Exercise 4.1.16 (Prove $\lfloor 2\alpha \rfloor + \lfloor 2\beta \rfloor \geq \lfloor \alpha \rfloor+ \lfloor \beta \rfloor + \lfloor \alpha + \beta \rfloor$ )

Proof.

(a)

We write $$\begin{array}{cc}\alpha =n+\nu ,\hfill & 0\le \nu <1\hfill \\ \beta =m+\mu ,\hfill & 0\le \mu <1,\hfill \end{array}$$

so that $n=\lfloor \alpha \rfloor ,m=\lfloor \beta \rfloor$.

• If $\nu +\mu <1$, then $\alpha +\beta =n+m+\nu +\mu$ so

  $$n+m\le \alpha +\beta <n+m+1.$$

  This gives $\lfloor \alpha +\beta \rfloor =n+m$, and by Theorem 4.1(4), $2\lfloor \alpha \rfloor \le \lfloor 2\alpha \rfloor$, thus

  $$\lfloor \alpha \rfloor +\lfloor \beta \rfloor +\lfloor \alpha +\beta \rfloor =2\lfloor \alpha \rfloor +2\lfloor \beta \rfloor \le \lfloor 2\alpha \rfloor +\lfloor 2\beta \rfloor .$$

• If $\nu +\mu \ge 1$, then

  $$n+m+1\le \alpha +\beta =n+m+\nu +\mu <n+m+2,$$

  so $\lfloor \alpha +\beta \rfloor =n+m+1$.

  Moreover $2\alpha =2n+2\nu ,2\beta =2m+2\mu$.

  Note that $2\lfloor \alpha \rfloor \le \lfloor 2\alpha \rfloor$ is always true, but here we can say a little more. Since $\nu +\mu \ge 1$, then $\nu \ge \frac{1}{2}$ or $\mu \ge \frac{1}{2}$ (perhaps both are true), thus

  $$2\alpha \ge 2n+1\text{ or }2\beta \ge 2n+1,$$

  therefore $\lfloor 2\alpha \rfloor \ge 2n+1$ or $\lfloor 2\beta \rfloor \ge 2n+1$ (and $\lfloor 2\alpha \rfloor \ge 2m,\lfloor 2\beta \rfloor \ge 2m$), so that in any case

  $$\lfloor 2\alpha \rfloor +\lfloor 2\beta \rfloor \ge 2n+2m+1.$$

  Then

  $$\lfloor \alpha \rfloor +\lfloor \beta \rfloor +\lfloor \alpha +\beta \rfloor =2n+2m+1\le \lfloor 2\alpha \rfloor +\lfloor 2\beta \rfloor .$$

In both cases, we obtain

$$\lfloor \alpha \rfloor +\lfloor \beta \rfloor +\lfloor \alpha +\beta \rfloor \le \lfloor 2\alpha \rfloor +\lfloor 2\beta \rfloor .$$

(b)

With the notations of part (a), $$\begin{array}{llll}\hfill \lfloor \alpha \rfloor +\lfloor \beta \rfloor +\lfloor 2\alpha +2\beta \rfloor & =3n+3m+\lfloor 2\nu +2\mu \rfloor ,& \hfill & \\ \hfill \lfloor 3\alpha \rfloor +\lfloor 3\beta \rfloor & =3n+3m+\lfloor 3\nu \rfloor +\lfloor 3\mu \rfloor .& \hfill & \end{array}$$

To build a counterexample, if suffices to find $\mu$ and $\nu$ such that

$$\lfloor 3\nu \rfloor +\lfloor 3\mu \rfloor <\lfloor 2\nu +2\mu \rfloor$$

Since $\mu =\nu =0.3$ satisfy these condition, an explicit counterexample is

$$\alpha =1.3,\beta =2.3.$$

Check:

$$\begin{array}{llll}\hfill \lfloor \alpha \rfloor +\lfloor \beta \rfloor +\lfloor 2\alpha +2\beta \rfloor & =\lfloor 1.3\rfloor +\lfloor 2.3\rfloor +\lfloor 7.2\rfloor & \hfill & \\ \hfill & =10& \hfill & \\ \hfill & >\lfloor 3\alpha \rfloor +\lfloor 3\beta \rfloor =\lfloor 3.9\rfloor +\lfloor 6.9\rfloor =9.& \hfill & \end{array}$$