Equivalence Relation on Natural Numbers such that Quotient is Power of Two

From ProofWiki
Jump to navigation Jump to search

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

\(\displaystyle x\) \(\alpha\) \(\displaystyle y\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle 2^n y\) for some $n \in \Z$
\(\displaystyle \leadsto \ \ \) \(\displaystyle y\) \(=\) \(\displaystyle 2^{-n} x\) dividing both sides by $2^{-n}$
\(\displaystyle \leadsto \ \ \) \(\displaystyle y\) \(\alpha\) \(\displaystyle x\) 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 defintion:

\(\displaystyle y\) \(=\) \(\displaystyle 2^m z\) for some $m \in \Z$
\(\displaystyle x\) \(=\) \(\displaystyle 2^n y\) for some $n \in \Z$
\(\displaystyle \) \(=\) \(\displaystyle 2^n \paren {2^m z}\)
\(\displaystyle \) \(=\) \(\displaystyle 2^{n + m} z\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(\alpha\) \(\displaystyle z\) 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