User:Espen180/Defining the Rationals

Let $$\mathbb{Z}$$ be the set of integers. We will consider the set $$Q=\{(a,b)\in \mathbb{Z}\times\mathbb{Z}\mid b\neq 0\} \subseteq \mathbb{Z}\times\mathbb{Z}$$. This set will be our "proto-rationals". Define the equivalence relation on $$Q$$ given by $$(a,b)\sim (c,d) \,\Leftrightarrow ad=bc$$. That this is an equivalence relation is obvious. To each equivalence class there is a set of equivalence classes. It is these equivalence classes which will constitute the rational numbers.

Let $$\mathbb{Q}=Q/\sim$$ and call $$\mathbb{Q}$$ the set of rational numbers. We will show that this $$\mathbb{Q}$$ has all the familiar properties of rationals. However, since we defined it using equivalence classes, we will have to show invariance with respect to the representative element at each step. Denote the equivalence class of $$(a,b)\in Q$$ by $$[a,b]$$. Before we begin, define the following operations on $$Q$$:


 * i) Addition: If $$(a,b),(c,d)\in Q$$, define $$(a,b)+(c,d)=(ad+bc,bd)$$.


 * ii) Multiplication: If $$(a,b),(c,d)\in Q$$, define $$(a,b)(c,d)=(ac,bd)$$.

$$Q$$ is obviously closed under these operations.


 * 1. Let $$(a,b)\in [x,y]$$ and $$(c,d)\in [z,w]$$. Then $$(a,b)(c,d)=(ac,bd)\in [xz,yw]$$.

Proof: Since $$ay=bx$$ and $$cw=dz$$, then $$acyw-bdxz=aycw-(bx)(dz)=aycw-aycw=0$$, so $$(ac)(yw)=(bd)(xz)$$, showing the theorem.


 * 2. Let $$(a,b)\in[x,y]$$ and $$(c,d)\in[z,w]$$, then $$(a,b)+(c,d)=(ad+bc,bd)\in[xw+yz,yw]$$.

Proof: Since $$ay=bx$$ and $$cw=dz$$, then $$(ad+bc)(yw)=adyw+bcyw=(ay)dw+(cw)by=bxdw+dzby=(xw+zy)(bd)$$, showing the theorem.

Also, since the above arguments used arbitrary representatives for the equivalence classes, addition and multiplication on equivalence classes is well-defined. Thus we can transfer, or project, the operations onto $$\mathbb{Q}$$. In the following, write $$[x,y]=\frac{x}{y}$$. Then, if $$\frac{x}{y},\frac{z}{w}\in\mathbb{Q}$$, we define


 * i) their sum $$\frac{x}{y}+\frac{z}{w}=\frac{xw+yz}{yw}\in\mathbb{Q}$$, and


 * ii) their product as $$\frac{x}{y}\cdot\frac{z}{w}=\frac{xz}{yw}\in\mathbb{Q}$$.

Under these operations it can readily be checked that $$\mathbb{Q}$$ is a commutative ring. It has additive identity $$\frac{0}{1}$$ and multiplicative identity $$\frac{1}{1}$$, written $$0$$ and $$1$$, respectively. In general, we write any element $$\frac{a}{1}\in \mathbb{Q}$$ simply as $$a$$.

We will now show that it is a field.

Let $$\frac{x}{y}\in\mathbb{Q}$$ such that $$x\neq 0$$. Then there exists an element $$\frac{y}{x}\in \mathbb{Q}$$ such that $$\frac{x}{y}=\frac{y}{x}=\frac{xy}{xy}=\frac{1}{1}=1$$, the multiplicative identity, showing that multiplicative inverses exist for nonzero elements, and therefore that it is a field, as promised. With this we have shown that $$\mathbb{Q}$$ has all the properties of the rationals and therefore is the set of all rational numbers.