Explain complex numbers to be like i'm a highschool dropout that found algebra hard

explain complex numbers to be like i'm a highschool dropout that found algebra hard

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extra component in other dimension

Square roots of rationals do not generally exist in the sense of calculation. But you can model them as if they existed.

For instance, there is no rational which when squared is equal to 2. So when you do arithmetic with the square root of 2 you just have to carry it around as an uncomputed term. But there's nothing special about doing this with 2. You could also do it with -1.

Think of it like this: counting numbers you learned in elementary school:
>0 1 2 3 4 5 6 (...)
Okay, now say there is an extra dimension at the other side of the 0, we call them negative numbers, denoted by a - character in front and they represent shortages for example. Our number line now becomes:
>(...) -3 -2 -1 0 1 2 3 (...)
Hmm, quite strange. When we take a square root, we can only really take it from a number >=0, since what number squared could ever equal a negative number, if (-1)*(-1)=1? Alright, let's yet again define a new number i^2 = -1 such that now, we can also take square roots of our newly acquired negative friends. Lets represent these new numbers as being orthogonal to the numbers we previously know, since it's basically like you're representing two different types of quantities (those that square to positive and those that square to negative). We now acquire
>....i....
>-1 0 1
>...-i....
and we are at what you now already know. So really, if you are comfortable with us expanding positive numbers to the negative numbers, you should feel comfortable expanding the real valued squares to the negative valued squares. Turns out this expansion has some really cool and useful properties, which is why you need to know about them.

cartesian product of Reals by themselves
another way to put it, extra dimension

Rotating numbers

>cartesian product of Reals by themselves
this does not yield a unique algebra

You know how on a graph you have the x-axis and the y-axis? And how on a 3d map, you add a z-axis? Complex numbers are the "i-axis", they're another "direction" you can plot numbers on that don't represent a regular "dimension" like width or length or depth but rather an "imaginary" dimension that is orthogonal to the three regular spatial dimensions.

okay but where does it go after i? 0 goes to 1 and then 2, but where does 0 go after i?

not if you use polar coordinate system

2 really only means 2 quantities of unity (1), so 2 = 2*1. So then, 2 quantities of i will be 2*i = 2i.

so is i^4 = -2?

No, that would the complex equivalent of saying that 1^4 = 2. Note that i can be regarded as unity in the purely complex direction. Also, note that i^4 = i^2 * i^2 = {by definition} = (-1) * (-1) = 1.

only if you do work on the system

so what's the square root of -2?

Simply apply the algebra you already know (I will not prove this stuff here, for now just believe me when I say that pretty much all basic algebra behaves nicely when translated into the complex numbers):
[math]\sqrt{-2} = \sqrt{-1*2} = \sqrt{-1} * \sqrt{2} = \sqrt{i^2} * \sqrt{2} = i*\sqrt{2} = i \sqrt{2}[/math]
Note that we defined i^2 = -1, but you will quickly learn when doing a few examples that [math]\sqrt{-a} = i\sqrt{a} [/math], simply because it is cumbersome doing this expansion each and every time.

The only real answer is that [math]\sqrt{-2}=\sqrt{-2}[/math]. The only answer is to restate the question. There's no other way to answer this question like there is a way to answer the question "what is 2+3?"

What's interesting is that even though the only answer is a restatement of the question, you can still do algebra with these elements.

so you're saying i'd have to define i to be the square root of -2? i guess that makes sense but i still don't really get the point considering then it just seems to be a weird way of using variables

>i still don't really get the point
The main point of extending fields is to solve equations that couldn't be solved before.

Complex numbers are a field of numbers that are isomorphic to R^2. Complex numbers are more intricate than R^2 because you can multiply the vectors in C together and get another vector in C. In particular, the structure of the formula mirrors certain trigonometric identities, like cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) and sin(x+y) = sin(x)*cos(y) + cos(x)*sin(y), so one can show how formulas like e^i*pi = -1.

complex solutions to the equation x^2=-2 are i*sqrt(2) as well as -i*sqrt(2)

anyone saying anything else is a highschool dropout

The cartesian product [math]\mathbb{R}\times\mathbb{R}[/math] is isomorphic in set-theoretic terms but not as a field homomorphism.

complex numbers are the algebraic closure of the real numbers, i.e. what you get when you pretend that every nonconstant polynomial with real coefficients has a root, even if it doesn't have a real root.

complex numbers are the number plane, which is better than a number line, so now you can multiply numbers to rotate now while before you could only stretch and reflect.

Easy. If you have an algebraically closed field of 0 characteristic, for which the transcendence degree of its prime field has continuum cardinality, then you're dealing with the complex numbers up to an isomorphism.

You can define evaluation if you have a standard form like a + bi, or decimals for real numbers. Obviously the answer to "what is 2 + 3?" is 5.

>You can define evaluation if you have a standard form like a + bi
A rose by any other name would smell as sweet.

What's strange about this is that the root of positive numbers still lie on that original line, but the root of negative numbers requires something new.

It's not strange, it's the entire fantasy of the reals.

There's a rational number line which is very straightforward to define. Every field extension of the rationals adds "something new", as you put it. Even adding "infinite" extensions to the rationals can only get you to solve polynomial equations. The representations of the roots of polynomials with unsolvable groups are not particularly trivial, but even if I grant you all n-dimensional extensions you're still nowhere near done. An infinity of infinite-dimensional extensions are needed, all at right angles to your nice, cozy rational number line.

The only strange thing about this process is how lackadaisical people are about swallowing the existence of the "reals" as a one-dimensional representation of this unwieldy mess.

those fucks were made up by homosexual fuckers like euler and those fags
never should have been invented
useless numbers

It's just a 2D Real number bro, you can write a+bi as (a,b).

The operations (multiplication, addition) defined on (a,b) are the same as using Real number operations on a+bi.

>LaughingPhysicists.jpg

Strange ? Are you kidding me ? Mankind worked with lines for ages. Geometry made a number represented by a line. Then came the graduation.
Furthermore, R is a 1-dimensional space on himself or on Q. Very natural as well.

I'm surprised you find strange the line-representation of reals.

>or on Q
>R
>1-dimensional
excuse me?

>R is a 1-dimensional space on himself or on Q
Don't be silly, R is infinite dimensional over Q.

>I'm surprised you find strange the line-representation of reals.
I responded to someone else that found it strange, if you were reading for content instead of buzzwords you would know this.

I excuse you, don't worry

Yes, but can you show me any base of it please?

the real numbers were literally constructed to be that way you tard

brainlet detected. [math](a,b)\times(c,d)[/math] has a canonical definition [math](a,b)\times(c,d)=(a\times c, b \times d)[/math]. That's not true in the complex plain.

You do realize i forms a subgroup of the group of nonzero complex [math]\mathbb{C}-0[/math] of four elements under multiplication, there is no subgroup of order 4 in [math]\mathbb{R}-0 \times \mathbb{R}-0[/math] under multiplication?

Learn how to use algebra and stop misinforming people

And illusionists saw women in half, so long as you don't inspect their box.

C was constructed because your IQ would not fit in R, not even as a negative IQ. Your intelligence is imaginary.
Fuckyoumulticocksuckerfaggot.

>hasn't taken analysis.jpg

In fact, you're not saying anything.
You're just messing around, showing us the uselessness of your math-style-loghorroeas.

The canonical injective subgroup of the riemanian form in half-integer powers is not informative on anything you claim to be saying.
Abelian sub-groups of roots 1-symmetrical from -1 are taking C-multiplicative continuons, compact banach-wise over-groups as basis, therefore introducing the 4-root of unit as a half-sesquilinear form from this space to (Z/4Z, +, *).

I have taken analysis. I've done researches on numerical analysis. You are not teaching me analysis, son. Go jerk off somewhere else and stop saying untrue things. I hate untrue or approximate statements.

>canonical
stopped reading right there

>if you use this unique system then you have a unique system
Wew lad, ur a fukin genius

should've read further
many folks fool geniuses
u part of many

youtu.be/0gyw6JJ7h2I

[math] i = /sqrt{-1} [/math]
Then you can figure out:
[math] i^3 = i*i*i = -i [/math]
WRONG
[math] i*i*i = captain [/math]

b i...

sqrt(-1) = i

sqrt(-1) = - sqrt(1) = -1 you dumbass
sqrt is linear function

They don't exist!

It most definitely is not

y axis represent complex value(bi) and x real (a)

It most definitely is
it's even a linear form which kernel forms an hyper-plan

you can do it with Im horizontal and Re vertical as well faggot

So are you saying that [math]\sqrt{2}=2\sqrt{1}=2[/math]?

I am saying sqrt(lambda M) = kappa sqrt(M)
with kappa=sqrt(lambda)
and sqrt(M)*sqrt(M) = M

That's not a linear function then, a linear function [math]f[/math] has the properties
[eqn]f(\lambda x)=\lambda f(x)[/eqn][eqn]f(x+y)=f(x)+f(y)[/eqn]Square root does neither of those things.

Don't know what happened with the Latex there

Why can't I get this right?

Write z1 and z2 in exponential form, then it's obvious

Hey i was like you m8, so i actually made a Complex number calculator.

This channel is the best thing ive found for math-newbs seeking to redeem their wicked ways:

/watch?v=kpywdu1afas

>inb4 qwawtrzjedoawdpsåfzæzæ,,,---------

you forgot "forall x ..., forall lambda, ..."
don't forget that sqrt(1+x) = 1+0,5x
gotcha

>you forgot "forall x ..., forall lambda, ..."
That was implicit

>don't forget that sqrt(1+x) = 1+0,5x
>gotcha
What the fuck are you on about

well quantification omitted can lead to gigantic errors I'd rather read a little mistake in a calculation than a statement unquantified

>first semester knowledge
>"learn some algebra"

topkek

>(a,b)×(c,d) has a canonical definition (a,b)×(c,d)=(a×c,b×d)

It doesn't. There is no canonical definition for the 2D cross product.

>You do realize i forms a subgroup of the group of nonzero complex C−0 of four elements under multiplication, there is no subgroup of order 4 in R−0×R−0 under multiplication?

Yes, obviously. But what does that have to do with anything? I really don't get what you're trying to say.

why do you think you deserve an easy explanation catered to your needs? What did you do to earn it?

Your product of z1z2 is wrong. (a+bi)(c+di) = ac - bd + bci + adi

no, y axis is rationals times the square root of 2 and the x axis are rationals

>the 2D cross product
could you give the definition of a n-D cross product please

complex numbers
don't exist
no number
exist
leave this
and go
cut some wood
kiddo

Tensor product m8

top kek

Whereof one cannot speak, thereof one must be silent.

you know what "definition" means right?
now define n-D cross product please.
show us the spaces you work in, the objects, and so on.
go on, kiddo. don't be a lazy fuck.