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.


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 $|\alpha| = -\alpha \geq 0$


  • $\begingroup$ Could you show us a more complicated example that doesn't have a simple proof like this one? $\endgroup$ – barrycarter Nov 3 '14 at 19:58
  • $\begingroup$ @barrycarter, why? Apparently I cannot show this simple proof, see Case 2 where I get stuck. In general I am looking for ideas how to make case 2 work and then abstract the remaining to do from there. $\endgroup$ – Spaced Nov 3 '14 at 20:12
  • $\begingroup$ OK, I might be misunderstanding the question, but if |a|=0 then a=+0 or a=-0, which are the same thing. I don't see this as a rational number question. It's true for natural numbers, integers, real numbers, and complex numbers as well. $\endgroup$ – barrycarter Nov 3 '14 at 20:23
  • $\begingroup$ @barrycarter, yes but it is part of this exercise to show this precise statement for rational numbers. If you see something I don't see ignore the AND in the question (for readability purposes I will put brackets around it) and look for the positive definiteness of |a| $\endgroup$ – Spaced Nov 3 '14 at 20:27
  • $\begingroup$ @Spaced Yes this is a nice exercise, to prove the properties of $\lvert \cdot \rvert$ from the definition of the rationals. Their definition is based on pairs of integers, some of which can be negative. So '-' is induced from usual negation of integers. You might have to use that fact at some point. $\endgroup$ – user4894 Nov 3 '14 at 20:35

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,


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 $a$.


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