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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.

DeepSea
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Trevor Thomas
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    Also ${e^{e^{.^{.^{.}}}}}$ with $n$ exponentiations, for all natural $n$. – YoTengoUnLCD Apr 24 '16 at 03:37
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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. – Stack Exchange Broke The Law Apr 24 '16 at 03:54
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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
  • @Lennart If $e^e$ is algebraic then $e$ is transcendental, yes. But we already know that $e$ is transcendental, so there's no contradiction. Of course we can compute that $e^e$ is still not an integer. But a big enough tower of $e$'s makes that impossible. – Noam D. Elkies Apr 25 '16 at 13:22
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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
  • @immibis -- I think a definition of "write" that the OP would accept would be either "provide an algorithm that would generate any finite number of members" or "provide an algorithm that would detect whether any given number is a member". – Michael Lorton Apr 25 '16 at 18:51

6 Answers6

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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).

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

DeepSea
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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 08:02
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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.

Noam D. Elkies
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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 09:41
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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
  • @NoamD.Elkies You still need the axiom of dependent choice. – Najib Idrissi Apr 25 '16 at 11:18
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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
  • @NoamD.Elkies How do you write down (or describe) a set containing all the $\alpha_i$? – Najib Idrissi Apr 25 '16 at 13:48
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    Inductively (a two-level induction, over $i$ and over the digits of $\alpha_i$). Note too that each digit of $\alpha_i$ requires only the first few digits of each $\alpha_j$ with $j<i$. – Noam D. Elkies Apr 25 '16 at 16:25
  • @NoamD.Elkies It does not make sense to write a set "inductively" in ZF. I think you could manage with the axiom of countable choice (and not dependent choice as I said earlier). – Najib Idrissi Apr 26 '16 at 14:40
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    Not sure what the issue is. There's an algorithm for writing the $n$-th digit of $\alpha_i$ as a function of $i$ and $n$; it's even polynomial time in $n+i$. That lets me define a sequence ${\alpha_i}_{i=1}^\infty$ without making any choices along the way, even if ZF doesn't let me call it a set for some reason.. – Noam D. Elkies Apr 26 '16 at 16:38
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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.

GEdgar
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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)

lionyouko
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