$A$ is a set of integers and $\lvert A \rvert = 501$.
For any $a \in A$, $1 \le a \le 1000$.
Prove that there are integers $a,b \in A$, such that $a \vert b$. (a divides b)
The question doesn't specify it, but I assume that $a \neq b$.
(Hint: Note that every integer $x$ can be uniquely presented as $x = y · 2 ^k$ where $k > 0$ and $y$ is odd. Consider the function $x \to y$)
Can I get some help?
We know that each $n \in \mathbb{N}$ can be written $n = y \cdot 2^k$ with unique $y$ odd and $k \in \mathbb{N}$. As this $y$ is unique (can be seen as the odd part of $n$), one can properly define the function $\phi : n \mapsto y = \dfrac{n}{2^{v_2(n)}}$.
What we can do is to show that $\phi$ is not injective on $A$.
Let us consider $B$ the set of odd integers between $1$ and $1000$. The cardinal of $B$ is $499$. Indeed, for $k \in \mathbb{N}, 0 \leq 2k+1 \leq 999$ is equivalent to $0 \leq k \leq 498$, so it gives $498 - 0 + 1 = 499$ possible values for $k$.
So $\phi : A \mapsto B$ is well-defined and cannot be injective because $|A| > |B|$. So you have two elements of $A$ with the same odd part. The smaller one clearly divides the other.
