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

Question

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.

Answer 1

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).

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

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)

Answer 6

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.