fractional part function

If x >= 1, let n-1 <= x < n for some integer n >= 2.

{x} + {1/x} = 1
x - (n - 1) + 1/x = 1
x - n + 1/x = 0
x^2 - nx + 1 = 0
x = [n +/- √(n^2 - 4)] / 2


If x <= -1, let -n <= x < -n + 1 for some integer n >= 2.

{x} + {1/x} = 1
x - (-n) + 1/x + 1 = 1
x + n + 1/x = 0
x^2 + nx + 1 = 0
x = [-n +/- √(n^2 - 4)] / 2


For 0 < x < 1 and -1 < x < 0, we have similar results(you can try).


Solution for {x} + {1/x} = 1:
x = [+/-n +/- √(n^2 - 4)] / 2 for ineger n >= 2

(note: when n = 2, x = 1 or -1)

2009-04-01 16:23:11 補充：
'ineger' in last two line should be integer

2009-04-01 19:41:40 補充：
CORRECTION (x <> +/- 1)

When n = 2, x = 1 or -1 (rejected)

Solution for {x} + {1/x} = 1:
x = [+/-n +/- √(n^2 - 4)] / 2 for integer n >= 3

Can I answer this in English? The answer I think is

((n+1) + ((n+1)^2 - 4) ^ (1/2)) / 2
((n+1) - ((n+1)^2 - 4) ^ (1/2)) / 2
(-(n+1) + ((n+1)^2 - 4) ^ (1/2)) / 2
(-(n+1) - ((n+1)^2 - 4) ^ (1/2)) / 2
for n = 2, 3, 4, ....