 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  An Odd Function (Posted on 2006-12-28) Let f:R→R satisfy
1. f(a)≠0 for some a in R
2. f(xf(y))=yf(x) for all x,y in R
Prove that f(-x)=-f(x) for all x in R.

 Submitted by Bractals Rating: 3.8333 (6 votes) Solution: (Hide) ``` f(a) ≠ 0 for some a in R f(xf(y)) = yf(x) for all x,y in R First we find the value of f(0) f(0) = f(0f(0)) = 0f(0) = 0 (1) Then a property of f f(w) = f(z) ==> wf(a) = f(af(w)) = f(af(z)) = zf(a) ==> w = z Therefore, f is one-to-one (or injective). (2) Using (2) we have z ≠ 0 ==> f(zf(xy)) = xyf(z) = xf(zf(y)) = f(zf(y)f(x)) ==> zf(xy) = zf(x)f(y) ==> f(xy) = f(x)f(y) (3) Using (1), (2), and (3) we have x = y = 1 ==> f(1)^2 = f(1)f(1) = f(1*1) = f(1) ==> f(1) = 0 or 1 ==> f(1) = 1 (4) Using (2), (3), and (4) we have x = y = -1 ==> f(-1)^2 = f(-1)f(-1) = f((-1)*(-1)) = f(1) = 1 ==> f(-1) = 1 or -1 ==> f(-1) = -1 (5) Finally, using (3) and (5) we have f(-x) = f(-1x) = f(-1)f(x) = -1f(x) = -f(x) ``` Comments: ( You must be logged in to post comments.)
 Subject Author Date x and y are independent Gamer 2006-12-31 01:49:07 thanx Gamer Ferdinand 2006-12-30 19:49:46 re: explanation of expression (1) Gamer 2006-12-30 18:55:12 problem with expression (1) Ferdinand 2006-12-30 18:10:07 re(4): A final input Bractals 2006-12-30 11:46:27 re(3): A final input JLo 2006-12-30 06:58:23 re(3): A final input JLo 2006-12-30 06:35:23 Four derived restrictions Gamer 2006-12-30 01:56:55 re: A final input Gamer 2006-12-30 00:01:06 re(2): A final input Bractals 2006-12-29 23:52:02 y is independent Gamer 2006-12-29 23:15:06 Please define y Diane 2006-12-29 21:58:21 re(2): A final input Bractals 2006-12-29 21:28:17 re: A final input JLo 2006-12-29 17:14:14 A final input Gamer 2006-12-29 14:11:48 All nice solutions JLo 2006-12-29 08:44:43 Similar Solution Gamer 2006-12-29 00:14:29 Unique solution JLo 2006-12-28 19:26:48 Please log in:
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