$\lfloor na\rfloor+\lfloor nb\rfloor=\lfloor nc\rfloor+\lfloor nd\rfloor$

Question

If $a,b,c,d$ are positive irrational number such that $$a+b=c+d$$

Prove that

$$\lfloor na\rfloor+\lfloor nb\rfloor=\lfloor nc\rfloor+\lfloor nd\rfloor\quad\forall \,\,n\in\mathbb N$$

My working:

$$\left(\lfloor na\rfloor+\lfloor nb\rfloor\right)-\left(\lfloor nc\rfloor+\lfloor nd\rfloor\right)=\left(\{nc\}+\{nd\}\right)-\left(\{na\}+\{nb\}\right)$$ $$\implies \left(\{nc\}+\{nd\}\right)-\left(\{na\}+\{nb\}\right)=\pm1,0$$

Answer 1

I have found an image of the book in question:

By looking at the proposed solution and by considering my counterexample in the comments above, we recognize that there must be something wrong. In particular, there appears to be an additional missing assumption that was left out of the problem statement... namely that $a+b$ is an integer.

Without that assumption, the line "$[na]+[nb]=nk - (\{na\}+\{nb\})$ showing that $(\{na\}+\{nb\})$ is an integer" is unjustified, but by including that $k$ is an integer then this would have been okay. Everything else that follows in the book's proof would have been justified.

Again, as written without that assumption the claim is false. A counterexample is $a=b=0.6101101110\dots$ and $c=0.0101101110\dots, d=1.2101101110\dots$ and considering $n=1$