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?

  • $\begingroup$ Show that $(a,b)\equiv (-a,-b)$, and thus that $[a,b]=[-a,-b]$. What's the problem? $\endgroup$ – Arturo Magidin Nov 27 '18 at 22:07
  • $\begingroup$ @ArturoMagidin how should I start to prove this? $\endgroup$ – NewMoon Nov 27 '18 at 22:09
  • $\begingroup$ @ArturoMagidin It maybe clear but I couldn't start to prove $\endgroup$ – NewMoon Nov 27 '18 at 22:10
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    $\begingroup$ 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. $\endgroup$ – Arturo Magidin Nov 27 '18 at 22:11
  • $\begingroup$ 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 $b\in\mathbb{Z}\setminus\{0\}$. $\endgroup$ – José Carlos Santos Nov 27 '18 at 22:11

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