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?!
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.
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)
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