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The norm of an algebraic number $\alpha$ is the product of its conjugates, $N(\alpha)$.

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 $|N(x-\alpha*y)| < C *X^{n-\gamma}$ $|x-\alpha*y|$ where C is some constant and $\gamma \geq 1$

In the case of $\gamma=1$ it can since: $|N(x-\alpha*y)|=\prod (x-\alpha_i*y)|=|x-\alpha*y|X^{n-1} \prod (x/X-\alpha_i*y/X) \leq |x-\alpha*y| \prod(1|+|\alpha_{i}|)X^{n-1}$

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.

I should add that the constant C should be effective, as it is in the case of $\gamma=1$.

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To make my questions a bit more clear: In the case of $\gamma=1$, the inequality follows immediately, I would like a proof that works for any $\gamma$, given the lower bound on the absolute value of $x-\alpha*y|$ I think there should be some way to use a lower bound on the absolute value of an algebraic number and transform it into an upper bound on the norm as is the case for $\gamma=1$. – Kale May 30 '12 at 22:23
So, you're asking whether a weaker hypothesis can imply a stronger conclusion? – Gerry Myerson May 30 '12 at 22:37
Gerry Myserson: I'm not sure what is stronger/weaker here. I would like to use a stronger Liouiville's inequality to imply an upper bound of a certain form on the norm of the algebraic number $x-\alpha*y$ (in absolute value). Basically I want $\prod (x-\alpha_i*y) (where i ranges over all conjugates of $\alpha$ other than itself to be $ \leq C*X^(\gamma=1)$ given that the stronger Liouville inequality $|x-\alpha*y| \leq c *X^{\gamma-1}$ does hold. – Kale May 30 '12 at 22:44
If $\gamma\gt1$, then $|x-\alpha y|\gt cX^{n-\gamma}$ is a weaker hypothesis than $|x-\alpha y|\gt cX^{n-1}$, while $|N(x-\alpha y|\lt CX^{n-\gamma}|x-\alpha y|$ is a stronger conclusion than $|N(x-\alpha y|\lt CX^{n-1}|x-\alpha y|$. Also, your comment is unreadable for TeXnical reasons. – Gerry Myerson May 31 '12 at 7:17
Isn't the power of X supposed to be in the denominator in Liouville bound (or have a negative exponent)? Also isn't the logic sorta flowing backwards here? The Liouville bound in a way is a result of the fact that the norm of an algebraic integer is an integer. Say, if $|m−n\sqrt2|$ is small, say much smaller than 1, then $|m+n\sqrt2|$ is approximately $2|n|\sqrt2$. As the norm $(m−n\sqrt2)(m+n\sqrt2)$ has absolute value $\ge1$, then we get the Liouville bound $$|m−n\sqrt2|>\frac{C}n.$$ Probably I misunderstood something about what you want to show – Jyrki Lahtonen May 31 '12 at 9:25

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