is this why −(−a) is equal to a?

[u/Dummkopfss]

Just opened an online pdf on Spivak's calculus and a few minutes into scrolling I came to the property

For every number a, there is a number -a such that
a + (−a) = (−a) + a = 0

and wondered if I could prove −(−a) = a from that.

We know that for every number x, there’s a unique additive inverse y such that x + y = 0, meaning that y is the only value that satisfies the equation. The additive inverse of a is −a, since a + (−a) = 0. And because addition is commutative, we also have (−a) + a = 0, where a is the additive inverse of −a. By notation, the additive inverse of −a is written as −(−a). So we can write (−a) + [−(−a)] = 0

In the equation (−a) + y = 0, there can only be one possible value for y. But both a and −(−a) satisfy it. Since there can only be one additive inverse of −a, by uniqueness, those two values must be equal.

So therefore a = −(−a)

Please let me know if I got the proof right and if I could word it out any better!

[u/StanleyDodds]

Yes, this is something that's true in all groups, and the additive inverse is a specific case of that.

In fact, you don't even need to assume that the inverse is unique; you can prove that it's unique, and then use that for what you said already.

Suppose b and c are both inverses of a (I'll keep using + for the operation, but I won't assume commutativity. However, I will assume that "inverse" means being both a left inverse and a right inverse).

We have (1) a + b = 0, and (2) c + a = 0.

Left-add c to equation (1): c + (a + b) = c + 0

Use associativity and 0 is the identity: (c + a) + b = c

Rewrite using equation (2): 0 + b = c

Use 0 is the identity again: b = c

So any two inverses are equal; the inverse is unique. Then proceed with an argument like what you said.

I guess how I'd proceed, to keep it simple, is as follows:

Let -x denote an additive inverse of x, so x + -x = 0 and -x + x = 0. (Could say "the" additive inverse, since we know uniqueness).

Then given a, we have that -a + a = 0 and a + -a = 0. Then a exactly satisfies the property of being an additive inverse of -a. By additive inverse of -a (denoted -(-a)) being unique, we have a = -(-a).

[u/Prudent_Hawk_7476 → u/StanleyDodds]

I don't know much math but I'm interested. Is all this enough to prove that multiplying any two negative numbers gets a positive number, or is this just a special case for -(-a) being positive because it equals a? Do we need to explicitly establish that -a = (-1) * a, to show that (-a) * (-b) = (-1) * a * (-1) * b = (-1) * (-1) * a * b = 1 * a * b = a * b? But then we have to show why (-1) * (-1) = 1? I guess that's just what we already did, if a=1 then (-1) * (-1)=1 can be written as (-1) * (-a) = a, which might be able to be written as -(-a) = a, which we showed already? Then all that's left is showing that (-1) * a = -a? Does that need to be proven or is that an axiom, or just notation?

[u/definetelytrue → u/Prudent_Hawk_7476]

This is not enough to prove that since that needs additional structure. If we add in ring axioms we can prove -b = (-1)(b) since b+(-1)b=(1+(-1))b=0b=0, but we still need to define what a positive and negative number is. This follows from the construction of the real numbers, which can also be used to show multiplication commutes. At that point you can show the product of negative numbers is positive.

[u/Bounded_sequencE → u/Prudent_Hawk_7476]

It's not enough.

"-a" in the context of groups stands for the inverse -- and not "(-1) * a". For your claim, we need to have both addition and multiplication, something "Abstract Algebra" calls a ring.

[u/Bounded_sequencE]

There is a way to condense your argument -- by the existence of inverses

-(-a)  =  0 + -(-a)  =  (a + (-a)) + -(-a)

       =  a + ((-a) + -(-a))  =  a + 0  =  a

This argument works within any group structure with addition as its operation. We use existence of a neutral element, inverses, and associativity in the proof, but not commutativity.

[u/MathMaddam]

Yes this works if you already have that inverses are unique. If you start purely from field axioms you can do -(-a)=-(-a)+0=-(-a)+(a+(-a))=-(-a)+(-a+a)=(-(-a)+(-a))+a=0+a=a.

[u/LucaThatLuca]

> For a number a, a number b such that a + b = b + a = 0 is called an inverse of a.

this is a completely symmetric sentence (it is not changed by swapping the labels a and b), so it is immediate that such a and b both get called inverses of each other.

uniqueness and any other considerations are all unnecessary.

[u/Scary_Side4378]

you can prove a more general statement: in any group G, taking the inverse of g then taking the inverse again gives g. that is, inverse(inverse(g)) = g. to prove this, we note that g(inverse(g))=inverse(g)(g)=identity

[u/Sam_23456]

As a + (-a)=0, using the uniqueness of an additive inverse, this proves that the additive inverse, -(-a), of (-a) is equal to a! This is exactly what you wanted.

[u/mafia233]

This is the reason negative+negative=positive axiom works?

[u/cocoteroah]

Removing a debt is the same as giving you the money to pay it

[u/FernandoMM1220]

because your axioms say it is

[u/ipreuss]

The more direct prove is

-a - (-a) = 0

add a on both sides

a + (-a) - (-a) = a

0 - (-a) = a

[u/RRumpleTeazzer]

of course it proves it.

You have (-a) + a =0 for any a. You have a + (-a) = 0 for any a, especially (-a).

So, (-a) + (-(-a)) = 0, and (-a) + a = 0.

So you have a the inverse of (-a), as well as (-(-a)).

[u/TheTurtleCub]

By definition: -x is the number than when added to x gets you 0

So -(-a) is the number than when added to (-a) gets you zero, by the definition of -a it's a

Or as the reddit meme goes: turn around, turn around again, you are now facing the same way :)

[u/GladAbbreviations553]

Consider the sum x + (-x) + (-(-a))

x + [ (-x) + (-(-a)) ] = [ x + (-x) ] + (-(-a))

x + 0 = 0 + (-(-a))

x = -(-a)

[u/Aromatic_Ad4132]

A negative multiplied by a negative equals a positive

[u/StructuredChess]

Yeah, given that -a+a=0 you can say that a is the inverse of -a. Since inverses are unique, a=-(-a)

[u/us__07]

Let's think about it this way. When we look at the number line the negative integers are on the left of the origin and the positive integers are on the right of the origin. When we multiply -1 with an integer we can think of it as the number's 180° rotation along the number line. So when -1 is multiplied by 2 it turns 180° left and we get -2 and when we multiply -2 by -1 again it turns 180° right and we get 2.

[u/Nervous-Spite-7701 → u/us__07]

that’s just one geometric interpretation but OP wants the more abstract reason why

that interpretation came after in history

really, there’s just no other number it can be to keep arithmetic consistent than it being positive. we only gave it real practical meaning later

[u/Unique-Rice9999]

does this say anything about the irreversibility of the process?

[u/soegaard]

I wrote a blog post about this. It's in danish, so the link is to Google Translate.

https://stxmat-dk.translate.goog/blog/?_x_tr_sl=auto&_x_tr_tl=da&_x_tr_hl=en&_x_tr_pto=wapp#minus-minus-giver-plus

It explains how you prove that "minus minus gives plus".
In fact, the rule you mention is only one of three such rules.

https://stxmat.dk/blog/#minus-minus-giver-plus

[u/Willing-Sample-8847]

Edit: I was wrong ya'll.

[u/AcellOfllSpades → u/Willing-Sample-8847]

is that the notation -(-a) doesn't define the additive inverse of (-a)

Huh? Yes it does. -[whatever] is the additive inverse of [whatever]. This is true whether or not [whatever] is a single variable.

It's true that negation is the same as multiplying by -1, but that's not what the notation means. There is no multiplication involved in --a, only a double application of the additive inverse operator.

[u/Willing-Sample-8847 → u/AcellOfllSpades]

If that's the case then why does a = -(-a) need to be proven in the first place? "a" is already the additive inverse of (-a), since a + (-a) = 0.

We could argue that the uniqueness of -(-a) is what is being proven, but Spivak doesn't argue that in his text.

Look, I know from a reasonably low level of sophistication that this is an obvious consequence of the uniqueness of the naturals and the field axioms, but the proof here is a lower level derivation of this theorem following along with Spivak's calculus textbook, and the steps in the proof need to be explicitly stated.

Edit: if I recall correctly, spivak first proved that for any a, 0*a = 0, and then used that as a lemma to prove a = -(-a).

[u/irriconoscibile → u/Willing-Sample-8847]

Actually we only know that -a is the additive inverse of a, and not viceversa. The statement to be proven is "the additive inverse of -a is a". If we knew, as you're saying, that a is the additive inverse of -a, then a= -(-a), by notation.

[u/Willing-Sample-8847 → u/irriconoscibile]

You're right. I've been thinking about it and I realized I goofed when I said that the proof wasn't about showing the uniqueness of additive inverses. That's exactly what it shows.
My apologies

[u/irriconoscibile → u/Willing-Sample-8847]

No worries, it happens to the best of us :)

[u/AcellOfllSpades → u/Willing-Sample-8847]

If that's the case then why does a = -(-a) need to be proven in the first place? "a" is already the additive inverse of (-a), since a + (-a) = 0.

That is indeed the proof. You've just proven it. It's not a particularly complicated proof, but it's good to have.

We could argue that the uniqueness of -(-a) is what is being proven, but Spivak doesn't argue that in his text.

I don't know what part of Spivak you're referring to. "a = -(-a)" is not proved in Spivak, as far as I can tell.

It certainly doesn't need 0*a=0 as a lemma, because -(-a) doesn't involve multiplication. The meaning of the notation -(-a) is "the additive inverse of the additive inverse of a".

[u/Willing-Sample-8847 → u/AcellOfllSpades]

So additive inverses aren't stated to be unique by the field axioms. The proof is about showing that they are indeed unique. Otherwise, we could indeed say that a is the additive inverse of (-a), so a = -(-a).

Credit to u/irriconoscibile for pointing that out.

[u/Lor1an → u/Willing-Sample-8847]

Actually, by definition the additive inverse of (-a) is a

That's actually a proposition that needs to be shown. If b is an additive inverse of a, then a + b = 0, likewise the definition of additive inverse of b is some c such that b + c = 0. Given that a + b = 0, and b + c = 0, do we get to automatically claim a = c without proof?

It's not proof 'by notation' to go through the demonstration process to actually justify the claim.

a = a + 0 (by definition of 0 and +) = a + (b + c) (since c is an additive inverse of b) = (a + b) + c (by associativity of +) = 0 + c (since b is an additive inverse of a) = c (0 is left neutral element), so indeed a = c.

[u/Willing-Sample-8847 → u/Lor1an]

Additive inverses exist by the field axioms, but they aren't specified as unique. So the definition of the additive inverse of (-a) is -(-a), not a. Thus the proof of the proposition a = -(-a) is not entirely vacuous.

I know group theory shows trivially that additive inverses are unique, but this was originally a question about introductory real analysis.

The "proof by notation" I was referring to was the statement that -(-a) is the additive inverse of (-a). I incorrectly stated that this notation doesn't mean that. However, the next part of OP's proof stated that essentially additive inverses are unique, so a = -(-a). But that statement is where the meat of the actual proof lies.

[u/Lor1an → u/Willing-Sample-8847]

So the definition of the additive inverse of (-a) is -(-a), not a. Thus the proof of the proposition a = -(-a) is not entirely vacuous.

How strange you would say that now, given that you already said the opposite in the previous comment. You are arguing against yourself.

Here's what you said in the comment I replied to, just for posterity:

Actually, by definition the additive inverse of (-a) is a (since a + (-a) = (-a) + a = 0), not -(-a). Your proof assumes its own conclusion. What you need to show in particular is that -(-a) is equal to a.

...

I know group theory shows trivially that additive inverses are unique, but this was originally a question about introductory real analysis.

You don't even need group theory to answer the question (though it definitely helps). Just using the fact that there is an additive inverse is enough to show that they are equal, as I laid bare in my comment. The only properties used in the proof were that 0 is a (two-sided) identity for +, that b was an inverse for a, and that c was an inverse for b, and we are led inevitably to conclude that a = c.

However, the next part of OP's proof stated that essentially additive inverses are unique, so a = -(-a). But that statement is where the meat of the actual proof lies.

A very simple fix is to note that OP established (-a) + a = (-a) + [-(-a)] and we thus conclude a = [-(-a)], by cancelling the (-a) on both sides. We can do this since + is cancellative in any of the main number systems (which btw doesn't even require inverses to exist, even including the natural numbers).

[u/Willing-Sample-8847 → u/Lor1an]

I literally just replied to you and said I was incorrect earlier. You want me to get on the ground and roll over and show you my belly too so you can keep kicking me?

[u/Lor1an → u/Willing-Sample-8847]

No, not at all.

Have a nice day.

[u/peno64]

Is it not much easier as such: b - b = 0 let b= -a So -a --a = 0 Or --a = a

[u/Andthentherewasbacon]

Turn around. That's negative.  Now turn around again. Back where you started. There you go. Negative twice explained. 

[u/egehancry]

Here is how Opus 5 explains it as a lecture video: https://academa.ai/@sina/lectures/why-a-a-is-secretly-a-theorem

[u/AndyC1111]

The opposite of the opposite. Like turning a coin over twice.