Proving $\lfloor 2x \rfloor = \lfloor x \rfloor + \lfloor x+0.5\rfloor$ I can intuitively see that this is true, but I'm having a very hard time proving it. I'm actually not even quite sure where to begin. I tried using the inequalities that define the floor function, and I'm pretty sure this is the way to go, but I'm failing to make any kind of progress.
I appreciate any help.
 A: Hint: split it in to cases $x\in[n,n+\frac{1}{2})$ and $x\in[n+\frac{1}{2},n+1)$ for some integer $n$.
A: Case 1: The fractional part of $x$ is less than $1/2$, i.e. $x=n+a$ where $n$ is an integer and  $0\le a<1/2$.  Then $\lfloor 2x\rfloor = \lfloor 2n+2a \rfloor = 2n$.  But also $\lfloor x\rfloor=n$ and $\lfloor x +0.5 \rfloor = n$.  Since $2n=n+n$, the equality holds.
Case 2: The fractional part of $x$ is more than $1/2$, i.e. $x=n+a$ where $n$ is an integer and $1/2\le a<1$. I leave the remainder as an exercise; let me know if you need more help.
A: You can "draw" an integer in and out of the floor function ($\lfloor x+n\rfloor=\lfloor x\rfloor+n$). Then it suffices to establish the property for $x\in[0,\frac12)$ and $x\in[\frac12,1)$, because the three floors remain constant in these intervals.


*

*$x\in[0,\frac12)\implies0=0+0$,

*$x\in[\frac12,1)\implies1=0+1$.
Simple as that.
A: Here is an alternative approach, which does not require a case split, and instead uses a form of induction.  I found this approach (through Wikipedia) in Matsuoka, Yoshio (1964), "Classroom Notes: On a Proof of Hermite's Identity", The American Mathematical Monthly 71 (10): 1115 (doi: 10.2307/2311413).$
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\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\floor}[1]{\lfloor#1\rfloor}
\newcommand{\then}{\Rightarrow}
\newcommand{\when}{\Leftarrow}
\newcommand{\true}{\text{true}}
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$
We are asked to prove $$
\tag{0}
\floor{2x} = \floor{x} + \floor{x + \tfrac 1 2}
$$ for all real $\;x\;$.  Looking at the graph of either side, we see that there is a kind of 'period' of length $\;\tfrac 1 2\;$ involved, specifically the intervals $[0,\tfrac 1 2)$, $[\tfrac 1 2, 1)$, etc.  Abbreviating $\ref 0$ by $\;P(x)\;$, this suggests we investigate $\;P(x+\tfrac 1 2)\;$:
$$\calc
    P(x + \tfrac 1 2)
\op=\hint{expand abbreviation $\;P\;$; simplify}
    \floor{2x + 1} = \floor{x + \tfrac 1 2} + \floor{x + 1}
\op=\hint{move integer $1$ out of floor, twice; subtract $1$ from both sides}
    \floor{2x} = \floor{x + \tfrac 1 2} + \floor{x}
\op=\hint{reorder RHS; abbreviation $\;P\;$}
    P(x)
\endcalc$$
So we've proven $$
\tag 1
\langle \forall x :: P(x) \;\equiv\; P(x + \tfrac 1 2) \rangle
$$ which is a kind of 'induction step'. If we can additionally prove the 'base case' $$
\tag 2
\langle \forall x : 0 \le x \lt \tfrac 1 2 : P(x) \rangle
$$ then by induction our goal $\;\langle \forall x :: P(x) \rangle\;$ follows.  Fortunately $\ref 2$ is easy to prove: assuming $\;0 \le x \lt \tfrac 1 2\;$, we have
$$\calc
    P(x)
\op=\hint{abbreviation $\;P\;$}
    \floor{2x} = \floor{x} + \floor{x + \tfrac 1 2}
\op=\hints{by the assumption, all three parts are in the [0,1) interval:}
        \hints{LHS: $\;0 \le 2x \lt 1\;$, first part of RHS: $\;0 \le x \lt \tfrac 1 2 \lt 1\;$,}
        \hints{second part of RHS: $\;0 \lt \tfrac 1 2 \le x + \tfrac 1 2 \lt 1\;$;}
        \hint{definition of floor}
    0 = 0 + 0
\op=\hint{arithmetic}
    \true
\endcalc$$
This completes the proof.
