Question

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Answer 1

$\displaystyle\it\\\left\{x+n\right\}=x,~\forall~n\in\mathbb{Z}.\\ne~propunem~sa~si~demonstram~aceasta~proprietate:\\\left\{x+n\right\}=(x+n)-[x+n]=x+n-[x]-n=x-[x]=\left\{x\right\}.\\---------------------------\\f\bigg(x+\frac{1}{2}\bigg)=\left\{2\bigg(x+\frac{1}{2}\bigg)\right\}=\left\{ 2x+2\frac{1}{2}\right\}=\left\{2x+1\right\}=\\(2x+1)-[2x+1]=2x+1-[2x]-1=2x-[2x]=\boxed{\it \left\{2x\right\}}.$