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The definition of Liouville number $l$ is, for every positive integer $n$, there exist integers $p,q$, such that $$\left\vert l-\frac{p}q\right\vert<\frac1{q^n}$$ is satisfied.

However, in an accepted and upvoted answer on MSE, it seems that one can choose $p,q$ arbitrarily. So, the definition of Liouville number should have ’for any’ replacing ’there exist’.

What confused me more is that the Liouville approximation theorem states that:

For any algebraic number $x$ of degree $n>2$, a rational approximation $\frac{p}q$ to $x$ must satisfy $$\left\vert x-\frac{p}q\right\vert >\frac1{q^n} $$

This makes me guess that the inequality for Liouville numbers is satisfied for any $(p,q)$.

Which is correct and which is wrong?

Szeto
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  • That doesn't make sense. Obviously there are rational numbers $\frac pq$ which are very far from $l$. – lulu Jun 15 '18 at 23:36
  • @lulu So $(p,q)$ is up to our choice? – Szeto Jun 15 '18 at 23:37
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    Not following. Please write a complete sentence. – lulu Jun 15 '18 at 23:38
  • Your Liouville approximation theorem shows that an (irrational) algebraic number is not a Liouville number. – GEdgar Jun 15 '18 at 23:38
  • @lulu when we know a number is Louiville, is the first inequality true for any (p,q), if we know some bounds of distance between p and q? – Szeto Jun 15 '18 at 23:40
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    Sorry to be pedantic, but that's not complete. What bounds is it that you know? – lulu Jun 15 '18 at 23:42
  • @lulu More directly, do you think the answer I quoted is correct? – Szeto Jun 15 '18 at 23:43
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    The answer you quoted doesn't appear to say anything like what you said. The definition of a Liousville number is not ambiguous. $\alpha \in \mathbb R$ is a Liouville number iff for every positive integer $n$ there exists integers $p,q$ with $q>1$ and $0<|\alpha -\frac pq|< \frac 1{q^n}$ – lulu Jun 15 '18 at 23:45
  • @lulu i am particularly uncomfortable with the fourth paragraph ‘Now choose p=n...’ How come one can choose p, q arbitrarily? – Szeto Jun 15 '18 at 23:50
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    Write to the author and ask them directly. – lulu Jun 15 '18 at 23:52

1 Answers1

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The first definition you gave is correct. I recommend you re-read what you linked. He said that pi is not a liouville number. The negation of there exists is for all.

Matthew Liu
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