# A question construction of rational numbers

In $$\mathbb{Z}$$ we can do addition, substraction and multiplication, but not division. For example we cannot divide $$2$$ by $$3$$ (i.e. the equation $$3x=2$$ has no solution in $$\mathbb{Z}$$, because $$\dfrac{2}{3}$$ is not in $$\mathbb{Z}$$. To remedy this situation, we will extend the structure $$\mathbb{Z}$$ to a larger set where not only we can do addition, substraction and multiplication compatible with the similar operations in $$\mathbb{Z}$$, but also division.

We would like to invent numbers that will stand for $$\dfrac{2}{3}$$ e.g. If we attempt to denote $$\dfrac{2}{3}$$ by the pair $$(2,3)$$ then we should we have $$(2,3)=(4,6)$$ and as we know this is not the case. We overcome this difficulty by noting that $$\dfrac{a}{b}=\dfrac{c}{d}$$ if and only if $$ad=bc$$. Accordingly, On set $$\mathbb{Z}\times(\mathbb{Z}\setminus\left\{0\right\})$$, we define the following relation:

$$(x,y)\equiv (z,t)\iff xt=yz$$

Theorem 1. Show that this is an equivalence relation. Define $$\mathbb{Q}=\mathbb{Z}\times(\mathbb{Z}\setminus\left\{0\right\})/_\equiv$$.

Theorem 2. Let $$(x,y),(z,t),(x',y'),(z',t')\in\mathbb{Z}\times(\mathbb{Z}\setminus\left\{0\right\})$$. If $$(x,y)\equiv (x',y')$$ and $$(z,t)\equiv (z',t')$$, then

$$i.$$ $$(xt+yz, yt)\equiv (x't'+y'z', y't')$$,

$$ii.$$ $$(xz,yt)\equiv (x'z',y't')$$.

Now define

$$[a,b]+[c,d]=[ad+bc, bd],$$ $$[a,b][c,d]=[ac,bd].$$

Theorem 3. Show that any $$\alpha\in\mathbb{Q}$$ can be written as $$[a,b]$$ for $$b>0$$ by taking $$[-a,-b]$$ if necessary.

My question: I showed Theorem 1 and Theorem 2 but I couldn't Theorem 3, can you help?

• Show that $(a,b)\equiv (-a,-b)$, and thus that $[a,b]=[-a,-b]$. What's the problem? – Arturo Magidin Nov 27 '18 at 22:07
• @ArturoMagidin how should I start to prove this? – NewMoon Nov 27 '18 at 22:09
• @ArturoMagidin It maybe clear but I couldn't start to prove – NewMoon Nov 27 '18 at 22:10
• How do you prove that $(a,b)\equiv (-a,-b)$? You use the definition of $\equiv$ and you verify that this pair (of ordered pairs) satisfies it. – Arturo Magidin Nov 27 '18 at 22:11
• You should start by noting that each element of $\mathbb Q$ is of the form $[a,b]$ for some $a\in\mathbb Z$ and some $b\in\mathbb{Z}\setminus\{0\}$. – José Carlos Santos Nov 27 '18 at 22:11