Is i2 = -1 a convention in math or it can be proved?

Why do some define i to be equal to (0,1) in IR^2 ?

Euler is the one who defined it like that right?

read a book or something

>IR

not taking your time to format your post correctly with latex should be a bannable offense. I'm tired of faggots typing 10 lines of equations without latex

math is a religion. it all builds on some convenient axioms.

definition

>make a symbol for the sole purpose of representing a concept
>" prove this symbol represents what it was made to represent"
Are you retarded?

Yes. It can be proven that [math] i [/math] exists. Just see the formal definition of the complex numbers.

And, in general, it can be proven that for any given field and a polynomial with coefficients from it, there exists another field containing the original one in which that polynomial has a root.

>i=sqrt(-1)

into the trash it goes.
only proper definition is i^2 = -1. if you disagree you belong to /pol/

it's a definition
the nice thing about it is that it creates consistent effects
example: draw 2+5i onto the complex plane
i(2+5i) = 2i -5
i(2i-5) = -2-5i, which is the opposite of 2+5i
so, multiplying with i twice is the same as multilying with -1

>so, multiplying with i twice is the same as multilying with -1
woah so that is the power of associativity

Guys you can't just say I DEFINE an element i such that i^2=-1. That's pure nosense.
You must prove that x^2+1 has a root in a field containing R.

I'm confused right now!!
But I clearly take your side, that seems mathematical to me

prove -1^2=1

You are the fucking retarded here, I'm talking about the pre-define that symbol of yours.

>" prove this symbol represents what it was made to represent"
who fucking said that, symbols are just symbols we are not supposed to prove anything about them, in physics they refer to i as j so fuck symbols

no you dork, that's the power of the definition
If you tried something similar with a x/0=s
definition, you would get nowhere

>So since it's a definition then it's not gonna be proved?
The idea is you come up with a symbol for something so you can use that symbol in place of the stuff the symbol is standing in for. You don't prove the symbol can stand in for the stuff it's standing in for because the act of making it stand in for that stuff is just a notation decision. There's no way for a decision to start having a label for something turn out to be false. Like if you get your kid a puppy and name him Spot, it's inane to say that decision is either true or false. It's not a truth issue at all, it's a naming convention.

>i=sqrt(-1)
>only proper definition is i^2 = -1. if you disagree you belong to /pol/
>I'm confused right now!!
He's just being an autist. There's nothing untrue about that other user's equation, but he doesn't like it because ACKTUALLY the square root of negative one can also be negative i.

1+1=2

Let R be ring.

For all a in R:
a0 = a(0+0) = a0 + a0, thus a0=0

For all a,b in R:
(-a)b + ab = (-a+a)b = 0b=0, which means that (-a)b = -(ab)
therefore (-a)(-b)=-(-(ab))=ab

If the ring has unity, call it 1, then
(-1)(-1) = 1*1 =1

I should have proven multiplication by 0 from the left to go full autistic but whatever.

>He's just being an autist. There's nothing untrue about that other user's equation, but he doesn't like it because ACKTUALLY the square root of negative one can also be negative i.

1 = sqrt(1) = sqrt(-1 * -1) = sqrt(-1)*sqrt(-1) = -1

>1 = sqrt(1) = sqrt(-1 * -1) = sqrt(-1)*sqrt(-1) = -1
That mistake has nothing to do with what I wrote. The square root of negative one can be either i or -i. Look it up.

>you can't say I DEFINE
Ok
try "Suppose there exists" instead

Now hold on for a second. You wouldn't happen to be browsing Veeky Forums before graduating from high school, would you?

You have to prove the existence: Two of the ways you can do it are :

by adding additional structure on [math] \mathbb{R} ^2 [/math] by defining [math] (a,b) \cdot (c,d) = (ac-bd, ad+bc) [/math] and identifying [math] \mathbb{R} [/math] with [math] (a,0) [/math] via isomorphism. [math] i [/math] is then [math] (0,1) [/math]
(this way is retarded, but it's legit)

or

by taking the quotient [math] \mathbb{R} / \langle x^2+1 \rangle [/math] and identifying [math] \mathbb{R} [/math] with [math] c + \langle x^2+1 \rangle [/math] via isomorphism. Then, [math] i [/math] is [math] x + \langle x^2+1 \rangle [math]
(that's the superior way)

fuck me then! why on earth I'm assuming every one on Veeky Forums is not a high school-er then maybe I'm wrong

You are fucking confusing things around here, go back and take a fucking logic and set theory course before using those terms

> [math] \mathbb{R} / \langle x^2+1 \rangle [/math]
[math] \mathbb{R} [x] / \langle x^2+1 \rangle [/math]

> x + \langle x^2+1 \rangle
[math] x + \langle x^2+1 \rangle [/math]

>You have to prove the existence:
no you don't

>he can't even point out what's wrong with the logic

>take a course
Undergrads detected.

>Undergrads detected.
I'm always assuming undergrads and above to be the dominant here in Veeky Forums

>no you don't
yes, you have you fucking retard

>being completely incapable of abstract thought

the definition of i is that i^2 = -1 meaning i is both positive and negative sqrt(-1)

>being a highschooler trying to lecture others about things he doesn't know

>looking down on "high schoolers"
know how i know you're a shitty undergrad?

>You are the fucking retarded here
This says more than I ever can

fucking retard, square roots can't be multivalued, that's not a well-defined function. Use branch cuts.

prove 1
Go on, I'm waiting.

>square roots can't be multivalued
lmgtfy.com/?q=complex square roots of −1

>there is no "!" in there, that was actually a question fewgot

if you really fucking knew what
>well-defined function
then you'd really could have understood what he really meant here

>i don't know shit about mathematics but here's a screenshot of wikipedia: the post

>(this way is retarded, but it's legit)
>the only set theoretically correct construction of [math]\mathbb{C}[/math] as a field is retarded
hello, undergrad

rings are commutative by definition, user

>rings are commutative by definition
no its not, addition is, but not multiplication, example: ring of n by n matrices

convention. it's literally a made up thing, just like pi

Ring and CRing isn't the same, user.

literally kill yourself if you think the definition of i is i = sqrt(-1)

i(0) = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 -10 - 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1)

sqrt(i*-55) = -(-1)^(3/4)sqrt(55)

You mean the symbol of pi

the concept of pi

Pi is the ratio of a circle's circumference to its dia, pi is a special number, like the exponent of nature logarithm. It's nothing like i. I only replaces sqrt-1 and has no geometric significance.

2 is just a concept made up to solve 1+1=x

>t. engineer
Algebraic curves.

>I only replaces sqrt-1 and has no geometric significance.
retard

there exist no circle who circumference to diameter ratio is equal to pi. mathematicians just made up the rules of what is a circle and build on top of those made up concepts

2 does not exist beacause there is no 2 things that are exactly the same
mathematicians just made up the rules of what is 2 and built on top of those made up concepts

this is true, but I don't see where's the problem

>typing 10 lines of equations without latex
[math]R[/math]is was one symbol tho

>implying OP wasn't talking about Gaussian integers

fpbp

Philosophically, you fag, a circle is a transcendent figure, that does not exist within our vicinity (dimension). It is necessary to test a perfect approximation so we can develop mathematics. Applied and pure.

Hence,
It is convenient to trade the concept of a circle to, with the faintest tidbit of faith, understand the geometric implications, because that stuff clearly MATTERS.
I know it's a concept, I was speaking to you given that we have agreed to the premises and the basic philosophy of mathematics. That's it's a language demonstrating not necessarily pragmatic values.

Literally me, mechanical, what's your point?

Non-excited electrons are exactly the same.

The point is that, as an engineer, you don't get to do the math where you get to see things in different context than 'muh application, muh intuition'. That you can't see imaginary unit having geometric meaning or interpretation doesn't mean there is no, algebraic curves being a topic where one can easily notice this. It also serves as a nice and easy introduction to algebraic geometry, where you can further see the geometric meaning of many concepts that you would never guess had any geometric meaning at all.
If you want to know more, Fulton is a very good book for exactly the topic.

How does religion builds itself on some conventional axioms?

Easy
[sqrt]-1[sqrd]

[math] i [/math] is not any more special than [math] 2i [math] or [math] \frac{1+i}{ \sqrt{2} } [/math] (ok, maybe the latter and [math] i [/math] are a bit more special cause they are on the unit circle, but you get the point).
It's not really like [math] \pi [/math] .

>implying numbers must have geometric significance to mean anything
>not only failing to understand number theory, but also geometry itself

It's not about material items.

First, assume that it is and work through problems involving complex numbers. You will then understand the significance of the 'initial' presupposition.

[math]\pi[/math] is not any more special than [math]2\pi[/math] or [math]\frac{1+\pi}{\sqrt{2}}[/math]. It's not really like [math]i[/math].
You're trying to tell me "ignore anything you know about [math]i[/math], just for the sake of argument". Well i'm sorry, but if you don't see how [math]i[/math] is special because you're not educated in that area of math, or you've never realized it, i'll have to ignore that request. Is it easier to see the geometric meaning behind [math]\pi[/math]? Sure. The book i suggested is definitely readable by an engineer and has several exercises where you specifically need to use the geometric meaning (some of them anyway) of [math]i[/math].

...

I am not the guy you were having a conversation with. I am a math undergrad currently fishing my studies.
I'm saying that it's not really i that's special, it's the complex plane in general. While π is just weirder, it's not algebraic like i is and... don't know how to properly phrase it.

Try to pinpoint what exactly is special about it. That would help. Never understood the fascination of people with it, though it's prevalent in the community. I've certainly seen imaginary unit give rise to much weirder structures.

It's actually the other way around. -1 = i^2 is the definition; sqrt(-1) is the consequence.

[spoiler]Dumbass. [/spoiler]

Are you thinking of the term "transcendental"?

the life of pi

how is (a,b)(c,d) defined? I've never seen that notation before

my bad i'm retarded
i'll go get some sleep now

Ofc I know π is transcendental, but this doesn't mean much. "Almost all" numbers are transcendental.

>Literally me, mechanical, what's your point?
that you don't know shit about math and should, therefore, shut the fuck up about that which you are ignorant, faggot

>[math]R[/math]
i want to murder you.

basically what said

for [math](a,b) \in \mathbb{R}^2[/math] and [math](c,d) \in \mathbb{R}^2[/math] we define the complex product as follows:
[math](a, b)(c, d) := (ac - bd, ad + bc)[/math]

you can verify yourself that this is equivalent to multiplying [math](a + bi)(c + di)[/math] using distribution and treating [math]i[/math] as if it were [math]\sqrt{-1}[/math].

How is the other way "set-theoretically incorrect"?

It isn't, it's actually more rigorous. He's just larping.

Implied by the fundamental theorem of algebra. X^2+1=0 has a solution.

>circly raisining
ftoa assumes existence of C and i=sqt(-1)

because no one thinks of C as a quotient field, everyone thinks of C as ordered pairs of real numbers, which is what it really is structurally and graphically

>brainlet doesn't know set theory
>thinks math phd student is larping
lmoa

wrong, it assumes existence of [math]\mathbb{C}[/math], standard definition of [math]i[/math] isn't through square root, in fact defining just [math]i=\sqrt{-1}[/math] doesn't allow you to describe the structure of [math]\mathbb{C}[/math], it's just plain wrong.
There are two solutions to [math]x=\sqrt{-1}[/math] in [math]\mathbb{C}[/math]

>There are two solutions to x=sqrt(−1) in C
Wrong. There are infinite solutions to x=(a+bi)^(c+di) in C. In fact any complex number can be expressed in infinitely many equivalent ways.