Find out the number of integers

Let $⌊x⌋ = n$ and $\{x\}\ = r$ . Then, applying Simon's favorite factoring trick,

$nr - 2n = r - 1$

$n(r-2) - r + 1 + (1) = 0 + (1)$

$n(r-2) - (r-2) = 1$

$(n-1)(r-2) = 1$

Now, considering the range of $r$ , we know that $0 ≤ r < 1$ .

So, $-2 ≤ r-2 < -1$ .

Multiplying the inequality by $n-1$ , with $n≠1$ , we get

$-2(n-1) ≤ (r-2)(n-1) < -(n-1)$

Dealing with both sides of the inequality separately, we have the following for the first part of the inequality:

$-2(n-1) ≤ (r-2)(n-1)$

$-2(n-1) ≤ 1$

$2(n-1) ≥ -1$

$(n-1) ≥ -1/2$ or $n ≥ 1/2$

Now for the second part of the inequality:

$1 < -(n-1)$

$-1 > (n-1)$ or $n<0$ , which is a contradiction to the previous result obtained.

As $n$ cannot be greater than $1/2$ and less than $0$ at the same time, we conclude there are no solutions for $n$ and so there are no solutions for $x$ .

x integer --> {x} = 0 --> [x] = 1/2 impossible. thus no integer solutions.

x non-integer --> [x] = 1 - 1/(2 - {x}) which cannot be an integer since 2 - {x} € (1/2 , 1). thus no non-integer solutions.

No solutions

By simple manipulation we can write fractional part of x in terms of the floor function..........Then, applying the inequality of it being between 0 and 1, we can see that no solution exists....