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

Solution 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).

Solution 2:

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

Solution 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.

Solution 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.

Solution 5:

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.