Exercise 4.1.27* (Generalization of Problem 16)

Proof. (We don’t use the hint in this solution.)

$\text{(⇐)}$

By Problem 16, we know that for all $\alpha \in ℝ,\beta \in ℝ$, $$\lfloor 2\alpha \rfloor +\lfloor 2\beta \rfloor \ge \lfloor \alpha \rfloor +\lfloor \beta \rfloor +\lfloor \alpha +\beta \rfloor .$$

If we replace $\alpha$ and $\beta$ by $\mathit{j\alpha }$ and $\mathit{j\beta }$, where $j$ is a positive integer, we obtain

$$\lfloor 2\mathit{j\alpha }\rfloor +\lfloor 2\mathit{j\beta }\rfloor \ge \lfloor \mathit{j\alpha }\rfloor +\lfloor \mathit{j\beta }\rfloor +\lfloor \mathit{j\alpha }+\mathit{j\beta }\rfloor .$$

Therefore, if $j=k$, then

$$\lfloor (j+k)\alpha \rfloor +\lfloor (j+k)\beta \rfloor \ge \lfloor \mathit{j\alpha }\rfloor +\lfloor \mathit{j\beta }\rfloor +\lfloor \mathit{k\alpha }+\mathit{k\beta }\rfloor .$$

$\text{(⇒)}$

Suppose now that $j\ne k$. We want to prove that there exist $\alpha$ and $\beta$ such that $$\lfloor (j+k)\alpha \rfloor +\lfloor (j+k)\beta \rfloor <\lfloor \mathit{j\alpha }\rfloor +\lfloor \mathit{j\beta }\rfloor +\lfloor \mathit{k\alpha }+\mathit{k\beta }\rfloor .$$

Since $j$ and $k$ play a symmetric role, we can assume that $j<k$ (the other case $j>k$ is treated similarly). Then

$$\frac{1}{2k}<\frac{1}{j+k},$$

so there exists some real number $\alpha$ such that

$$\begin{array}{lll}\hfill \frac{1}{2k}<\alpha <\frac{1}{j+k}<\frac{1}{j},& & \hfill \text{(1)}\end{array}$$

for instance, $\alpha =\frac{1}{2}\left(\frac{1}{2k}+\frac{1}{j+k}\right)$.

Take $\beta =\alpha$. Since $0<\alpha <1∕j$ and $0<\beta <1∕j$, then $0<\mathit{j\alpha }<1,0<\mathit{j\beta }<1$, so we obtain

$$\begin{array}{lll}\hfill \lfloor \mathit{j\alpha }\rfloor =\lfloor \mathit{j\beta }\rfloor =0.& & \hfill \text{(2)}\end{array}$$

Since $0<(j+k)\alpha <1$ and $0<(j+k)\beta <1$, then

$$\begin{array}{lll}\hfill \lfloor (j+k)\alpha \rfloor =\lfloor (j+k)\beta \rfloor =0.& & \hfill \text{(3)}\end{array}$$

Finally $2\mathit{k\alpha }\ge 1$, thus $\lfloor 2\mathit{k\alpha }\rfloor \ge 1$, and $\alpha =\beta$, so

$$\begin{array}{lll}\hfill \lfloor \mathit{k\alpha }+\mathit{k\beta }\rfloor \ge 1.& & \hfill \text{(4)}\end{array}$$

By equations (2), (3) and (4),

$$\lfloor (j+k)\alpha \rfloor +\lfloor (j+k)\beta \rfloor <\lfloor \mathit{j\alpha }\rfloor +\lfloor \mathit{j\beta }\rfloor +\lfloor \mathit{k\alpha }+\mathit{k\beta }\rfloor .$$

This shows that

$$j\ne k\Rightarrow (\exists <!--\; FUNCTION\; APPLICATION\; -->\alpha \in ℝ,\exists <!--\; FUNCTION\; APPLICATION\; -->\beta \in ℝ,\lfloor (j+k)\alpha \rfloor +\lfloor (j+k)\beta \rfloor <\lfloor \mathit{j\alpha }\rfloor +\lfloor \mathit{j\beta }\rfloor +\lfloor \mathit{k\alpha }+\mathit{k\beta }\rfloor ),$$

or equivalently

$$(\forall <!--\; FUNCTION\; APPLICATION\; -->\alpha \in ℝ,\forall <!--\; FUNCTION\; APPLICATION\; -->\beta \in ℝ,\lfloor (j+k)\alpha \rfloor +\lfloor (j+k)\beta \rfloor \ge \lfloor \mathit{j\alpha }\rfloor +\lfloor \mathit{j\beta }\rfloor +\lfloor \mathit{k\alpha }+\mathit{k\beta }\rfloor )\Rightarrow j=k.$$