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Lemma: let $\alpha \in \mathbb{R}^+$ and $p_1,p_2,\dots, q_1, q_2, \ldots \in \mathbb{N}$ such that $\left|\alpha q_n - p_n \right| \neq 0$ for all $n \in \mathbb{N}$ and $$ \lim_{n \rightarrow \infty} p_n = \lim_{n \rightarrow \infty} q_n = \infty\,$$ If $ \lim_{n \rightarrow \infty} \left| \alpha q_n - p_n \right| = 0.\,$ then $α$ is irrational.

The proof proceeds by contradiction, (assume $\alpha=a/b$, derive nonsense, hence $\alpha$ irrational)

What if $\alpha$ depends on some variables let's say $n$ and $a$ (like $\sqrt[n]{a}$ although this is a bad example as that number can be rational anyway)

The question is: is there some version of that lemma where $\alpha$ isn't assumed to be constant but can depend on two or more variables? (so that we can choose $p_n$ and $q_n$ that holds for all $m$)

Also are there some references that discuss extensively that lemma?


i mean can we generalize the problem so that, for example instead of taking $α=2^{1/2}$ and making an analytic proof of the irrationality of $2^{1/2}$ using that lemma, we could generalize that and take $α=2^{1/m}$ where $m∈\Bbb N\setminus\{0\}$, and show its irrationality using a more extended and general version of the lemma?

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If $\alpha(t)$ is a nonconstant continuous real-valued function of a real variable $t$, it will be rational for some $t$ by the Intermediate Value Theorem. So I don't see what you hope to achieve by such a version of the lemma.

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  • $\begingroup$ what if we put some constraints on $t$? or if $α(t)$ is a function from the naturals to themselves $\mathbb{N\longrightarrow N}$ $\endgroup$ – user153330 Mar 14 '15 at 15:02
  • $\begingroup$ what if $\alpha(t)$ isn't real-valued function? $\endgroup$ – user153330 Mar 15 '15 at 17:08
  • $\begingroup$ If $\alpha(t)$ isn't real, then it certainly isn't a rational number. $\endgroup$ – Robert Israel Mar 16 '15 at 15:05
  • $\begingroup$ i meant what if α(t) wasnt continuous, and was only from the natural numbers to the natural numbers or from the rationals to the rationals? $\endgroup$ – user153330 Mar 19 '15 at 15:07
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This lemma depends crucially on the fact that $\alpha$ does not depend on $n$ (and is completely agnostic to whether $\alpha$ depends on some "external" variables). In particular, one should note that we have this condition: $$|\alpha q_n-p_n|\neq 0$$ which is really the heart of the proof and excludes a trivial case where we take $\alpha=\frac{p}q$ and $p_n=np$ and $q_n=nq$ (or anything similar). However, notice that if we jiggle $\alpha$ around by any amount, we can make it so that $|\alpha_n p_n-q_n|\neq 0$ even when $\alpha_n$ converges to $\alpha$. This is a rather crucial problem, because it means that, we cannot even state this for $\alpha_n$ converging "fast enough". In a sense, we need to extract infinitely much information from a single $\alpha$ to determine if it is rational (hence all the limits in the definition) and moving $\alpha$ around is not conducive to that. (You can also think of this as stemming from the fact that both rationals and irrationals are dense in the reals)

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  • $\begingroup$ i mean can we generalize the problem so that, for example instead of taking $\alpha=2^{1/2}$ and making an $\sf{analytic}$ proof of the irrationality of $2^{1/2}$ using that lemma, we could generalize that and take $\alpha=2^{1/m}$ where $m\in\Bbb N\setminus\{0\}$, and show its irrationality using a more extended and general version of the lemma? $\endgroup$ – user153330 Sep 18 '15 at 20:33
  • $\begingroup$ hope you get the idea i'm trying to get to $\endgroup$ – user153330 Sep 18 '15 at 20:45

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