An equation with the floor function

Notation. $x=n+r$ with $n=\lfloor x\rfloor$, $0\le r<1$. The equation is
$$n+(n+r)=5\iff2n+r=5.$$

Consequences. Since $0\le r<1$, we have $5-1<2n\le5$, i.e. $2< n\le2.5$. Since $n\in\mathbb{Z}$, $n=2$.

Then. $r=5-2n=5-4=1$. But $r<1$, contradiction. So no solution.

Conclusion. No real solution.