > f(c) - f(d).

*Last edited by Ricky (2006-03-14 03:28:26)*

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

Pages: **1**