Equivalence Relation on Natural Numbers such that Quotient is Power of Two
Theorem
Let $\alpha$ denote the relation defined on the natural numbers $\N$ by:
- $\forall x, y \in \N: x \mathrel \alpha y \iff \exists n \in \Z: x = 2^n y$
Then $\alpha$ is an equivalence relation.
Equivalence Class Contains 1 Odd Number
Let $\eqclass n \alpha$ be the $\alpha$-equivalence class of a natural number $n$.
Then $\eqclass n \alpha$ contains exactly $1$ odd number.
Equivalence Class of Prime
Let $\eqclass p \alpha$ be the $\alpha$-equivalence class of a prime number $p$.
Then $\eqclass p \alpha$ contains no other prime number other than $p$.
Smallest Equivalence Class with no Prime
Let $\eqclass x \alpha$ denote the $\alpha$-equivalence class of a natural number $x$.
Let $r$ be the smallest natural number such that $\eqclass r \alpha$ contains no prime number.
Then $r = 9$.
One of Pair of Equivalent Elements is Divisor of the Other
Let $c, d \in \N$ such that $c \mathrel \alpha d$.
Then either:
- $c \divides d$
or:
- $d \divides c$
where $\divides$ denotes divisibility.
Proof
Checking in turn each of the criteria for equivalence:
Reflexivity
We have that for all $x \in \N$:
- $x = 2^0 x$
and as $0 \in \Z$ it follows that:
- $x \mathrel \alpha x$
Thus $\alpha$ is seen to be reflexive.
$\Box$
Symmetry
\(\ds x\) | \(\alpha\) | \(\ds y\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds 2^n y\) | for some $n \in \Z$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds 2^{-n} x\) | dividing both sides by $2^{-n}$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(\alpha\) | \(\ds x\) | as $-n \in \Z$ |
Thus $\alpha$ is seen to be symmetric.
$\Box$
Transitivity
Let $x \mathrel \alpha y$ and $y \mathrel \alpha z$.
Then by definition:
\(\ds y\) | \(=\) | \(\ds 2^m z\) | for some $m \in \Z$ | |||||||||||
\(\ds x\) | \(=\) | \(\ds 2^n y\) | for some $n \in \Z$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 2^n \paren {2^m z}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2^{n + m} z\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\alpha\) | \(\ds z\) | as $n + m \in \Z$ |
Thus $\alpha$ is seen to be transitive.
$\Box$
$\alpha$ has been shown to be reflexive, symmetric and transitive.
Hence by definition it is an equivalence relation.
$\blacksquare$
Sources
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 6$. Indexed families; partitions; equivalence relations: Exercise $8$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $3$: Equivalence Relations and Equivalence Classes: Exercise $5$