I am trying to solve this below problem from Norman Bigg's Discrete Mathematics textbook, but cannot reconcile his solution with my work.
Let $X$ be a subset of $\{1, 2, \ldots 2n\}$ and $Y$ be the set of odd numbers $\{1, 3, \ldots, 2n-1\}$. Define a function $f: X \to Y$ by the rule \begin{equation} f(x) = \text{the greatest member of $Y$ that exactly divides $x$}. \end{equation} Show that if $|X| > n + 1$, the n $f$ is not an injection, and deduce in this case that $X$ contains distinct numbers $x_1$ and $x_2$ such that $x_1$ is a multiple of $x_2$.
Here's what I have so far.
Define a function $X \to Y$ by the given rule. Clearly, for any $n$, $|Y|=n$, so $|X| > |Y|$, and $f$ cannot be injective by the Pigeonhole Principle, meaning that there are two elements in $X$, $x_1$ and $x_2$, such that $f(x_1) = f(x_2) = y$. If $x_1 = x_2$ for any such $x_1$ and $x_2$, $f$ would be injective, so we must be able to find $x_1$ and $x_2$ such that $x_1 \neq x_2$. Since $y$ divides both $x_1$ and $x_2$, we must have \begin{equation} x_1 = ay \ \text{and} \ x_2 = by, \end{equation} for $a, b, \in \mathbb{N}$. But $x_1$ and $x_2$ are even, by the definition of $X$, it must be the case that $a$ and $b$ are even, since $y$ are not. Hence, \begin{equation} x_1 = 2cy \ \text{and} \ x_2 = 2dy, \end{equation} for $c, d \in \mathbb{N}$.
Without sacrificing generality, take $x_1 > x_2$. Hence, $a > b$. It suffices to demonstrate that $b | a$.
This is the point at which I am unable to complete the argument. The solutions manual frames the argument quite a bit differently, arguing that we can write $x_1 = 2^{m_1} y$ and $x_2 = 2^{m_2} y$ for naturals $m_1, m_2$. I do not understand why we can write $x_1$ and $x_2$ with powers of $2$, instead of multiples. The one assumption I haven't used is that $y$ is odd: substituting in some form of $2j - 1$ for $y$ doesn't seem to yield much good, though.
Any help with this would be greatly appreciated.
