0
$\begingroup$

Consider set $T$, a subset of $\mathbb{Q}$, structured as functions in the form of a quotient. Specifically,

$$T=\left\{\frac{N(m)}{D(n)}\right\}$$

Where $N$ and $D$ are functions defined on integers that allow $T$ to be a subset of $\mathbb{Q}$.

The problem is, for chosen values of $n$ and $m$, the elements of $T$ are not always fully reduced fractions. For example, in $T=\left\{\frac{m}{n^2+n+5}\right\}$, the element $\left\{\frac{5}{4^2+4+5}\right\}\in T$ but $\gcd(5,25)\neq 1$

I want split $T$ into a countable union of finite sets defined for only relatively prime numerator and denominator values. For example,

$\left\{\frac{m}{6n+1}\right\}=\bigcup\limits_{n\in\mathbb{Z}}\left\{\left.\frac{m}{6n+1}\right|\gcd(m,6n+1)=1\right\}\bigcup\limits_{n\in\mathbb{Z}}\left\{\left.\frac{m}{6n+5}\right|\gcd(m,6n+5)=1\right\}=\left\{\left.\frac{m}{6n+1}\right|\gcd(m,6n+1)=1\right\}\bigcup\left\{\left.\frac{m}{6n+5}\right|\gcd(m,6n+5)=1\right\}$

Inorder to do so I must find a way to list the numerator and denominator of simplified fractions in $T$ from the outputs of functions $D$ and $M$.

I determined the denominators of all reduced fractions in $T$ by listing the divisors of all outputs of $D$. This is true for all the simplified elements as long as the numerator and denominator of $T$ do not share common factors, such as $\left\{\frac{m^2}{n^2}\right\}=\left\{\left(\frac{m}{n}\right)^{2}\right\}$.

The list has non-repeated divisors organized from least to greatest. Mathematically I will show this as $\bigcup\limits_{z\in\mathbb{N}}d_{n}^{+}[D(n)]$, where $d_z^{+}[x]$ takes the zth smallest positive and non-repeated divisor of the outputs of $x$.

For example, if $T=\left\{\frac{m}{2^n+n}\right\}$, since the numerator and denominator do not share common factors and $D(n)=2^n+n$, the denominator values of reduced fractions in $T$ is

$$\bigcup_{z\in\mathbb{N}}d_z^{+}[2^n+n]=\left\{1,2,3,4,5,6,7,8,...\right\}$$

We get this list by doing the following. Say we take the first positive outputs of $2^n+n$

$$\left\{1,3,6,11,20,37,70,...\right\}$$

The divisors of each output are repesented in brackets

$$\left\{\left\{1\right\},\left\{1,3\right\},\left\{1,3,2,6 \right\}, \left\{1,11\right\},\left\{20,1,4,5,10,2\right\},\left\{1,37\right\},\left\{1,70,2,35,7,10\right\}....\right\}$$

We can rearrange these divisors, without repeated values, from least to greatest

$$\left\{1,2,3,4,5,6,7,10,11,20,35,37...\right\}$$

The more outputs taken, the closer the list of divisors becomes $\mathbb{N} \ \backslash \left\{0\right\}$

$$\left\{1,2,3,4,5,6,7,8,9,10,11...\right\}$$

Hence $d_{1}^{+}[2^n+n]=1$, $d_{2}^{+}[2^n+n]=2$, $d_{3}^{+}[2^n+n]=3$ and so on.

Infact, I generalized

$\left\{\frac{m}{D(n)}\right\}=\bigcup\limits_{z\in\mathbb{N}}\left\{\left.\frac{m}{d_z^{+}[D(n)]}\right|\gcd{\left(m,d_z^{+}[D(n)]\right)}=1\right\}$

When $N(m)$ is linear, I am able to determine the numerator values of simplfied elements in $T$ by listing divisors of the outputs of $N$.

$\left\{\frac{am+b}{D(n)}\right\}=\bigcup\limits_{z\in\mathbb{N}}\bigcup\limits_{s\in\mathbb{Z}}\left\{\left.\frac{d_s^{+}[am+b]}{d_z^{+}[D(n)]}\right|\gcd{\left(d_s^{+}[am+b],d_z^{+}[D(n)]\right)}=1\right\}$

This leads to equations such as

$\left\{\frac{4m+1}{2^n+n}\right\}=\left\{\left.\frac{2m+1}{n}\right|\gcd{(2m+1,n)=1}\right\}$

$\left\{\frac{m^2}{n^3}\right\}=\left\{\left.\frac{m}{n}\right|\gcd{(m,n)}=1\right\}$

$\left\{\frac{6m+1}{6n+1}\right\}=\left\{\left.\frac{6m+1}{6n+1}\right|\gcd{(6m+1,6n+1)=1}\right\}\bigcup\left\{\left.\frac{6m+5}{6n+5}\right|\gcd{(6m+5,6n+5)=1}\right\}$

$\left\{\frac{4m+1}{n^2+n+1}\right\}=\bigcup\limits_{z\in\mathbb{Z}}\left\{\left.\frac{4m+1}{\text{d}_{z}^{+}[n^2+n+1]}\right|\gcd(4m+1,\text{d}_{z}^{+}[n^2+n+1])=1\right\}$

The problem is when $N(m)$ is not linear, I cannot take non-repeated divisors of $N$ to determine the numerator values of reduced elements in $T$.

For example, if we set $T=\left\{\frac{m^2+1}{n^2+n+1}\right\}$, the element $\left\{29/7\right\}\not\in T$; however, 29 is a divisor of an output of $N(m)=m^2+1$. In other words, $d_9[m^2+1]=29$ but $\left\{29/7\right\}\not\in T$

In conclusion, is there a way of determining the numerator of reduced fractions of any set $T$ from the outputs of non-linear $N$?

$\endgroup$
  • $\begingroup$ What is your concrete problem and functions $D(n),N(m)$ ? $\endgroup$ – reuns Aug 15 '17 at 22:14
  • $\begingroup$ I want to split $T$ into a collection of finite sets with relatively prime numerator and denominator values. The finite sets can be combined into countalby infinite sets in the examples I listed.The functions $D(n)$ and $N(m)$ can be any function that allows $T\subseteq\mathbb{Q}$ $\endgroup$ – Arbuja Aug 15 '17 at 22:18
  • $\begingroup$ In general there is not much more to say. So what is your concrete problem, what are your concrete functions $N(m),D(n)$ ? $\endgroup$ – reuns Aug 15 '17 at 22:25
  • $\begingroup$ Are you still considering $(N(m),D(n)) = (am+b, cn+d)$ and things related ? What is your concrete problem and functions ? $\endgroup$ – reuns Aug 15 '17 at 22:48
  • $\begingroup$ @reuns I looked up concrete problems. I am still unclear what it means mathematically. However, I'll consider $(am^2+bm+c, dn^2+ex+f)$ and $(a^x,b^x)$. $\endgroup$ – Arbuja Aug 16 '17 at 2:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.