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An Odd Function (Posted on 2006-12-28) Difficulty: 3 of 5
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 independentGamer2006-12-31 01:49:07
thanx GamerFerdinand2006-12-30 19:49:46
re: explanation of expression (1)Gamer2006-12-30 18:55:12
problem with expression (1)Ferdinand2006-12-30 18:10:07
re(4): A final inputBractals2006-12-30 11:46:27
re(3): A final inputJLo2006-12-30 06:58:23
re(3): A final inputJLo2006-12-30 06:35:23
Four derived restrictionsGamer2006-12-30 01:56:55
re: A final inputGamer2006-12-30 00:01:06
re(2): A final inputBractals2006-12-29 23:52:02
y is independentGamer2006-12-29 23:15:06
Please define yDiane2006-12-29 21:58:21
re(2): A final inputBractals2006-12-29 21:28:17
re: A final inputJLo2006-12-29 17:14:14
A final inputGamer2006-12-29 14:11:48
All nice solutionsJLo2006-12-29 08:44:43
SolutionSimilar SolutionGamer2006-12-29 00:14:29
SolutionUnique solutionJLo2006-12-28 19:26:48
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