Is there a way to write an infinite set that contains only irrational numbers without integer multiples?

The infinite set must not contain integer multiples of any other members of that set. For example,$\pi$ is a member, but we cannot have $2\pi, 3\pi$, and so on. Same applies for any other irrational number in the set.

Also, that infinite set must be equinumerous to $\mathbb{N}$ (natural numbers). This seems intuitive to me, as there are many ways to line up infinite sets with $\mathbb{N}$. But I am having trouble thinking of such an infinite sets regarding only irrationals.

Thanks.

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    Also $\{e^{e^{.^{.^{.}}}}\}$ with $n$ exponentiations, for all natural $n$. – YoTengoUnLCD Apr 24 '16 at 3:37
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    By one meaning of "write", you can't write any infinite set. By another meaning, you already did, in the question title. – immibis Apr 24 '16 at 3:54
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    @YoTengoUnLCD the construction surely works, but I don't think it's known that e.g. $e^{e^{e^{e^{e^{e^{e^{e^e}}}}}}}$ is not an integer, nor an integer multiple of $e^e$. – Noam D. Elkies Apr 24 '16 at 17:41
  • @immibis You need to prove such a set exist if you mean to use the description in the title. – Najib Idrissi Apr 25 '16 at 11:17
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    @NoamD.Elkies More importantly, would the integer generated by a power tower of $e$ be named after the discoverer, or called "the squealing integer"? – Yakk Apr 25 '16 at 17:49

The set of square roots of prime numbers: $$\{\sqrt{2},\sqrt{3},\sqrt{5},\ldots,\sqrt{p},\ldots\}$$ is an example of such a set.

Assume $\sqrt{a}=k\sqrt{b}$ for some integer $k$. Then $a=k^2b$ so we get that $k=\sqrt{a/b}$ which is impossible if both $a$ and $b$ are prime as that ratio will never be a perfect square (or even an integer).

The set $S = \{ n+\sqrt{2}: n \in \mathbb{N} \}$ also satisfies the above condition.

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    Indeed for any irrational $x$, you have $S = \{ n+x: n \in \mathbb{Z} \}$ satisfying the condition – Henry Apr 25 '16 at 8:02

Plenty of choices: in addition to the two noted already there's $$ \{ \pi, \pi^2, \pi^3, \pi^4, \ldots \}, $$ $$ \{2^{1/2}, 2^{1/3}, 2^{1/4}, 2^{1/5}, \ldots \}, $$ and even $$ \{ 2\sqrt2, \, 3\sqrt2, \, 5\sqrt2, \, 7\sqrt2, \, 11\sqrt2, \, \ldots \} $$ (with prime multipliers), since you didn't disallow rational multiples.

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    In the same spirit as the last, $\{\,q\sqrt 2\mid q\in \Bbb Q\cap [1,2)\,\}$ – Hagen von Eitzen Apr 24 '16 at 18:52

Select an irrational $\alpha_1$. Select any $\alpha_2 \in \mathbb{R} \setminus (\mathbb{Q} \cup \mathbb{Z} \alpha_1)$, which is guaranteed to be non-empty as $\mathbb{R}$ is uncountable. Induct.

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    It does use axiom of choice however! – Rico Apr 24 '16 at 9:41
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    @Rico only needs the weaker axiom of dependent choice, surely. ;) – Stan Liou Apr 24 '16 at 10:08
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    Actually Cantor's diagonal argument does give an explicit construction at each step. – Noam D. Elkies Apr 24 '16 at 15:14
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    More generally, you're relying on the lemma that for any finite set of irrationals, the set of the their integer multiples does not cover the set of irrationals. But this is obvious: the set of integer multiples of any finite set is discrete, but the irrationals are dense. – Jack M Apr 24 '16 at 22:50
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    @NajibIdrissi I don't think so, because I can use a standard ordering of ${\bf Q}$ and interleave it with standard orderings of the ${\bf Z}\alpha_i$ chosen so far. – Noam D. Elkies Apr 25 '16 at 13:24

What about: all irrationals in $[2,3]$ ... None of them is an integer multiple of another.

Oops, that's uncountable.

How about all numbers in $[2,3]$ that are rational multiples of $\sqrt{2}$? Again, none of them is an integer multiple of another.

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    That works, but why $[2,3]$ rather than $[1,2]$? – Noam D. Elkies Apr 24 '16 at 15:13
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    Or "any countable subset of $[2,3]$ – Stella Biderman Apr 24 '16 at 17:37
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    @NoamD.Elkies Probably because 2 is integer multiple of 1. – Somnium Apr 25 '16 at 9:59
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    But 1 and 2 are rational, so wouldn't be chosen in any case. – Noam D. Elkies Apr 25 '16 at 13:25

In the same spirit as answered before, you probably have noted that any irrational known can establish the set desired, since the fundamental condition maintain. Not so fast, I know this is obvious, but you can simply take only one irrational and take of the first decimal place of it and make a new member, and them the same thing for the next. Since you have chosen such number that have not an end, of course for each decimal place its infinitude has, you can add a member with that infinitude minus position-1 related decimal.

(I am from stack overflow, this just shows up on my feed)

  • ops: do not know if it will be enumerable. – lionyouko Apr 24 '16 at 5:59
  • this is in fact ennumerable. Anything constructed by an iterated sequence like this is ennumerable – Stella Biderman Apr 28 '16 at 19:37

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