The integer part of $x+1/2$ expressed in terms of integer parts of $2x$ and $x$

Question

Show that for all $x\in\mathbb{R}$ $$E\left(x+\frac{1}{2}\right)=E(2x)-E(x)$$ where $E(x)$ is the integer part of $x$, that is the number $p\in\mathbb{Z}$ such that $p\le x <p+1$.

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Progress: so far I know that $p+\frac{1}{2}<x+\frac{1}{2}<p+\frac{3}{2}$, and I'm stuck here.

Answer 1

Observe few properties about $E(x)$:

1. $E(x)=E(E(x))$
2. $\forall x\in\Bbb{R}, \exists \text{ }0\le r<1\text{ such that }x=E(x)+r$

So by considering the 2 possible cases:

For $\frac12\le r<1$, we have $1\le \frac12+r<\frac32$, so

\begin{align} E(x+\frac12)&=E(E(x)+r+\frac12)\\&=E(E(x)+1)\\&=E(x)+1 \end{align}

While

\begin{align} E(2x)-E(x)&=E(2E(x)+2r)-E(E(x)+r)\\&=E(2E(x)+1)-E(E(x))\\&=2E(x)+1-E(x)\\&=E(x)+1\\&=E(x+\frac12) \end{align}

Similar for $0\le r<\frac12$.