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/Willing-Sample-8847]

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.

Edit: thanks for the downvotes.

Proof "by notation" isn't valid. You can't prove -(-a) = a by assuming -(-a) is the additive inverse of (-a). If you still think I'm incorrect, check Spivak's own proof in the book. He doesn't prove a = -(-a) by assuming -(-a) is the additive inverse of (-a). He does it directly using the field axioms.

The reason in particular that you can't do this "by notation" is that the notation -(-a) doesn't define the additive inverse of (-a). It defines the additive inverse of "a" multiplied by the additive inverse of "1".

[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/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/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/Andthentherewasbacon]

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

[u/peno64]

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

[u/AndyC1111]

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