New answers tagged


So perhaps this is a counterexample you are looking for? Take the equation $$4x_1 + 5x_2 + 9x_3 = 0.$$ We have a solution $x = (3,3,-3)$ which satisfies $\max|x_i| \leq 122^{1/4} \approx 3.68$. The next linearly independent solution of smallest height seems to be $y = (6,-3,-1)$, but $$\max|x_i| \max|y_i| = 3 \cdot 6 = 18 > \sqrt{122} \approx 11.04.$$


Fix $\epsilon>0$. Fix an integer $N>1/\epsilon$. Consider the fractional parts of the numbers $k\omega, 0<k\le N$. These are commonly denoted $$ \{k\omega\}:=k\omega-\lfloor k\omega\rfloor. $$ Because $\omega$ is irrational the numbers $\{k\omega\}\in(0,1)$, $k=1,2,\ldots,N$, are all distinct. Because there are $N$ of them some two of them, say $\{...


It boils down to showing that given any $\epsilon > 0$ there are integers $m, n$ such that $$0 < m + n\omega < \epsilon\tag{1}$$ Clearly this means we need to find good rational approximations to irrational $\omega$. This follows easily from a theorem that there are infinitely many approximations of the form $p/q$ with $$\left|\omega - \frac{p}{q}\...


Hint In this context ($\omega$ irrational) it can be shown that the set $\{n\omega-\lfloor n\omega\rfloor\mid n\in\mathbb Z\}$ is a dense subset of $(0,1)$. Also it can be shown to be a subset of $A$.


What $\inf A =0$ means is that, for any $\epsilon >0$ there is some multiple of $\omega$ that is within distance $\epsilon$ of some integer. Hint 1: Pigeonhole principle. For simplicity assume $\omega <1$. Then each interval $[0,1], [1,2], [2,3]$ contains a multiple of $\omega$, call them $n \omega, m \omega$ and $r\omega$. Each of those three ...


The irrationality measure of $\pi$ is less than $8$. This means that $$ \Bigl|\pi-\frac{n}{m}\Bigr|>\frac{1}{m^8} $$ for all $n$ and all $m$ suficiente large. Thus, for $m$ large enough, $$ \epsilon(m)\ge\frac{1}{m^8}. $$


We denote $d(y, A) = \inf \{\mid a - y \mid,\ a \in A \}$. If you consider $x$ an algebraic number of degree $2$, then there exists $C>0$ s.t. $\forall (p,q) \in \mathbb{Z}\times \mathbb{N}^*, \mid x - \frac{p}{q} \mid \ge \frac{C}{q^2}$ Hence $$\forall n \in \mathbb{N}^*,\ d(nx,\ \mathbb{Z}) \ge \frac{C}{n}$$ Thus $d(n\pi x, \pi \mathbb{Z}) \ge \frac{\...

Top 50 recent answers are included