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I would like to know if the following proof for showing that $a = |a|$ implies $0 \leq a$ for all $a \in \mathbb R$ is correct, using the axioms of the inequality operation "$\leq$".

"Let $a \in \mathbb R$, and suppose that $a = |a|$. Assume by contradiction that $a \lt 0$. Then by definition, $|a| = -a$. Thus, it must be that $a = -a$. But this is false since $a \neq 0$. Thus, it must be that $0 \leq a$".

Thanks in advance.

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    $\begingroup$ You did it the "hard" way, as @Anand showed it but you are correct. $\endgroup$ – nicomezi Jul 22 '19 at 4:58
  • $\begingroup$ Precisely what tools are you expecting to be able to use? How are you defining the absolute value? What are the axioms of the inequality operation "$\le$"? Your proof looks fine to me. It is also possible that your proof is about the simplest possible (and not "the hard way"), depending on what tools you are permitting yourself to use. $\endgroup$ – Xander Henderson Jul 22 '19 at 5:14
  • $\begingroup$ His proof uses implicitely the usual definition of the absolute value ( with "$\le$"), which is the same as Anand used. Anand used that $|a| \ge 0$, which can indeed deserve more details in some context, but can be proven directly and within a line if the $\le$ axioms are considered known (which is the case here). So I disagree when you say that this could be the simplest proof since you can prove the statement using exactly the same tools with approximately the same "length" and without any logic trick such as contradiction, as Anand showed it. @XanderHenderson $\endgroup$ – nicomezi Jul 22 '19 at 8:02
  • $\begingroup$ @nicomezi The "usual" definition with which I am familiar is that $$|a| = \begin{cases} a & \text{if $a \ge 0$, and} \\ -a & \text{if $a < 0$.}\end{cases} $$ From this definition, it is not immediate that $|a| \ge 0$. This follows from having shown that if $a<0$, then $-a > 0$. If this is one of the results which the asker is permitting themselves to use, then the proof given by Anand is fine (though Anand does not seem to acknowledge that this is a result of the definition, not the definition itself). $\endgroup$ – Xander Henderson Jul 22 '19 at 13:05
  • $\begingroup$ As to a direct proof that $|a| \ge 0$ for all $a$, this is why I asked the original asker to give more context. Specifically, in this case, to edit the axioms of the "$\le$" relation into their question. The actual question being asked here is so close to the axioms that the asker really needs to make explicit exactly what tools and results are already known to them. $\endgroup$ – Xander Henderson Jul 22 '19 at 13:12
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We begin by noting that $|a|\ge 0$ and thus, $a=|a|\ge 0\implies a\ge 0$. This completes the proof.

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  • $\begingroup$ How do you know that $|a| \ge 0$? $\endgroup$ – Xander Henderson Jul 22 '19 at 5:08
  • $\begingroup$ This is the definition of $|a|$. Indeed, for any vector $v$ in any dimension, $|v|$ represents its distance from the origin and distance is always positive. To be more precise$$v=\hat{v}|v|$$ $\endgroup$ – Anand Jul 22 '19 at 5:44
  • $\begingroup$ The definition of $|a|$ is that $|a|\ge 0$? No... this is not correct. Again, how do you know that $|a| \ge 0$? The "usual" definition with which I am familiar (and which typically applies to the kinds of courses where this question might come up) is piecewise. The definition of the magnitude of a vector is probably not relevant, but even in that setting, it is necessary to show that the distance function is nonnegative (which follows from the fact that $x^2 \ge 0$ for all $x\in\mathbb{R}$). $\endgroup$ – Xander Henderson Jul 22 '19 at 13:08
  • $\begingroup$ @XanderHenderson Go by definition! $|a|$ for some real number $a$ is its distance from the origin (on the number line). $\endgroup$ – Anand Jul 22 '19 at 16:44
  • $\begingroup$ Your definition of the absolute value requires that we answer a further question, i.e. what is the "distance" between two real numbers? Now your proof requires the introduction of a metric, which seems overly complicated. So, again, you need to be careful about defining your terms. My guess is that we have $$|a| := \begin{cases}a&\text{if $a\ge 0$, and}\\-a&\text{if $a<0$.}\end{cases}$$ From this definition, showing that $|a|\ge 0$ for all $a$ is not hard, but it does require some argument. $\endgroup$ – Xander Henderson Jul 22 '19 at 20:42

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