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

Question

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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 $$v\pmatrix{\frac{a}{b}}=f(a)-f(b).$$

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QUESTION:

Is $v$ well-defined?

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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 $X_{/\sim}$.

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Answer 1

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 $$\frac{a}{b}$$

Answer 2

$v$ 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.

Answer 3

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 ($\mathbb Q$).