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## New answers tagged diophantine-approximation

1

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 $$\max|x_i| \max|y_i| = 3 \cdot 6 = 18 > \sqrt{122} \approx 11.04.$$

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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 $\{... 0 It boils down to showing that given any$\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 $$\left|\omega - \frac{p}{q}\... 1 Hint In this context (\omega irrational) it can be shown that the set \{n\omega-\lfloor n\omega\rfloor\mid n\in\mathbb Z\} is a dense subset of (0,1). Also it can be shown to be a subset of A. -1 What \inf A =0 means is that, for any \epsilon >0 there is some multiple of \omega that is within distance \epsilon of some integer. Hint 1: Pigeonhole principle. For simplicity assume \omega <1. Then each interval [0,1], [1,2], [2,3] contains a multiple of \omega, call them n \omega, m \omega and r\omega. Each of those three ... 2 The irrationality measure of \pi is less than 8. This means that$$ \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}. $$1 We denote d(y, A) = \inf \{\mid a - y \mid,\ a \in A \}. If you consider x an algebraic number of degree 2, then there exists C>0 s.t. \forall (p,q) \in \mathbb{Z}\times \mathbb{N}^*, \mid x - \frac{p}{q} \mid \ge \frac{C}{q^2} Hence$$\forall n \in \mathbb{N}^*,\ d(nx,\ \mathbb{Z}) \ge \frac{C}{n}$$Thus$d(n\pi x, \pi \mathbb{Z}) \ge \frac{\...

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