Given irrational numbers $0<\gamma_1<\gamma_2<1,$ real numbers $0\leq a<b\leq 1,$ does $\exists n$ such that $\{n\gamma_1\}, \{n\gamma_2\}\in(a,b)?$

Question

For this question, $\{x\}$ means the fractional part of the real number $x.$

Given irrational numbers $0<\gamma_1<\gamma_2<1,\ $ real numbers $a,b$ with $0\leq a<b\leq 1,$ does there exist $n\in\mathbb{N}$ such that $\ a< \left\{ n\gamma_1 \right \} < b\ \ $ and $\ \ a< \left\{ n\gamma_2 \right \} < b\ ?$

My thoughts: For irrational $\gamma$ we have that $\{\ \{\gamma n\}:\ n\in\mathbb{N}\}$ is dense (in fact, equidistributed) in $[0,1].$ But I don't see how to use this.

Any thoughts/hints/ideas?

Answer 1

Not always .For example,if you choose $\gamma_1=\pi/8,\gamma_2=\pi/4$.The point set $\{(\{n\gamma_1\},\{n\gamma_2\}):n\in \mathbb{N}\}\subset \mathbb{R}^2$ is dense on

$$ C:y=\left\{\begin{align}&2x,&0\le x<\frac{1}{2}\\ & 2x-1,&\frac{1}{2}\le x <1 \end{align}\right.. $$ However,if you choose $a=3/8,b=5/8$,The square $[a,b]\times [a,b]$ does not intersect with $C$.