Discussion between user 170039 and Thomas

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15:14

A couple of view points. The definition. The definition of $\lvert x \rvert$ is $$ \lvert x \rvert = \begin{cases} x & \text{ for } x\geq 0 \\ -x & \text{ for } x< 0 \end{cases} $$ So according to the definition, $\lvert 0 \lvert = 0$ and $-0 = 0$. and so indeed $\lvert 0 \rvert = - 0$ (they ...

user131753

What exactly is the definition of $-0$?

Thomas

@user170039: The definition of $-x$ is an element $y$ such that $x + y = 0$. This is exactly satisfied by what we usually denote by $-x$.

user131753

How do you construct $\mathbb{N}$? From $\mathbb{R}$ or using Peano Axioms?

Thomas

@user170039: The second viewpoint is just from group theory where we consider the group of real numbers.

user131753

Which "second viewpoint" are you referring to?

Thomas

15:14

I present two "view points" in my answer. The first is the "definition". One can simply answer the question by saying that by definition $0$ is a solution to the equation. The second view point is the one from group theory where we have a precise definition of what $-x$ means in an additive group. There are probably other ways to think about it.

so?

user131753

But the definition of $x$ is not precise as presented.

user131753

It is,

Thomas

Obviously the group theory answer requires more of the OP and that is why I wrote that I wasn't sure that it was helpful.

user131753

$$|x|=\begin{cases}x& \text{for}\ x>0\\0& \text{for}\ x=0\\-x& \text{for}\ x<0\end{cases}$$

Thomas

user131753

15:18

What is your question?

Thomas

I don't have a question. I don't understand your comment.

user131753

This is the only precise definition of $|x|$ I know. Observe that if you construct $\mathbb{N}$ (and eventually $\mathbb{R}$) from Peano Axioms, then also this definition makes sense.

Thomas

Have you studied abstract algebra?

user131753

Sure.

Thomas

And so you know what a group is?

user131753

15:21

Sure.

Thomas

Ok, so the real numbers under addition is a group. Here $-0$ is just the inverse of the element element $0$. $0$ is the identity element in the group.

user131753

Sure. But then my question is how do you define a real number?

Thomas

That is a good question. And it might/ might not be helpful to the OP if you wrote an answer on the construction of the real numbers. My only point was that viewing the real numbers as a group, $-0$ is the inverse of $0$. Now no matter what, you will always end up having to talk about what the real numbers are. But I don't think this is helpful to the OP. But then again, I don't know.

I don't think the OP really wants to know about how to construct the real numbers. I don't think the OP is looking for that kind of depth.

palsch

@Thomas :-)

Thomas

Hi palsch. Was my answer helpful?

user131753

15:32

Your answer seems to be ok if you assume that $\mathbb{R}$ is a field with order axioms and completeness axiom and if you can construct $\mathbb{N}$ from that (Apostol discusses such a construction via the notion of inductive set) then $-0$ makes sense because then by field axioms for every $x\in\mathbb{R}$ there exists an unique $-x\in\mathbb{R}$.

Thomas

And again, I don't think the OP would find it helpful to include this in an answer. I even think the group theory answer is a bit of a stretch. The OP has currently accepted an answer that says more or less nothing, so ... But, please provide your answer. Maybe if the OP would provide some more context/background it would be easier to provide a helpful answer.

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