algebra precalculus - A problem about floor function

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:

1. 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 $n$.
   
   




2. 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.




3. I'd like a proof with algebraic construction.

Thanks

Answer

Look at the following figure of the checkered $(x,y)$-plane with gridlines at integer $x$'s and $y$'s:

The points $z_n:=(n\ a,n\ b)$ $\ (n\in{\mathbb Z})$ are marked. We will show that at least one $z_n$ with $n>0$ falls in the interior of a grey triangle, which means $\{n a\}+\{n b\}>1$.

We distinguish the following two cases:

(i) $b/a$ is rational. $-$ In this case there are $p$, $q\in{\rm N}_{\geq1}$ with ${\rm gcd}(p,q)=1$ and a $\lambda>0$ with $a=p\lambda$, $b=q\lambda$. All points $z_n$ are lying on the line

$$\ell:\quad t\mapsto z(t)=(pt, qt)\qquad(-\inftyin fact $z_n=z(n\lambda)$. This line passes through the points $(j p, j q)$ $\ (j\in{\mathbb Z})$ and actually is a closed curve $\gamma$ on the torus $T$ obtained by identifying points $(x,y)$ and $(x',y')$ with $x\equiv x'$ $\ ({\rm mod}\ p)$ and $y\equiv y'$ $\ ({\rm mod}\ q)$. Let $\pi: \ {\mathbb R}^2\to T$ denote the corresponding projection. There is a finite set of segments $\sigma_k\subset\gamma=\pi(\ell)$ colored in grey, which makes up for half of the length of $\gamma$.

(i.i) If $\lambda$ is rational then the $z_n$ project onto only finitely many equally spaced points on $\gamma$, and some $z_n$ with $n>0$ will project onto the same point as $z_{-1}$ and therefore will lie in the interior of a grey triangle. (This case has already been dealt with by the OP.)

(i.ii) If $\lambda$ is irrational then the points $\pi(z_n)$ are dense on $\gamma$, and some of them will lie in the interior of one of the grey intervals of $\gamma$. It follows that the corresponding $z_n$ are lying in a grey triangle.

(ii) $b/a=:m$ is irrational. $-$ Put $\delta:=1-a-b$ and consider again the line
$$\ell: \quad t \mapsto z(t):=(t, m t)\qquad(-\infty(now with a different parametrization). The $z_n$ are equally spaced on $\ell$ with distance $d:=\sqrt{a^2+b^2}$. The line $\ell$ intersects the vertical gridlines $x=k\in{\mathbb Z}$ in the points $(k, m k)$ whose ordinates $y_k:=m k$ are irrational for $k\ne0$. Therefore the $y_k$ are dense ${\rm mod}\ 1$. There is a $k_0>0$ such that $1-\{y_{k_0}\}<\delta$. It is easy to see that the segment $\sigma\subset\ell$ with endpoints $(k_0-a, y_{k_0}-b)$ and $(k_0, y_{k_0})$ is lying completely in the grey triangle to the left of $(k_0, y_{k_0})$. Since $\sigma$ has length $d$ there is at least one $z_n\in\sigma$, and this $z_n$ lies in the interior of a grey triangle.