I have a rather conceptual question about showing certain small lemmas regarding the absolute value function on $\mathbb{Q}$. I want to only give one example:
Let $a,b \in \mathbb{Q}$ and $|.|$ denote the absolute value function, show that $$|a| \geq 0 \text{ (and } |a| =0 \iff a =0 )$$
So in order to tackle this problem, here are the definitions. We defined the rational numbers through the equivalence classes of all tuples $(x,y), (u,v) \in \mathbb{Z} \times \mathbb{Z} \setminus \lbrace 0 \rbrace$ such that $$(x,y) \sim (u,v) : \iff x v = yu $$
So for a rational number we can define $$\frac{a}{b}:= [a,b]_{\sim_\mathbb{Q}} \text{ for } (a,b) \in \mathbb{Z} \times \mathbb{Z} \setminus \lbrace 0 \rbrace $$
Finally we have the absolute value function defined as for $\alpha \in \mathbb{Z}$ $$ | \alpha | := \begin{cases} \alpha \text{ if } \alpha \in \mathbb{Q}_{+} \\ - \alpha \text{ else} \end{cases}$$ Where $\mathbb{Q}_+:= \lbrace m/n \mid m,n \in \mathbb{N} \text{ with } n \neq 0\rbrace$
Note: In class we did not define what $- \alpha$ means, I suppose it makes most sense to set for $\alpha=[a,b]$ with $(a,b) \in \mathbb{Z} \times \mathbb{Z} \setminus \lbrace 0 \rbrace$ that $-\alpha := [-a,b]$ because now $\alpha + (-\alpha)=0_{\mathbb{Q}}$. I hope that this is correct.
Finally:
My approach: I suppose I have to work with all the definitions in order to rigorously show the statement. Hence let $\alpha= [a,b]$ with $(a.b) \in \mathbb{Z} \times \mathbb{Z} \setminus \lbrace 0 \rbrace$.
Case 1: Assume that $\alpha \in \mathbb{Q_+}$ then we know that $|\alpha|=\alpha$ and that $a \in \mathbb{N}$ and $b \neq 0$. $$\alpha \geq0 \text{ means } [a,b] \geq [0,1] \iff a \cdot 1 \geq b \cdot 0 \\ \iff a \geq 0 \text{ which is true} $$
Case 2: Assume that $\alpha \notin \mathbb{Q}_+$, so we have $|\alpha|=-\alpha=[-a,b]$ $$|\alpha| \geq 0 \iff -\alpha \geq 0 \iff [-a,b] \geq [0,1] \iff -a \cdot 1 \geq b \cdot 0 \\ \iff a \leq 0 $$ Which is not necessarily true.
2nd approach: Let $\alpha = [a,b]$ and $\gamma = [c,d]$ again with b,d non zero of course. and a,b,c,d being integers. Thus I can say $$\alpha \leq \gamma \iff [a,b] \leq [c,d] \iff ad \leq bc \iff -ad \geq -bc \\ \iff [-a,b] \geq [-b,c] \iff - \alpha \geq - \gamma $$ Which seems to be possible because I am working over $\mathbb{Z}$ thus I can say.
If $\alpha \geq 0 \implies |\alpha| = \alpha > 0$ and similarly
if $\alpha \leq 0 \implies -\alpha \geq 0$ but $|\alpha| = -\alpha \geq 0$
Better?
If you are having any difficulties, step back and define the absolute value for $ \mathbb{Z}$ in the same way. Then check your list of properties there. If it is true there, then note that when selecting your rational number representative,
$[a,b]_{\sim_\mathbb{Q}}$
you can always take $b$ to be positive. With this choice, you will see that using the definition of absolute value, all the 'action' occurs in the numerator $a$.