Reversing Irrationality (1 Viewer)

Sy123 · Dec 12, 2012

So we all know that irrational numbers can be expressed as an infinite construction of integers in various forms.

I.e:

$1+\frac{1}{1+\frac{1}{1+1\frac{1}{1+...}}}= \phi$

$e=\sum_{n=0}^{\infty} \frac{1}{n!}$

$\frac{\pi^2}{6}=\sum_{n=1}^{\infty} \frac{1}{n^2}$

and so on.

Is it possible to construct something of an infinite series partially composed of irrational numbers, to form a rational?

i.e. $\sum_{n=1}^{\infty} a^\frac{1}{n} = k$

a is irrational and k is rational (this is just an example of such a way to see an infinite series composed of irrational numbers. It could be a continued fraction, infinite square roots etc.)

If this is possible, is it also possible for transcendental numbers (which I assume would be harder to do this if applicable to irrationals in the first place)

(also no trivial stuff like e-e=0, e is irrational etc)

And yes you can technically for example rearrange the Basel problem to make 6 the subject and etc. But I'm looking for an actual series.

Carrotsticks · Dec 12, 2012

$\\ \sum_{k=0}^{\infty} \sqrt{2} (1-\sqrt{2})^k = \frac{\sqrt{2}}{1-(1-\sqrt{2})} = \frac{\sqrt{2}}{\sqrt{2}} = 1$

The irrationals do not form a closed set and the metric space of the irrationals is incomplete, so summing an infinite number of them can yield a value in Q.

Rezen · Dec 12, 2012

I'm unsure of what exactly your asking. Is it possible to form an infinite series of irrational numbers such that the series equals a rational number? Then the answer is yes. A 'nontrivial' construction is as follows:

$\text{let } \sum_{k=0}^{\infty}a_k \text{ be a converging series that converges to a rational number such that each } a_k >0. \text{ We then create a new series where we rewrite each rational } a_k = b + (a_k - b) \text{ where b is a non-negative irrational number less than } a_k. \text{ the new series converges}$

The above is not exactly rigorous but should give you the idea of the construction.

Sy123 · Dec 12, 2012

Carrotsticks said:
$\\ \sum_{k=0}^{\infty} \sqrt{2} (1-\sqrt{2})^k = \frac{\sqrt{2}}{1-(1-\sqrt{2})} = \frac{\sqrt{2}}{\sqrt{2}} = 1$

The irrationals do not form a closed set and the metric space of the irrationals is incomplete, so summing an infinite number of them can yield a value in Q.

Well that just fell apart lol.
Thanks

Carrotsticks · Dec 16, 2012

Regarding this, I just remembered something decently relevant from the 2011 2U Maths HSC.

$\\ \sum_{k=2}^{n} \frac{1}{\sqrt{k-1}+\sqrt{k}} = \sqrt{n} -1 $ via a Telescoping Sum, and we observe that if $ n $ is a perfect square, then the sum of $ n $ irrational numbers yields not only a rational number, but an integer.$$

Sy123 · Dec 16, 2012

Carrotsticks said:
Regarding this, I just remembered something decently relevant from the 2011 2U Maths HSC.

$\\ \sum_{k=2}^{n} \frac{1}{\sqrt{k-1}+\sqrt{k}} = \sqrt{n} -1 $ via a Telescoping Sum, and we observe that if $ n $ is a perfect square, then the sum of $ n $ irrational numbers yields not only a rational number, but an integer.$$

Oh that is pretty cool. Nice.

But can we generate something like this for transcendental numbers, do they require something more complicated, the only examples that were given involved square roots which are not transcendental.
With the exception of, for example:

$\sum_{n=1}^{\infty}e/2 (1-e/2)^ n$

Which converges to a rational. Is there anything like that telescoping sum you posted, whereby we can end up with a rational from a transcendental number.?

Carrotsticks · Dec 16, 2012

Well, the self-cancelling only occurs for finite sums, so in order to do what you said, you would need to construct a partial sum and then take a limit as n -> infinity, and have it converge to a rational number. It most certainly CAN be done, because all transcendentals are irrational, so it follows that a sum of infinite of them CAN yield a value in Q.

As for an example.. I cannot think of one over the top of my head ATM.

seanieg89 · Dec 16, 2012

Rezen's construction provides a counterexample for transcendental numbers too.

Reversing Irrationality (1 Viewer)

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