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Concerning the criteria for irrationality:

Theorem says "A number $\alpha$ is irrational if and only if for every $\epsilon >0$ there exist integers $h$ and $q$ such that $0 < | q\alpha - h | < \epsilon$."

So if we read the "if" part of the theorem it says that "if for every $\epsilon >0$ there exist integers $h, q$ such that $0 < | \alpha - h | < \frac{\epsilon}{q}$, then $\alpha$ is irrational, i.e $\alpha$ is not $\frac{m}{n}$ for any integers $m and n$." Ok so far?

Now I say the following: Let $\alpha$ be indeed a rational number say $\frac{m}{n}$, then for every $\epsilon >0$, we can let $\frac{h}{q} = \frac{\left(m + \frac{\epsilon}{n}\right)}{n}$, so that $|\alpha - \frac{h}{q}| = |\frac{m}{n} - \frac{\left(m + \frac{\epsilon}{n}\right)}{n}| = | \frac{\epsilon}{n^2} | < \frac{\epsilon}{n} = \frac{\epsilon}{q}$, since $n = q$ and $m = h$.

Why do I find this counter example to the theorem? Maybe my counter example is flawed. To me the theorem is saying that a number is irrational iff it can be approximated arbitrarily by rational numbers, but rational numbers can also be arbitrarily approximated by rational numbers so why is the theorem stating that?

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    $\begingroup$ The flaw is that you define $h$ as $m+\varepsilon/n$, which is not necessarily an integer. $\endgroup$ – Tomas Nov 22 '13 at 18:28
  • $\begingroup$ @Paulo Find $h$ and $q$ given $\alpha = 1/2$ and $\epsilon = 1/2$ then. $\endgroup$ – NovaDenizen Nov 22 '13 at 20:08
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The flaw is that you define $h$ as $m+\varepsilon/n$, which is not necessarily an integer.

Also, the theorem does not imply that the rational numbers can not be arbitrarily approximated by rational numbers, but that they can not be arbitrarily approximated with "small" denominators. This is because - as you already noted - the error in the approximation is required to be smaller than $\varepsilon/q$ where $q$ is the denominator of the approximating rational.

Note, that if $\alpha=\frac ab$, then $$\left|q\cdot\frac{a}{b}-h\right|=\left|\frac{aq-hb}{b}\right|$$ and $aq-hb\in\mathbb Z$. So this expression is either zero or bigger than $\frac 1{|b|}$. This implies that there is no approximation for $\varepsilon<\frac 1{|b|}$ which is not exact.

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Your logic says that since $\frac{h}{q} = \frac{m+\frac{\epsilon }{n}}{n}$, then $q = n$. This is not necessarily true. Suppose it were. Then this would imply that $h=m+\frac{\epsilon}{n}$. But there are infinitely many $\epsilon$ such that $m+\frac{\epsilon}{n}$ is not an integer, a condition imposed on $h$ by hypothesis.

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  • $\begingroup$ Thank you very much also! Great and enlightening. $\endgroup$ – Paulo Nov 22 '13 at 19:27

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