# An Equivalence Relation: Introspection into a Particular Well-Defined Quotient

## DATA:

Let $f:\mathbb{Z}\setminus \{0\}\rightarrow \mathbb{N}$ be a function defined by $$f(n) = \{k~:~n=2^km,~m\in \cal{O}\},$$ where $\cal{O}$ is the set of odd integers.

Let $v:\mathbb{Q}\setminus \{0\}\rightarrow \mathbb{Z}$ be a function defined by $$v\pmatrix{\frac{a}{b}}=f(a)-f(b).$$

## QUESTION:

Is $v$ well-defined?

## KNOWN:

Let $X$ be a set and $\sim$ be an equivalence relation on $X$. Let $f:X\rightarrow Y$. If $\forall x,x'\in X$ we have that $x\sim x' \implies f(x)=f(x')$, then $f$ defines a function $X_{/\sim}\rightarrow Y$ by $[x] \mapsto f(x)$. In this case, we say $f$ is "well defined" on the quotient $X_{/\sim}$.

• The function $f$ (as currently defined) is a function to the power set $\mathcal{P}(\mathbb{N})$, not $\mathbb{N}$ itself. Commented Jul 15, 2013 at 0:41
• @ZevChonoles But the set is a singleton... so it kinda makes sense. A slight modification is necessary.
– Pedro
Commented Jul 15, 2013 at 0:42
• @Peter: Certainly, I understand what's intended; just wanted to point out an improvement that Trancot should make to the question. Commented Jul 15, 2013 at 0:46
• @Trancot: What equivalence relation are you referring to? Commented Jul 15, 2013 at 0:47
• @ZevChonoles I think I'm referring to the equivalence relation over $\mathbb{Q}$ that requires that $1/2$ is related to $2/4$ and so on, right? Commented Jul 15, 2013 at 1:05

$v$ is well defined if it's independant of the representation of the fraction $\frac{a}{b}$

Let $a=2^km$ and $b=2^lm'$ and $d=2^s m''$ then $$v\left(\frac{a}{b}\right)=f(a)-f(b)=k-l$$ and $$v\left(\frac{da}{db}\right)=f(da)-f(db)=(k+s)-(l+s)=k-l$$ so $v$ is well defined.

• What do you mean by "independent of the representation of"? Commented Jul 15, 2013 at 1:00
• We can write the fraction $\frac{a}{b}$ in the form $\frac{ka}{kb}$ for $k\neq 0$ so if $v(\frac{a}{b})\neq v(\frac{ka}{kb}))$ then $v$ isn't well defined.
– user63181
Commented Jul 15, 2013 at 1:02
• So this is not well-defined? Commented Jul 15, 2013 at 1:03
• Just to be clear, for future people looking here, what are $l, s$, etc.? Commented Jul 15, 2013 at 1:15
• $l,s$, etc are known from the context and are defined as shown in you question.
– user63181
Commented Jul 15, 2013 at 1:41

Recall that we obtain $\Bbb Q$ by quoting the set $\Bbb Z\times (\Bbb Z-\{0\})$ with the equivalence relation $$(a,b)\sim (a',b')\iff ab'=a'b$$

This hints that we should see $v$ as a map $$\nu:\Bbb Z\times (\Bbb Z-\{0\})\to\Bbb N$$ defined as $$\nu(a,b)=f(a)-f(b)$$

and we ought to prove (or disprove) that $ab'=a'b\implies \nu(a,b)=\nu(a',b')$.

Note that if $m$ is odd, $$\nu(mn,mk)=\nu(n,k)$$ since $\text{odd}\times \text{odd}=\text{odd}$. Similarily, if $m=2^j$ is even, $$\nu(2^jn,2^jk)=j+f(n)-(j+f(k))=f(n)-f(k)=\nu(n,k)$$

Since this considers all possible alterations on the pair $n,k$, we conclude $\nu$ is well-defined.

OBS $\nu(a,b)$ simply returns the exponent of $2$ (negative or positive) in $$\frac{a}{b}$$

• May I quote you? ;-) Commented Jul 15, 2013 at 0:52
• @amWhy Feel like choping people into disjoint pieces?
– Pedro
Commented Jul 15, 2013 at 0:54
• chopping Not at all...just thought someone needed to ask ;-) Commented Jul 15, 2013 at 0:55
• @amWhy I don't understand!
– Pedro
Commented Jul 15, 2013 at 1:08

Suppose that $a/b=c/d$ $\Leftrightarrow$ $ad=bc$. Write $a=2^mm'$, $b=2^nn'$, $c=2^pp'$, $d=2^qq'$, where $m',n',p',q'$ are odd integers. We have $2^{m+q}m'q'=2^{n+p}n'p'$. From the uniqueness of the decomposition of integers as a product of prime numbers (up to a sign) we get $m+q=n+p$, and therefore $v(a/b)=m-n=p-q=v(c/d)$.

Remark. This is the usual way to extend a valuation function on an integral domain (in this case $\mathbb Z$) to a valuation (the $2$-adic valuation) on its field of fractions ($\mathbb Q$).