The function f(x) = x - [x] has period of 1

Proof:

Let T be a positive real number. Again let f(x) be periodic with period T.

Then, f(x + T) = f(x) ∀ x ∈ R

=> x + T - [x + T] = x - [x] ∀ x ∈ R

=> T = x - [x] - x + [x + T] ∀ x ∈ R

=> [x + T] - [x] = T ∀ x ∈ R

=> T = 1, 2, 3, 4,............

Thus there exist T > 0 such that f(x + T) = f(x) ∀ x ∈ R.

So, f(x) is a periodic function.

Now, the smallest value of T satisfying f(x + T) = f(x) ∀ x ∈ R is 1.

Hence, the function f(x) = x - [x] has period of 1