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Let $a_1$ and $a_2$ be irrational numbers. Prove that there exist an infinite number of triplets $\left(p_1,p_2,p\right)$ where $p_1$,$p_2$,$p$ are integers and $p>0$ such that $$\left|a_i-\frac{p_i}{p}\right|< \frac{1}{\sqrt {p^3}}$$ for each $i=1,2$.

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  • $\begingroup$ I tried using Dirichlet's Theorem ... $\endgroup$ – alexb Nov 28 '15 at 11:38
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    $\begingroup$ Could you find a better title for your question than just a random number of initial words from the question text? It is not supposed to be an incipit. And especially not a misleading one -- you're not actually asking anyone to prove that infinite numbers exist! $\endgroup$ – hmakholm left over Monica Nov 28 '15 at 11:57
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Let $d\in\mathbb{N}$ be arbitrary. Suppose that we have $d$ real (not necessarily irrational) numbers $a_1,a_2,\ldots, a_k$. For a natural number $n$, consider the set $T_n\subseteq [0,1)^d$ consisting of $d$-tuples $\big(\left\{ka_i\right\}\big)_{i=1}^d$ where $k=0,1,2,\ldots,n^d$. Here, $\{x\}=x-\lfloor x\rfloor$ is the fractional part of a real number $x$. Subdivide $[0,1)^d$ into $n^d$ congruent $d$-hypercubes of the form $\prod_{i=1}^d\,\left[\frac{t_i-1}{n},\frac{t_i}{n}\right)$, where $t_1,t_2,\ldots,t_d\in\{1,2,\ldots,n\}$, and apply the Pigeonhole Principle. We can show then that there are points corresponding to two distinct values of $k$, say $k=k_1$ and $k=k_2$, in $T_n$ whose $i$-th coordinates are at distance less than $\frac1n$ for every $i=1,2,\ldots,d$. Without loss of generality, assume $k_1<k_2$.

Let $p:=k_2-k_1\leq n^d$. Take $p_i$ to be the integer closest to $pa_i$ for each $i=1,2,\ldots,d$. Ergo, $$\left|pa_i-p_i\right|=\big|\left\{k_2a_i\right\}-\left\{k_1a_i\right\}\big|<\frac{1}{n}$$ for every $i=1,2,\ldots,d$. Then, $$\left|a_i-\frac{p_i}{p}\right|<\frac1{pn}\leq\frac1{p^{1+\frac1d}}$$ for each $i=1,2,\ldots,d$. By arbitrarily changing the value of $n$, we then see that there are infinitely many tuples of integers $\left(p_1,p_2,\ldots,p_d,p\right)$ with $p>0$ such that $$\left|a_i-\frac{p_i}{p}\right|<\frac1{p^{1+\frac1d}}$$ for each $i=1,2,\ldots,d$.

Question: Does there exist a sequence $\left\{a_i\right\}_{i\in\mathbb{N}}$ of real numbers such that

(1) there are only finitely many $p\in\mathbb{N}$ such that the following inequality holds: $\sup\Big\{\min\big\{\left\{pa_i\right\},1-\left\{pa_i\right\}\big\}\,\Big|\,i=1,2,\ldots\Big\}\leq\frac{1}{p}$?

(2) there is no $p\in\mathbb{N}$ such that $\sup\Big\{\min\big\{\left\{pa_i\right\},1-\left\{pa_i\right\}\big\}\,\Big|\,i=1,2,\ldots\Big\}\leq\frac{1}{p}$?

What if the $a_i$'s are required to be irrational?

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