Equivalence Relation on Natural Numbers such that Quotient is Power of Two/Equivalence Class Contains 1 Odd Number

Theorem

Let $\alpha$ denote the relation defined on the natural numbers $\N$ by:

$\forall x, y \in \N: x \mathrel \alpha y \iff \exists n \in \Z: x = 2^n y$

We have that $\alpha$ is an equivalence relation.

Let $\eqclass n \alpha$ be the $\alpha$-equivalence class of a natural number $n$.

Then $\eqclass n \alpha$ contains exactly $1$ odd number.

Let $x$ be a natural number whose $\alpha$-equivalence class is $\eqclass x \alpha$.

If $x$ is odd, then $\eqclass x \alpha$ contains that odd number $x$

We have that $x$ is of the form $x = 2^r y$ for some $r \in \Z$ where $y$ is an odd number.

Thus by definition $y \in \eqclass x \alpha$ is an odd number which is contained in $\eqclass x \alpha$.

Thus it has been shown that an arbitrary $\eqclass x \alpha$ contains at least one odd number.

It remains to be demonstrated that $\eqclass x \alpha$ contains no more than $1$ odd number.

Aiming for a contradiction, suppose $\eqclass n \alpha$ contains the odd numbers $p$ and $q$ such that $p \ne q$.

Then:

$p = 2^n q$

for some $n \in \Z$

If $n = 0$ then $p = q$ which contradicts $q \ne p$.

Thus $n \ne 0$.

Let $n > 0$.

Then:

$p = r q$

for $r \in \Z$ such that $r = 2^n$.

Thus $p$ is has $2$ as a divisor.

This contradicts the supposition that $p$ is odd.

Let $n < 0$.

Then:

$p = \dfrac q {2^m}$

where $m = -n$ and so $m > 0$.

Thus:

$q = r p$

for $r \in \Z$ such that $r = 2^m$.

Thus $q$ has $2$ as a divisor.

This contradicts the supposition that $q$ is odd.

Thus all cases of $n$ lead to a contradiction.

It follows that no $\alpha$-equivalence class can contain more than one odd number.

$\blacksquare$

1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 6$. Indexed families; partitions; equivalence relations: Exercise $8$