Why is

why is
-1^2 = +1 ?

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en.m.wikipedia.org/wiki/Unary_operation
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en.wikipedia.org/wiki/Negative_number#Negation
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(-1)(-1) = 1

-(1^2) = -1
(-1)^2 = 4

you can imagine the negative sign ( - ) as an instruction to go to the place on the number line that's opposite to zero (0) where you are now.

say -7 would be go to seven places to the left on the number line. -(-7) would be go seven places to the left, and then mirror that. (seven places to the right)

ergo -1 * -1 would be 1 times 1 and then reverse the number line twice. since the line is one dimensional, reversing it twice gives you the same thing as before

doesnt seem like a fundamental proof.

using binomial theorem, but probably still not fundamental enough
-1 * -1 = (1-2) * (1-2)
= 12 - 2*1*2 + 22
= 1

ok. the negative numbers are a field extension of the natural numbers to give solutions to polynomials like:
[math] x + 7 = 0 [/math]

so we take the polynomial
[math] x + 1 = 0 [/math]
and we manipulate it
[math] (x + 1) (x - 1) = 0 [/math]
[math] x^2 -1 = 0 [/math]
[math] x^2 = 1 [/math]
which clearly has solution -1 by construction.

1 + -1 = 0
(1 + -1)^2 = 0^2 = 0
1^2 + 2(1)(-1) + (-1)^2 = 0
1 + -2 + (-1)^2 = 0
therefore (-1)^2 = 1

I enjoyed your reasoning. Never thought about it like that, cheers.

It's such a fucking meme. No one here understands arithmetic. (-1)^2=-1-1=-2.

No problem, buddy.
this is the first time i've been forced to think about this myself too

This proof works not only in Z and its extensions, but any ring where 1 is the identity and 1 + -1 = 0, like Fp or matrices.

You're all retarded.

The - in this context is a unary operator such that - applied to a is -a.

Unary operators take precedence over all other operations. Therefore
-a^2 = (-a)^2 = -a*-a = a^2

There's nothing fundamental about it. It's not 0-a. It's not -1*a. It's its own object.

>true by definition
gee, thanks.

It's defined as the additive inverse, the number such that a + -a = 0, which is equivalent to saying -a = 0 - a. The next step is to actually prove that -(-a) = a, which you never did. Stop being stupid.

Wrong.
en.m.wikipedia.org/wiki/Unary_operation

Unary operators require nothing except the existence of elements in their sets.

I didn't claim -(-a) = a because I never used it.

I only applied the unary operator once.
Next time, please think before you speak.

>I didn't claim -(-a) = a because I never used it.
you claimed it implicitly.

you're too unfamiliar with proofs to participate in this discussion.

Yeah, I know what unary operation are.
>mobile link
The unary operation -, when applied to a, returns the value -a such that a + -a = 0. That's its definition (Herstein 26, Jacobson 31, though both use the equivalent multiplicative notation). In your "proof" you just put (-a)*(-a) = a^2 without proof, even though that's exactly what op wanted proven.

Seriously, you're just throwing around the term "unary operation" to cover for the fact that you're too incompetent at algebra to even prove -1 * -1 = 1.

I've never seen that definition, but you can use it, I guess.

Op didn't ask why it was true in general, he asked a specific question. Not the same thing. I wasn't trying to prove negative times negative is positive.

But I only applied the unary operator to each term once. Nowhere did I imply or use -(-a) = a.

I used a property that -a*-a = a^2.
It would only be an implication if the following holds:

-(-a) = a iff -1*-a = a

Which isn't necessarily true. Which is why I didn't use it.

>I've never seen that definition, but you can use it, I guess.
It's the standard algebraic definition. I'm not sure I've ever seen another definition.

>Op didn't ask why it was true in general, he asked a specific question. Not the same thing. I wasn't trying to prove negative times negative is positive.
You weren't trying to prove anything since you just assumed this is true.

>I used a property that -a*-a = a^2
Either that's part of the definition (hint: it's not) or you need to prove it, which you didn't. That's literally the whole point of OP's post. You can't just call something a "property" and move on.

-1^2 = -1 retard
duckduckgo.com/?q=-1^2 &t=hf&ia=calculator

It's not the definition on the wiki. Addition doesn't need to be defined at all for it to work.

Op asked why -1^2 = 1. He didn't ask why it's true in general. It's perfectly valid to assume -1*-1 = 1.

In fact, which things you choose as theorems and which things you take as axioms is mostly arbitrary.

If I assume -a * -a = a, there's no problem.
If op asked why this is the case, I would have proved it instead of assuming it.

But the ideas are distinct and assuming you know what someone means or thinks will get you in trouble.

You're better off answering literally. If that's not what they meant, they'll tell you.

(.)(.)=boobs

Lol!

Use BEDMAS

en.wikipedia.org/wiki/Negative_number#Negation

>If I assume -a * -a = a, there's no problem.
"Why does (-1)*(-1) = 1?"
"Because it does."
Very helpful.

>Addition doesn't need to be defined at all for it to work.
-1 isn't well defined without addition. You could create a new number with the definition that it does not equal 1 but its square does, but without addition you have no way to assert that it's equal to -1.

Well, if we're working with the set of integers, then addition doesn't need to be defined for the unary operator to work.
The set of integers can be constructed by addition, but doesn't have to be.

Like I said, I could prove that -1*-1 = 1. But that's not what op asked. Even if I did prove it, would I also have to prove addition? What I have to prove the numbers exist?

Again, how far you go depends on what you take as axioms.

I read ops question as
Why does -1^2 = -1*-1 and not -1*(1*1).

Then it's reasonable to assume that we already know a negative times a negative is positive, because it's not the conclusion of the argument given.

Stating it any other way does not directly answer op, so why argue that way?

I'm not the guy you're responding to, but OP is obviously asking why (-1)^2 under the Peano axioms gives us 1. Why the fuck would he ask "If the product of two negative numbers is positive, why is the product of THESE two negative numbers positive?"? That's retarded.

So yes, addition does have to be defined. The unary operator - is defined under the binary operator +, in fact it's defined as the number that must be added to x to give zero.

But like you said, OP didn't ask about -(-x), which actually makes our job easier, since we have a simpler case here, and less to write.

For OP, multiplication is recursively defined as follows:
[math]x\cdot S(y) = (x\cdot y) + x[/math]
[math]x\cdot0 = 0[/math]

While integers are defined as ordered pairs of natural numbers (x,y) such that x-y is equivalent. So -1 would be equivalent to (0,1). Multiplication for integers are defined as:
[math](x_1,y_1 )\cdot(x_2, y_2 ) = (x_1\cdot x_2+y_1\cdot y_2,x_1\cdot y_2+y_1\cdot x_2 )[/math]

This means that -1 * -1 = (0,1)*(0,1) =
(0*0+1*1,0*1+1*0) = (1,0) = 1.

HAHAHA how are negative numbers real hahaha nigga they don't even exist in real life HAHAHA

>Op asked why -1^2 = 1. He didn't ask why it's true in general. It's perfectly valid to assume -1*-1 = 1.
No it isn't, because you've assumed the answer to the question!
>OP: Why is (-1)^2 = 1?
Retard: (-1)^2 = (-1)*(-1)
>OP: Okay, but how does that answer my question?
Retard: Because I have made the assumption that (-1)(-1) = 1

You see how you've not explained anything, you've just shifted the question from "why does (-1)^2 = 1?" to "why does (-1)(-1) = 1?".
>If I assume -a * -a = a, there's no problem.
The problem persists: by making this assumption, you're not explaining why (-1)^2 = 1, you're just showing that it's a special case of a larger assumption but making no effort to explain why one would want to make that larger assumption, say because it follows from more general principles.

>In fact, which things you choose as theorems and which things you take as axioms is mostly arbitrary.
This is true but again it misses the point. If you assume that your unary operation is involutive, then yes (-1)(-1) = 1, by assumption, but you've not explained we should think of your unary operation as the negative of a number -- you would instead need to prove that -n = 0-n, and -1*n = -n, otherwise your unary operator could be any involution, and showing that for SOME involution (called $), (1$)(1$) = 1 doesn't answer OP's question.

[math]( \cdot ) ( \cdot )[/math]

...

i suppose second row is mathematically dishonest? should be
x^2 + (-1)^2

Like I said, I believe op was asking why -1^2 = -1*-1
Not why -1*-1 = 1

So there's nothing wrong with what I said.

fuck I almost proved that (-1)^2 = -1. And I would have gotten away with it too, if it weren't for you meddling kids!

Hope I didn't kill this thread with my Scooby Doo meme

This is the best solution

>a fundamental proof
No such thing. It is so because we, humans, define it as such. The reason we define it is to lighten mental workload for solving certain problems. That's all math really is, it's just gotten deep and complex enough that the beancounter kikes working on it lose track of the train of thought that brought them where they are in all the abstraction.

because the product is "ALL" of the two numbers. and -/+ are just directions, -/+ are just polarities, they don't have any intrinsic "something-ness" or "nothing-ness"

>(1-2) * (1-2)
>= 12 - 2*1*2 + 22
how does that work

There is literally nothing wrong with this

>(-1)^2 = 4
wat