Here's the problem
Suppose that $$ a, b \in \mathbb {R^+},\qquad 0 < a + b < 1 $$ Prove or disprove that $$ \exists n \in \mathbb{Z^+}: \left\{na\right\} + \left\{nb\right\} \ge 1$$ where $\{x\} = x - \lfloor x \rfloor$ and $\lfloor x \rfloor = \max\left\{ n | n \in \mathbb{Z}, n \le x \right\}$.
Postscripts:
The original problem in magazine ( ISSN 1005-6416 ) might be (For a long time, I can't remember clearly):
$a, b$ are irrationals, subjects to $\forall n \in \mathbb{Z^+}: \left\{ na \right\} + \left\{ nb \right\} \le 1$, prove that $a + b \in \mathbb{Z}$.
Because I've found that $a, b$ needn't have been irrationals, I insist that there be a general proof (I think it's algebrian). I think the steps are like this:
- Divides $\left\{ (x, y) | x, y \in \mathbb{R^+}, x + y < 1 \right\}$ into many (infinity) pieces.
- For each piece, we obtain a $n$.
It may be easy when $a, b \in \mathbb{Q}$. Here's the proof:
Let $f(n) = \left\{na\right\} + \left\{nb\right\}$. For each $x \not \in \mathbb{Z}$, we have $\left\{x\right\} + \left\{-x\right\} = 1$
So if $a, b \in \mathbb{Q}$, we have $f(1) + f(-1) = 2$
Note that: $\exists T \in \mathbb{Z^+}$ subjects to $\forall n \in \mathbb{Z}: f(n + T) = f(n)$
So $2 = f(1) + f(-1) = f(1) + f(2T-1) \le 2 \max\left(f(1), f(2T-1)\right)$
We've done.
I'd like a proof with algebraic construction.
Thanks