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This maybe a silly question, but I was thinking about it.

First we have the set $\mathbb{N}=\{0,1,2,3,...\}$ of natural numbers, then to create $\mathbb{Z}$ we do the following: For each natural number $n \in \mathbb{N}, n \geq 1$ we adjoin the number $−n$ to $\mathbb{N}$, so we have $\mathbb{Z} = \{...,-3,-2,-1,0,1,2,3,...\}$, and we add the property that $ n + -n= 0$ so each of $n$ and $-n$ is the additive inverse of the other.

Note that till now we have defined $-n$ only when $n$ is positive.

Now we want to prove that $-(-n) = n$, which then allows us to write $-n$ where $n$ is any integer, positive or negative.

In what follows : $a,b \geq 1$(since $-n$ is not defined yet when $n$ is negative):

First: $-a.b + a.b = (-a + a) .b = 0.b =0 $, it then follows that by adding $-(a.b)$ to both sides that $-a.b = -(a.b)$

Now: $-a.-b + -(a.b) = -a.-b + -a.b = -a(-b+b) = -a.0 = 0$, it follows that by adding $a.b$ to both sides that $-a.-b=a.b$

I thought we can prove that $-(-n) = n$ by using the above statements to have:

$-1.n=-n$

$-1.-n = -(1.-n) = -(-n)$

$-1.-n = 1.n = n$

and since the last two statement are equal, then we have $-(-n) = n$

So $-n$ is the additive inverse of $n$ regardless whether n is positive or negative.

My question: Am I over complicating things? Is there a better proof of this statement?

Loai Ghoraba
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  • Also: https://math.stackexchange.com/q/1102059/42969 – Martin R Dec 04 '23 at 10:35
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    Thanks for the links. I think they have the same problem I stated, that at first we have $-a$ is the number when added to the positive number $a$ it gives 0. The statement at first applies not to all numbers but to positive numbers $a$. – Loai Ghoraba Dec 04 '23 at 10:41
  • Both are additive inverses of $-n$ so they are equal by uniqueness of inverses - see the linked dupe (which explains a general way of discovering the uniqueness proof). – Bill Dubuque Dec 04 '23 at 17:04

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This is a good question.

You are showing that $-n=(-1)n$, which is interesting in itself but not needed here.

As you said, $-a$ is the number that added to $a$ gives you $0$. So the original equality $n+-n=0$, seen as $-n+n=0$, tells you that $n$ is the number that added to $-n$ gives you $0$. That is, $-(-n)=n$.

To make the argument complete one needs to show that additive inverses are unique (otherwise, it doesn't even make sense to write $-a$). If $a+b=a+c=0$, then $$ b=b+0=b+(a+c)=(b+a)+c=0+c=c. $$

Martin Argerami
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  • the problem with that as per my understanding, is that at first we have $-a$ is the number when added to the positive number $a$ it gives 0. The statement at first applies not to all numbers but to positive numbers $a$. – Loai Ghoraba Dec 04 '23 at 10:37
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    Well, that's the problem of not constructing things formally. If you just make up negative numbers and you tell me that the arithmetic rules don't apply to them, what do you expect? If you define $-a$ as you say and you stay there, then it doesn't even make sense to ask what $-b$ means when $b$ is negative. – Martin Argerami Dec 04 '23 at 10:41
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    Thanks, then the problem is with my construction. I'm a computer science graduate and wanted to understand these stuff better. – Loai Ghoraba Dec 04 '23 at 10:53
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    can you suggest a text that constructs various number systems formally but in an accessible manner ? or where are these stuff discussed generally? – Loai Ghoraba Dec 04 '23 at 11:05
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    Without much search, here is a short text that constructs the integers out of the naturals in a typical fashion. The key is to be comfortable with equivalence relations. – Martin Argerami Dec 04 '23 at 12:31
  • Thanks a lot, I will read it. – Loai Ghoraba Dec 04 '23 at 13:38
  • One last question: are such constructions discussed in real analysis, abstract algebra or what? – Loai Ghoraba Dec 04 '23 at 16:47
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Dec 04 '23 at 16:56
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    @Loai You can find a good exposition of constructions of the standard number systems in the book Numbers by Ebbinghaus et al. Iirc there are also good expositions in one of the volumes of Fundamentals of Mathematics by Behnke et al. $\ \ $ – Bill Dubuque Dec 04 '23 at 17:23
  • @BillDubuque thanks, I will have a look at it – Loai Ghoraba Dec 04 '23 at 17:54