Prove the following identity holds for all real numbers $x$ Prove the following identity holds for all real numbers $x$:
$$\lfloor4x\rfloor=\lfloor x \rfloor +\left\lfloor x+\frac14 \right\rfloor+\left\lfloor x+\frac24\right\rfloor+\left\lfloor x+\frac34\right\rfloor$$
I understand that $\lfloor4.3\rfloor$ would be $4$ and $\lfloor-2.4\rfloor$ would be $-3$
I am trying to prove this by cases. I think that I should prove each case first such as $\lfloor x\rfloor$ first and then $\left\lfloor x+\frac14\right\rfloor$ and on but I'm having trouble proving it in generality
 A: Hint: Separate all real numbers into a few groups, depending on the relation of their fractional part to the intervals formed by $0$, $\frac14$ , $\frac12$ , $\frac34$ and $1$.
A: Let $\{x\}=x-[x]$, (i.e., the fractional part of $x$).
First Case: if $\{x\}\in[0,0.25)$ then $$4\{x\}<1\Rightarrow [4x]=4[x],$$ and also $$[x]=[x+1/4]=[x+2/4]=[x+3/4],$$ so for this case we have $$[4x]=4[x]=[x]+[x+1/4]+[x+2/4]+[x+3/4].$$
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Second Case: if $\{x\}\in[0.25,0.5)$ then $$1\le4\{x\}<2\Rightarrow [4x]=4[x]+1,$$ and also $$[x]=[x+1/4]=[x+2/4],$$ but now $$[x+3/4]=[x]+1.$$ So for this case we have $$[4x]=4[x]+1=[x]+[x+1/4]+[x+2/4]+[x+3/4].$$
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I think you see the pattern and can complete the last two cases.
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EDIT: I added lots of spacing and made the equations display style for readability.
A: There is no need to consider different cases.
Observe that any real number $x$ admits the representation $\,x=n+k\,\frac14+r$, where $k$ is an integer within $[0,4)$ and $r$ is a real within $[0,\frac14)$.
Then prove that each of l.h.s. and r.h.s. of your identity equals $4n+k$.
A: Define
$$
f(x)=\lfloor x\rfloor+\left\lfloor x+\tfrac14\right\rfloor+\left\lfloor x+\tfrac12\right\rfloor+\left\lfloor x+\tfrac34\right\rfloor-\lfloor4x\rfloor\tag1
$$
Note that $f$ is periodic with period $\frac14$:
$$
\begin{align}
f\!\left(x+\tfrac14\right)
&=\left\lfloor x+\tfrac14\right\rfloor+\left\lfloor x+\tfrac12\right\rfloor+\left\lfloor x+\tfrac34\right\rfloor+\color{#C00}{\lfloor x+1\rfloor}\color{#090}{-\lfloor4x+1\rfloor}\\
&=\color{#C00}{\lfloor x\rfloor+1}+\left\lfloor x+\tfrac14\right\rfloor+\left\lfloor x+\tfrac12\right\rfloor+\left\lfloor x+\tfrac34\right\rfloor\color{#090}{-\lfloor4x\rfloor-1}\\
&=f(x)\tag2
\end{align}
$$
Furthermore, for $x\in\left[0,\frac14\right)$, each term in $(1)$ is $0$. Thus, $f(x)=0$ for x in a complete period.
So, we have that for all $x\in\mathbb{R}$, $f(x)=0$. That is,
$$
\lfloor x\rfloor+\left\lfloor x+\tfrac14\right\rfloor+\left\lfloor x+\tfrac12\right\rfloor+\left\lfloor x+\tfrac34\right\rfloor=\lfloor4x\rfloor\tag3
$$
