[x] + [x + 1/2] = [2x]

We need to prove that \([x] + [x + 1/2] = [2x]\). We can represent \(x\) as:

$$ x = [x] + \{x\} $$

where \(\{x\}\) is the fractional part. We will break the proof in two cases. One where \(\{x\} \lt 1/2\) and one where \(\{x\} \ge 1/2\).

If \(\{x\} \lt 1/2\), then the integer part of \(x\) won't increment when adding 1/2:

$$\begin{gather} [x + 1/2] = [x] \\ [x] + [x + 1/2] = 2[x] \end{gather}$$

Is \(2[x] = [2x]\)? Since \(2x = x+x\):

$$\begin{gather} x+x = [x] + \{x\} + [x] + \{x\} \implies \\ 2x = 2[x] + 2\{x\} \implies \\ [2x] = [2[x] + 2\{x\}] \end{gather}$$

Since \(0 \le \{x\} \lt 1/2\), then \(0 \le 2\{x\} \lt 1\), and since \(2[x]\) is an integer:

$$\begin{align} &[2x] = [2[x] + 2\{x\}] \implies \\ &[2x] = 2[x] + [2\{x\}] \\ &[2x] = 2[x] \end{align}$$

This means:

$$\begin{align} &[x] + [x + 1/2] = 2[x] \implies \\ &[x] + [x + 1/2] = [2x] \end{align}$$

Now let's consider the case where \(1/2 \le \{x\} \lt 1\). Since \(x = [x] + \{x\}\):

$$\begin{align} [x + 1/2] &= [[x] + \{x\} + 1/2] \\ &= [x] + [\{x\} + 1/2] \end{align}$$

Adding them gives:

$$ [x] + [x + 1/2]= 2[x] + [\{x\} + 1/2] $$

If \(1/2 \le \{x\} \lt 1\), then \(1 \le \{x\} + 1/2 \lt 1.5\). Let \(s\) be the fractional part of \(\{x\} + 1/2\). Since the integer part is 1:

$$\begin{align} 1 + s &= \{x\} + 1/2 \\ [x] + [x + 1/2] &= 2[x] + [1+s] = [x] + [x + 1/2]= 2[x] + 1 + [s] \\ &= 2[x] + 1 \end{align}$$

Is \([2x] = 2[x]+1\)? If \(1/2 \le \{x\} \lt 1\), then \(1 \le 2\{x\} \lt 2\). Let \(t\) be the fractional part of \(2{x}\). Since \([2x] = [2[x] + 2\{x\}]\):

$$\begin{align} [2x] &= [2[x] + 2\{x\}] = 2[x] + [2\{x\}] \\ &= 2[x] + [1 + t] = 2[x] + 1 + [t] \\ &= 2[x] + 1 \end{align}$$

If \([x] + [x + 1/2] = 2[x] + 1\) and \([2x] = 2[x]+1\):

$$\begin{align} [x] + [x + 1/2] = [2x] \end{align}$$