Method 1 : Draw graphs of |x – 1| and {x} and note values of x where intersection of two graphs occur.
Method 2: By algebra.
|x – 1| = {x} …...............(1)
Case 1 : Let x

1 & x

Z

{x} = 0
We can write (1) as x – 1 = 0 or x = 1
Taking intersection, we get x = 1 .…..........(2)
Case 2 : Let x > 1 & x

Z
(1) becomes x – 1 = {x} or [x] + {x} – 1 = {x}
i.e., [x] = 1

1

x
Taking intersection, we get 1
So (1) is 1 – x = 0 i.e. x = 1
Taking intersection, we get
x
…............(4)Case 4 : 0

Z
So (1) can be written as 1 – x = {x} or 1 – [x] – {x} = {x}
{x} = (1 – [x] ) / 2 We know that for 0

x

1, [x] = 0

{x} = (1 – 0) / 2 = ½ or
x = 1 / 2 …............(5)Case 5 : –1

x

1 – x = [x} or [x] = 1 – 2{x}
For –1

x
So, 1 – 2{x} = –1 or {x} = 1, which is not true.
x
…...........(6) Similarly, for negative x there is no solution of (1).
Taking union of (2), (3), (4), (5) & (6), we get
x
[1,2)
1/2 ,
which are the solutions of |x – 1| = {x}.
Hence, there are infinitely many solutions of |x – 1| = {x} .