Sharpenings of Liouville's inequality - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T13:59:26Z http://mathoverflow.net/feeds/question/98425 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98425/sharpenings-of-liouvilles-inequality Sharpenings of Liouville's inequality Kale 2012-05-30T22:12:51Z 2012-05-30T22:29:10Z <p>The norm of an algebraic number $\alpha$ is the product of its conjugates, $N(\alpha)$.</p> <p>Suppose that I have an inequality of the form $|x-\alpha*y| > c X^{n-\gamma}$ where $X=max{|x|,|y|}$ and c is some constnat, can this be used to find an upper bound on the norm of $x-\alpha*y$ of the form $|N(x-\alpha*y)| &lt; C *X^{n-\gamma}$ $|x-\alpha*y|$ where C is some constant and $\gamma \geq 1$</p> <p>In the case of $\gamma=1$ it can since: $|N(x-\alpha*y)|=\prod (x-\alpha_i*y)|=|x-\alpha*y|X^{n-1} \prod (x/X-\alpha_i*y/X) \leq |x-\alpha*y| \prod(1|+|\alpha_{i}|)X^{n-1}$ </p> <p>It's unclear to me how it might be possible to find a sharper upper bound on the norm of $x-\alpha*y$ by using a sharper exponent in liouiville-type inequalities.</p> <p>I should add that the constant C should be effective, as it is in the case of $\gamma=1$.</p>