TITLE: A problem about floor function
QUESTION [21 upvotes]: 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

REPLY [2 votes]: Case 1: $a,b$ rationals.
This case is simple. $a=\frac{m}{n}$ and $b=\frac{r}{p}$ with both fractions irreducible.
WLOG $n \geq p$. 
Subcase 1.1 *$n < p$*
Solve $rx \cong -1 \mod p$, which has the solution $x= x_0 +lp$.
Then for all $x$ we have $\{ xb \}=\frac{p-1}{p}$, and because $1 < n < p$ you can easely find such $x$ so that $xa \notin Z$, meaning $\{ xa \} >\frac{1}{n}$.
Since $\frac{p-1}{p}+\frac{1}{n} > \frac{p-1}{p}+\frac{1}{p} =1$ we are done.
Subcase 1.2 *$n = p$*
Same idea. 
Solve $rx \cong -1 \mod p$, which has the solution $x= x_0 +lp$.
The only possible issues are if $xm= 0, 1 \mod p$.
But this is not possible.
The first case is not possible since both $x,m$ are invertible, while in the second case you get
$$rx \cong -1 \mod p \Rightarrow r \cong -x^{-1} \mod p$$
$$mx \cong 1 \mod p  \Rightarrow m \cong x^{-1} \mod p $$
Thus, $r+m \cong 0 \mod p$ and hence $a+b= \frac{m+r}{p} \in Z$ which is a contradiction.
Case 2 A rational, b irrational. This case follows immediately from Dirichclet theorem.
Indeed, $a=\frac{m}{n}$ irreducible.
The equation $xm \cong -1 \mod p$ has solution $x= x_o+np$, and by Dirichlet theorem, since $b$ is irrational, the set $\{ (x_0+np)b\} $ is dense in $[0,1]$.
Pick some $k$ so that $\{ (x_0+kp)b\} > \frac{1}{n}$ and you are done, $(x_0+kp)$ satisfies your condition.
Case 3: $a,b$ irational. Then $b= \alpha a$.
Subcase 3.1 $\alpha$ irrational. See Jyrki's comment.
Subcase 3.2 $\alpha$ rational. $\alpha=\frac{m}{n}$ irreducible.
Fix some $\epsilon$ so that $(m+n)\epsilon <1$. Then we also have $m-1 < m(1-\epsilon) < m$.
By Dirichlet theorem, the set $\{nka\} $ is dense in $[0,1]$.
Pick some $k$ so that
$$\{ kna \} > 1- \epsilon \,.$$
Then 
$$kna = l+  1- \epsilon$$
for some integer $l$.
Hence
$$knb=kn\frac{m}{n}a=\frac{m(l+1-\epsilon)}{n}$$ 
Note that by our choise of $\epsilon$,
$$\frac{ml+m-1}{n} < \frac{m(l+1-\epsilon)}{n} < \frac{ml+m}{n} \,.$$
This shows that $[\frac{m(l+1-\epsilon)}{n}] \leq \frac{ml+m-1}{n}$ and hence
$$\{\frac{m(l+1-\epsilon)}{n}  \} >  \frac{1-m\epsilon}{n}$$
Thus
$$\{ kna \} + \{ knb \} >1- \epsilon+  \frac{1-m\epsilon}{n}  \geq 1$$
the last inequality following from $(m+n)\epsilon<1$.
P.S. The proof is extremely technical, there might be a mistake into it. I wrote but I would definitely not enjoy reading it :P