Set Theory

The Rational Numbers

Table of contents

Definitions:

Let \(\{(x_{0},y_{0}),(x_{1},y_{1})\}\subset\mathbb{Z}\times(\mathbb{Z}\setminus\{0\})\). Define:

1. \((x_{0},y_{0})+(x_{1},y_{1}):=(x_{0}\cdot y_{1}+x_{1}y_{0},y_{0}\cdot y_{1})\).
2. \((x_{0},y_{0})\cdot(x_{1},y_{1}):=(x_{0}\cdot x_{1},y_{0}\cdot y_{1})\).
3. If \(x_{0}\neq0\), then \((x_{0},y_{0})^{-1}:=(y_{0},x_{0})\).

Claim:

The relation \(\backsimeq:=\left\{ ((x_{0},y_{0}),(x_{1},y_{1})):x_{0}\cdot y_{1}=y_{0}\cdot x_{1}\right\}\) is an equivalence relation on \(\mathbb{Z}\times(\mathbb{Z}\setminus\{0\})\).

Proof:

Let \(\left\{(x_{0},y_{0}),(x_{1},y_{1}),(x_{2},y_{2})\right\}\subset\mathbb{Z}\times(\mathbb{Z}\setminus\{0\})\).

• Reflexivity: \(x_{0}\cdot y_{0}=y_{0}\cdot x_{0}\).
• Symmetry: \(x_{0}\cdot y_{1}=y_{0}\cdot x_{1}\Longleftrightarrow x_{1}\cdot y_{0}=y_{1}\cdot x_{0}\).
• Transitivity: if \(x_{0}\cdot y_{1}=y_{0}\cdot x_{1}\) and \(x_{1}\cdot y_{2}=y_{1}\cdot x_{2}\), then \[ \begin{align*} x_{0}\cdot y_{1}\cdot x_{1}\cdot y_{2} &=y_{0}\cdot x_{1}\cdot y_{1}\cdot x_{2}\\ x_{0}\cdot y_{2}&=y_{0}\cdot x_{2}. \end{align*} \]

Claims:

Let \(\{p,q\}\subset\mathbb{Z}\times(\mathbb{Z}\setminus\{0\})\), and let \(a\backsimeq p,b\backsimeq q\). Then:

1. \(a+b\backsimeq p+q\).
2. \(a\cdot b\backsimeq p\cdot q\).

Proofs:

Write \(p=(p_{0},p_{1}),q=(q_{0},q_{1}),a=(a_{0},a_{1}),b=(b_{0},b_{1})\). We are given that

\[ \begin{align*} a_{0}p_{1}&=p_{0}a_{1}\\ b_{0}q_{1}&=q_{0}b_{1}. \end{align*} \]

1. \[ \begin{align*} \left(a_{0}b_{1}+a_{1}b_{0}\right)p_{1}q_{1} &=a_{0}b_{1}p_{1}q_{1}+a_{1}b_{0}p_{1}q_{1}\\ &=(a_{0}p_{1})b_{1}q_{1}+(b_{0}q_{1})a_{1}p_{1}\\ &=(a_{1}p_{0})b_{1}q_{1}+(b_{1}q_{0})a_{1}p_{1}\\ &=\left(p_{0}q_{1}+p_{1}q_{0}\right)a_{1}b_{1}. \end{align*} \] Therefore \(\frac{a_{0}b_{1}+a_{1}b_{0}}{a_{1}b_{1}}\backsimeq\frac{p_{0}q_{1}+p_{1}q_{0}}{p_{1}q_{1}}\), and hence \(a+b\backsimeq p+q\).
2. Trivial.

Definition:

We denote the set \((\mathbb{Z}\times(\mathbb{Z}\setminus\{0\}))/\backsimeq\) by \(\mathbb{Q}\), and call it the rational numbers, or the rationals for short. We denote the class \([(x,y)]_{\backsimeq}\) by \(\frac{x}{y}\) or by \(\nicefrac{x}{y}\).

Definition:

Let \(q\in\mathbb{Q}\). Then \(q\) is called a rational number, or a rational for short.

Definitions:

Let \(\{\frac{x_{0}}{y_{0}},\frac{x_{1}}{y_{1}}\}\subset\mathbb{Q}\) be rational numbers. Then:

1. \(\frac{x_{0}}{y_{0}}+\frac{x_{1}}{y_{1}}:=\frac{x_{0}y_{1}+x_{1}y_{0}}{y_{0}y_{1}}\).
2. \(\frac{x_{0}}{y_{0}}\cdot\frac{x_{1}}{y_{1}}:=\frac{x_{0}x_{1}}{y_{0}y_{1}}\).
3. \(-\frac{x_{0}}{y_{0}}:=\frac{-x_{0}}{y_{0}}\).

Corollaries:

Let \(\frac{x}{y}\in\mathbb{Q}\), and let \(z\in\mathbb{Z}\setminus\{0\}\). Then:

1. \(\frac{x}{y}=\frac{z\cdot x}{z\cdot y}\).
2. \(-\frac{x}{y}=\frac{-x}{y}=\frac{x}{-y}\).
3. \(\frac{x}{y}=\frac{-x}{-y}\).

Corollary:

Let \(\{x,y\}\subset\mathbb{Z}\setminus\{0\}\). Then \(\frac{0}{x}=\frac{0}{y}\).

Corollary:

Let \(q\in\mathbb{Q}\). Then there exists \(x\in\mathbb{Z}\) and \(n\in\mathbb{N}\setminus\{0\}\) such that \(\frac{x}{n}=q\).

Corollaries:

Let \(\{x,y,z\}\subset\mathbb{Q}\). Then:

1. \(x+y=y+x\).
2. \((x+y)+z=x+(y+z)\), and we may write \(x+y+z\) briefly.
3. \(x+0=x\).
4. \(x-x=0\).
5. \(x\cdot y=y\cdot x\).
6. \((x\cdot y)\cdot z=x\cdot(y\cdot z)\), and we may write \(x\cdot y\cdot z\) briefly.
7. \(x\cdot0=0\).
8. \(x\cdot1=x\).
9. \(x\cdot(-1)=-x\).
10. \((-x)\cdot(-y)=x\cdot y\).
11. \((-x)\cdot y=x\cdot(-y)=-(x\cdot y)\).
12. \(x\cdot(y+z)=x\cdot y+x\cdot z\).

Definition:

\(\le:=\left\{(\frac{x_{0}}{y_{0}},\frac{x_{1}}{y_{1}})\in\mathbb{Q}\times\mathbb{Q}:(x_{0}\cdot y_{1}-x_{1}\cdot y_{0})(y_{0}\cdot y_{1})\le0\right\}\).

Claim:

\(\le\) is a total order relation on the rationals.

Proof:

Let \(\{\frac{x_{0}}{y_{0}},\frac{x_{1}}{y_{1}},\frac{x_{2}}{y_{2}}\}\subset\mathbb{Q}\).

• Reflexivity: \((x_{0}\cdot y_{0}-x_{0}\cdot y_{0})(y_{0}\cdot y_{0})=0\le0\).
• Antisymmetry: assume \((x_{0}\cdot y_{1}-x_{1}\cdot y_{0})(y_{0}\cdot y_{1})\le0\) and \((-x_{0}\cdot y_{1}+x_{1}\cdot y_{0})(y_{0}\cdot y_{1})\le0\). Then \(-(x_{0}\cdot y_{1}-x_{1}\cdot y_{0})(y_{0}\cdot y_{1})\le0\), so \((x_{0}\cdot y_{1}-x_{1}\cdot y_{0})(y_{0}\cdot y_{1})=0\). Hence \(x_{0}\cdot y_{1}=x_{1}\cdot y_{0}\), so \(\frac{x_{0}}{y_{0}}=\frac{x_{1}}{y_{1}}\).
• Transitivity: assume \((x_{0}\cdot y_{1}-x_{1}\cdot y_{0})(y_{0}\cdot y_{1})\le0\) and \((x_{1}\cdot y_{2}-x_{2}\cdot y_{1})(y_{1}\cdot y_{2})\le0\). Then \[ \begin{align*} (x_{0}\cdot y_{1}\cdot y_{2}-x_{1}\cdot y_{0}\cdot y_{2} +x_{1}\cdot y_{0}\cdot y_{2}-x_{2}\cdot y_{0}\cdot y_{1}) (y_{0}\cdot y_{1}\cdot y_{2})&\le0\\ (x_{0}\cdot y_{2}-x_{2}\cdot y_{0})(y_{0}\cdot y_{2})&\le0. \end{align*} \] Hence \(\frac{x_{0}}{y_{0}}\le\frac{x_{2}}{y_{2}}\).
• Totality: if \((x_{0}\cdot y_{1}-x_{1}\cdot y_{0})(y_{0}\cdot y_{1})\le0\), then \(\frac{x_{0}}{y_{0}}\le\frac{x_{1}}{y_{1}}\). Otherwise \(0<(x_{0}\cdot y_{1}-x_{1}\cdot y_{0})(y_{0}\cdot y_{1})\), hence \((-x_{0}\cdot y_{1}+x_{1}\cdot y_{0})(y_{0}\cdot y_{1})<0\), so \(\frac{x_{1}}{y_{1}}<\frac{x_{0}}{y_{0}}\).

Definition:

Let \(\{p,q\}\subset\mathbb{Q}\). If \(p\le q\) and \(p\neq q\), we write \(p<q\).

Corollaries:

Let \(\{x,y\}\subset\mathbb{Z}\). Then:

1. \(\frac{x}{1}+\frac{y}{1}=\frac{x+y}{1}\).
2. \(\frac{x}{1}\cdot\frac{y}{1}=\frac{x\cdot y}{1}\).
3. \(x\le y\Longleftrightarrow\frac{x}{1}\le\frac{y}{1}\).

Corollary:

Let \(q\in\mathbb{Q}\) with \(0<q\). Then there exist \(\{m,n\}\subset\mathbb{N}\setminus\{0\}\) such that \(\frac{m}{n}=q\).

Claims:

Let \(\{x,y,z\}\subset\mathbb{Q}\). Then:

1. \(x<y\Longleftrightarrow x+z<y+z\).
2. If \(z>0\), then \(x<y\Longleftrightarrow x\cdot z<y\cdot z\).
3. If \(z<0\), then \(x<y\Longleftrightarrow y\cdot z<x\cdot z\).

Proof:

Let \(x=\frac{a}{b},y=\frac{c}{d},z=\frac{e}{f}\), where \(b,d,f>0\). By definition of the order, \(x<y\Longleftrightarrow ad<cb\).

1. \(x+z=\frac{af+eb}{bf}\), and \(y+z=\frac{cf+ed}{df}\). Therefore \(x+z<y+z\Longleftrightarrow(af+eb)df<(cf+ed)bf\), that is, \(adf^{2}+ebdf<cbf^{2}+edbf\). Since \(ebdf=edbf\), we get \(adf^{2}<cbf^{2}\). Since \(f^{2}>0\), this is equivalent to \(ad<cb\), and therefore \(x+z<y+z\Longleftrightarrow x<y\).
2. Assume \(z>0\). Since \(f>0\), \(e>0\). Thus \(xz=\frac{ae}{bf}\) and \(yz=\frac{ce}{df}\). Hence \(xz<yz\Longleftrightarrow aedf<cebf\), that is, \(ad\cdot ef<cb\cdot ef\). Since \(ef>0\), we get \(ad<cb\), and therefore \(xz<yz\Longleftrightarrow x<y\).
3. Assume \(z<0\). Since \(f>0\), \(ef<0\), so \(e<0\). As before, \(yz<xz\Longleftrightarrow cebf<aedf\), that is, \(cb\cdot ef<ad\cdot ef\). Since \(ef<0\), multiplying by \(ef\) reverses the inequality, so this is equivalent to \(ad<cb\). Therefore \(yz<xz\Longleftrightarrow x<y\).

Notation:

Let \(x\in\mathbb{Z}\) be an integer. From now on we will no longer distinguish between \(x\) and \(\frac{x}{1}\). For all practical purposes, we assume \(\mathbb{Z}\subset\mathbb{Q}\).

Claim:

Let \(\{x,y\}\subset\mathbb{Z}\setminus\{0\}\), write \(n:=\gcd(x,y)\), and let \(\{a,b\}\subset\mathbb{Z}\setminus\{0\}\) satisfy \(an=x\) and \(bn=y\). Then \(\gcd(a,b)=1\).

Proof:

Write \(m:=\gcd(a,b)\). Then there exist \(\{p,q\}\subset\mathbb{Z}\setminus\{0\}\) such that \(pm=a\) and \(qm=b\). Hence \(pmn=x\) and \(qmn=y\). Since \(mn\mid x\) and \(mn\mid y\), by the definition of \(\gcd\), \(mn\mid n\). Therefore \(mn\le n\), and hence \(m=1\).

Corollaries (Canonical Representation of the Rationals):

Let \(q\in\mathbb{Q}\).

1. If \(0<q\), then there exist unique \(\{m,n\}\subset\mathbb{N}\setminus\{0\}\) such that \(\gcd(m,n)=1\) and \(\frac{m}{n}=q\).
2. If \(q<0\), then there exist unique \(\{m,n\}\subset\mathbb{N}\setminus\{0\}\) such that \(\gcd(m,n)=1\) and \(-\frac{m}{n}=q\).