1. State the problem: Prove that $$\left\lfloor\frac{x+2}{2}\right\rfloor-2\left\lfloor\frac{x+2}{4}\right\rfloor = 1-\left(\left\lfloor\frac{x}{2}\right\rfloor-2\left\lfloor\frac{x}{4}\right\rfloor\right)$$. 2. Strategy: Write $x=4k+r$ with $k\in\mathbb{Z}$ and $r\in\{0,1,2,3\}$, reduce both sides to expressions in $k$ and $r$, and then check the four residues. 3. Compute the basic floor values using $x=4k+r$. $$\left\lfloor\frac{x}{4}\right\rfloor = k$$ $$\left\lfloor\frac{x}{2}\right\rfloor = 2k + \left\lfloor\frac{r}{2}\right\rfloor$$ $$\left\lfloor\frac{x+2}{4}\right\rfloor = k + \left\lfloor\frac{r+2}{4}\right\rfloor$$ $$\left\lfloor\frac{x+2}{2}\right\rfloor = 2k + \left\lfloor\frac{r+2}{2}\right\rfloor$$ 4. Evaluate the left-hand side by substitution. $$\left\lfloor\frac{x+2}{2}\right\rfloor-2\left\lfloor\frac{x+2}{4}\right\rfloor = (2k + \left\lfloor\frac{r+2}{2}\right\rfloor) -2\left(k + \left\lfloor\frac{r+2}{4}\right\rfloor\right)$$ $$= \cancel{2k} + \left\lfloor\frac{r+2}{2}\right\rfloor -\cancel{2k} -2\left\lfloor\frac{r+2}{4}\right\rfloor$$ $$= \left\lfloor\frac{r+2}{2}\right\rfloor -2\left\lfloor\frac{r+2}{4}\right\rfloor$$ 5. Evaluate the right-hand side by substitution. $$1-\left(\left\lfloor\frac{x}{2}\right\rfloor-2\left\lfloor\frac{x}{4}\right\rfloor\right)=1-\left(2k + \left\lfloor\frac{r}{2}\right\rfloor -2k\right)$$ $$=1-\left(\cancel{2k} + \left\lfloor\frac{r}{2}\right\rfloor -\cancel{2k}\right)$$ $$=1-\left\lfloor\frac{r}{2}\right\rfloor$$ 6. Reduce to the four cases $r\in\{0,1,2,3\}$ and verify equality. r=0: $$\left\lfloor\frac{r+2}{2}\right\rfloor -2\left\lfloor\frac{r+2}{4}\right\rfloor=\left\lfloor\frac{2}{2}\right\rfloor -2\left\lfloor\frac{2}{4}\right\rfloor=1-0=1$$ $$1-\left\lfloor\frac{r}{2}\right\rfloor=1-\left\lfloor 0\right\rfloor=1$$ r=1: $$\left\lfloor\frac{3}{2}\right\rfloor -2\left\lfloor\frac{3}{4}\right\rfloor=1-0=1$$ $$1-\left\lfloor\frac{1}{2}\right\rfloor=1-0=1$$ r=2: $$\left\lfloor\frac{4}{2}\right\rfloor -2\left\lfloor\frac{4}{4}\right\rfloor=2-2\cdot 1=0$$ $$1-\left\lfloor\frac{2}{2}\right\rfloor=1-1=0$$ r=3: $$\left\lfloor\frac{5}{2}\right\rfloor -2\left\lfloor\frac{5}{4}\right\rfloor=2-2\cdot 1=0$$ $$1-\left\lfloor\frac{3}{2}\right\rfloor=1-1=0$$ 7. Conclusion: For each residue $r$ the reduced left-hand side equals the reduced right-hand side, so the identity holds for all real $x$.