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Assume we have a sequence of rational numbers $\left(\frac{p_n}{q_n}\right),$ where $\gcd(p_n,q_n)=1, \ \forall n \in \mathbb N$.

We know that $$\lim_{n\to\infty} \left(\frac{p_n}{q_n}\right)= x$$ and $$\lim_{n\to\infty} p_n = \lim_{n\to\infty} q_n = \infty.$$

Assume $x = \frac{M}{N},$ where $\gcd(M,N)=1$. Now, there exists such $k \in \mathbb N$ that $p_n>M$ and $q_n > N, \ \forall n\ge k$.

Does this prove that x is irrational?

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  • $\begingroup$ No, it doesn't, $x + \frac{1}{2^k}$ converges to $x$, and gets arbitrarily large numerators and denominators. $\endgroup$ – Daniel Fischer Aug 8 '13 at 23:35
  • $\begingroup$ @DanielFischer That doesn't prove x is rational either. So this is necessary, but not sufficient condition for irrationality? $\endgroup$ – Valtteri Aug 8 '13 at 23:39
  • $\begingroup$ Right. Necessary but not sufficient. You know that $x$ is irrational if the approximations are "good", e.g. $\lvert x - \frac{p_n}{q_n}\rvert < \frac{1}{2q_n^2}$ for all (actually, arbitrarily large suffices) $n$. $\endgroup$ – Daniel Fischer Aug 8 '13 at 23:43
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If $x = M/N$ with positive integers $M,N,m,n$, either $|x - m/n| = 0$ or $|x - m/n| \ge \dfrac{1}{Nn}$. On the other hand, if $x$ is irrational there are infinitely many pairs $(m,n)$ with $0 < |x - m/n| < \dfrac{1}{n^2}$. In this sense, irrational numbers can be very well approximated by rationals, but rational numbers can't be very well approximated by other rationals.

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Given $x = M/N$ as you say, good approximations parametrized by $n \in \mathbb N$ include $$ \frac{nM + 1}{nN + 1} $$

A (real) number is irrational if and only if the simple continued fraction for it is infinite. If the number is a "quadratic irrational" $(A + \sqrt B)/ C$ For integers $A,B,C,$ with $C \neq 0$ and $B$ positive but not a square, then the s.c.f. is eventually periodic and infinite.

http://en.wikipedia.org/wiki/Continued_fraction

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