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.
Best Answer
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).