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Anonymous Mon 21 Feb 07:01:09 2022 No.14251946 View Reply Original Report

Why does i^2 equal -1? By definition i = ?-1 which means i^2 can be written as ?-1 * ?-1 which would lead to ?1 and then positive 1. Am I overthinking the actual point of imaginary numbers?

Anonymous

Anonymous Mon 21 Feb 2022 07:04:24 No.14251950 Report

Quoted By:

>>14251946
You're right in your first definition. The algebra magic to get i*i = 1 comes from not including a +- sign

Anonymous

Anonymous Mon 21 Feb 2022 07:04:42 No.14251952 Report

Quoted By: >>14251957 >>14251959 >>14251968

>>14251946
>lead to ?1 and then positive 1.
what? Write an equation please

Anonymous

Anonymous Mon 21 Feb 2022 07:09:21 No.14251957 Report

Quoted By: >>14253530 >>14255180 >>14260497

>>14251952
i^2=i*sqrt(-1)
>bring the I into the sqrt
i^2 = sqrt(-1*i^2) = sqrt(-1*-1) = sqrt(1) = 1

Anonymous

Anonymous Mon 21 Feb 2022 07:09:50 No.14251959 Report

Quoted By:

>>14251952
sqrt(1) = sqrt(-1 * -1) = sqrt(-1) * sqrt(-1) = i
no plus minus sign though.

Anonymous

Anonymous Mon 21 Feb 2022 07:11:59 No.14251964 Report

Quoted By:

Did math just get btfo?

Anonymous

Anonymous Mon 21 Feb 2022 07:13:40 No.14251967 Report

Quoted By: >>14253505 >>14260537

>>14251946
>By definition i = ?-1
nope

Anonymous

Anonymous Mon 21 Feb 2022 07:14:16 No.14251968 Report

Quoted By:

>>14251952
An example of this I can think of is
(?x)^2
which can be expressed as either ?x * ?x or ?x^2 which both come out to x. The way I see it is I can implement it like this: ?(x*x). Back to the i scenario I implement it as ?(-1*-1) which is ?1 which is just positive one. Or is there some goofy rule I'm missing here?

Anonymous

Anonymous Mon 21 Feb 2022 07:17:47 No.14251972 Report

Quoted By: >>14252257 >>14257692

>>14251946
Square-root is not well-defined.
$\sqrt{1} = \pm 1$
$\sqrt{-1} = \pm i$

Anonymous

Anonymous Mon 21 Feb 2022 07:20:34 No.14251979 Report

Quoted By: >>14251982

>>14251946
by definition it's actually i^2 = 1

Anonymous

Anonymous Mon 21 Feb 2022 07:21:43 No.14251982 Report

Quoted By:

>>14251979
-1*

Anonymous

Anonymous Mon 21 Feb 2022 07:21:47 No.14251985 Report

Quoted By:

>>14251946
> By definition i = ?-1
no. the definition is $i^2 = -1$

> ?-1 * ?-1 which would lead to ?1
incorrect. taking the sqrt of a negative or complex number follows different rules.

Anonymous

Anonymous Mon 21 Feb 2022 09:53:18 No.14252203 Report

Quoted By:

>>14251946
i is necessarily flawed and most math professors I talked to admit it:
i2=-1
-i2=-1
therefore i is not a defined number like -1 is to 1 but is just a "new" number, a vector in any direction as long as it satisfies the right angle made with the real axis

Anonymous

Anonymous Mon 21 Feb 2022 10:21:31 No.14252257 Report

Quoted By: >>14255419

>>14251972
That would be correct if sqrt(x) isn't a function.

Anonymous

Anonymous Mon 21 Feb 2022 11:01:04 No.14252328 Report

Quoted By:

>>14251946
>By definition i = ?-1
that's not the definition of i

Anonymous

Anonymous Mon 21 Feb 2022 11:05:48 No.14252336 Report

Quoted By:

>>14251946
>which would lead to ?1 and then positive 1
prove it

Anonymous

Anonymous Mon 21 Feb 2022 16:37:37 No.14253025 Report

Quoted By:

>>14251946
>By definition i = ?-1
Whoever thought you this was wrong.

$i^2\equiv-1$ is the definition.

$i\neq\sqrt{-1}$ , and this is easily proven using the definition.

Anonymous

Anonymous Mon 21 Feb 2022 16:41:56 No.14253039 Report

Quoted By:

It's there to work with other numbers, working as x being manipulated by sqrt(-1). No one says it's real.

Anonymous

Anonymous Mon 21 Feb 2022 17:01:26 No.14253091 Report

Quoted By:

It's better to think in terms of the complex plane.

$i = e^{\frac{i \pi}{2}} \implies i^2 = e^{i \pi} = -1$

Anonymous

Anonymous Mon 21 Feb 2022 17:16:44 No.14253129 Report

Quoted By:

$<br />-1×-1 = 1<br />1×1 = 1<br />-1 = 1<br /><script type="math/tex">$

Anonymous

Anonymous Mon 21 Feb 2022 17:27:40 No.14253162 Report

Quoted By:

The complex conjugation is an automorphism, the "i" and "-i" are definitionally interchangebale, because they are both roots of the field extension of the reals by x^2+1=0.

Anonymous

Anonymous Mon 21 Feb 2022 18:16:41 No.14253295 Report

Quoted By: >>14253305

>>14251946
That's the definition of $i$ . $i$ is defined to be the solution to $x^2+1 =0$ .

Anonymous

Anonymous Mon 21 Feb 2022 18:18:01 No.14253305 Report

Quoted By:

>>14253295
Careful, this is ambiguous since $-i$ is also a solution.

Anonymous

Anonymous Mon 21 Feb 2022 19:20:57 No.14253505 Report

Quoted By: >>14254201

>>14251967
EE detected.

Anonymous

Anonymous Mon 21 Feb 2022 19:28:16 No.14253530 Report

Quoted By:

>>14251957
thats not how the order of operations works

Anonymous

Anonymous Mon 21 Feb 2022 19:48:18 No.14253581 Report

Quoted By:

>>14251946
This is true of natural numbers where the binary operation * is preserved under mappings (i.e. a homomorphism) but this is not true for all real numbers.

Anonymous

Anonymous Mon 21 Feb 2022 20:52:20 No.14253820 Report

Quoted By:

>>14251946
principle of explosion. It was a vacant position, and utilizing it allowed for greater degrees of freedom.

Anonymous

Anonymous Mon 21 Feb 2022 21:28:31 No.14253940 Report

Quoted By: >>14254047

>>14251946
Nobody tell OP about the split complex numbers, he'll melt down.

Anonymous

Anonymous Mon 21 Feb 2022 21:51:46 No.14254026 Report

Quoted By: >>14254131

>>14251946
Yes, the point of imaginary numbers is to set that arbitrary rule. Math is a fucking joke.

Anonymous

Anonymous Mon 21 Feb 2022 22:01:12 No.14254047 Report

Quoted By:

>>14253940
Wait until he sees triplex

Anonymous

Anonymous Mon 21 Feb 2022 22:43:38 No.14254131 Report

Quoted By: >>14254138

>>14254026
Did you not go past middle school? Imaginary numbers have countless applications. An algebraically closed field is arguably more natural than an incomplete subset of the reals.

Anonymous

Anonymous Mon 21 Feb 2022 22:46:18 No.14254138 Report

Quoted By: >>14254196

>>14254131
Complex numbers are not taught in US public schools.

Anonymous

Anonymous Mon 21 Feb 2022 23:13:51 No.14254196 Report

Quoted By:

>>14254138
I learned them in 9th grade at my public high school, although only in the context of complex roots of polynomial equations.

Anonymous

Anonymous Mon 21 Feb 2022 23:16:12 No.14254201 Report

Quoted By:

>>14253505
they're right though.

t. math

Anonymous

Anonymous Tue 22 Feb 2022 00:37:36 No.14254393 Report

Quoted By:

>>14251946
>which would lead to ?1 and then positive 1
?1 is positive or negative 1, not just positive 1.

Anonymous

Anonymous Tue 22 Feb 2022 03:50:38 No.14254823 Report

Quoted By:

What is the OP's logic? Am I right that a square root of 'n' is defined as the solution for the equation x^2 = n?

In that case, (sqrt(n))^2 = n which would mean that sqrt(-1) * sqrt(-1) = -1

Anonymous

Anonymous Tue 22 Feb 2022 04:25:48 No.14254891 Report

Quoted By:

>>14251946

Axiom.

Anonymous

Anonymous Tue 22 Feb 2022 04:29:30 No.14254898 Report

Quoted By:

>>14251946
IT JUST DOES, OK??

Anonymous

Anonymous Tue 22 Feb 2022 04:42:09 No.14254918 Report

Quoted By:

>the actual point of imaginary numbers
it is not uncommon to find the historical origin of math to precede its use case by hundreds of years. it is fair to think of both real numbers and imaginary numbers as different ways to measure some third thing, that third thing having both a real and imaginary component.
the "right" way to think about imaginary numbers is to be so familiar with them after studying communications, networks, linear systems, signals, and electronics (or more obvious subjects eg diff eq) that you no longer need a physical conceptualization of them.
let the imaginary numbers become as unordinary to you as the counting numbers, as familiar to your eyes as 0 or -64.

Anonymous

Anonymous Tue 22 Feb 2022 08:11:39 No.14255180 Report

Quoted By:

>>14251957
>bring the I into the sqrt
Prove you can do that.
Hint: you can't.

Anonymous

Anonymous Tue 22 Feb 2022 08:24:58 No.14255187 Report

Quoted By:

It sounds like the confusion comes from thinking square root has to be multiplicative. It doesn't, only the principal square root of positive numbers is. If x^2=a and y^2=b, it's not necessarily true that sqrt(ab)=xy, even if a and b are positive, because that doesn't necessarily mean x and y are positive, so we certainly shouldnt expect this of a supposed square root of -1. We simply call i such that i^2 = -1, there is no real choice to make it positive i, we just know there are two square roots of -1 called i and -i.
i is not the positive square root of -1, there is no such thing. So we can't define a multiplicative, complex valued function sqrt over all real numbers mapping -1 to i. We can just make some assignments in a field extension.

Anonymous

Anonymous Tue 22 Feb 2022 12:01:21 No.14255416 Report

Quoted By: >>14255425

>>14251946
You got it wrong. The definition is i^2 = -1. Only retards say i is the square root of -1. It's not. Roots of negative numbers are undefined.

Anonymous

Anonymous Tue 22 Feb 2022 12:05:04 No.14255419 Report

Quoted By:

>>14252257
It isn't. Until you cut the complex plain and get rid of the negative numbers. Then, it is.

Anonymous

Anonymous Tue 22 Feb 2022 12:07:42 No.14255425 Report

Quoted By: >>14261073

>>14255416
Define a square root and prove that i isn't a square root of -1.

Anonymous

Anonymous Tue 22 Feb 2022 13:03:43 No.14255519 Report

Quoted By:

sqrt(x*y) = sqrt(x) * sqrt(y) is only valid when x,y >= 0

Anonymous

Anonymous Wed 23 Feb 2022 04:14:41 No.14257692 Report

Quoted By:

>>14251972
radical means principal square root

Anonymous

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Anonymous Wed 23 Feb 2022 19:25:46 No.14259177 Report

Quoted By:

>>14251946
(1+i)(1-i)=(1)2-(i)2=1+1=2

Anonymous

Anonymous Wed 23 Feb 2022 22:52:41 No.14259779 Report

Quoted By:

>>14251946
OP, I suggest to read a book on complex analysis if u r really interested in this subject. It is basically about imaginary numbers and it uses a lot of contour integrals. Right now I am focus on set theory.

Anonymous

Anonymous Thu 24 Feb 2022 02:34:28 No.14260232 Report

Quoted By:

>>14251946
An array of numbers.

Anonymous

Anonymous Thu 24 Feb 2022 05:14:07 No.14260497 Report

Quoted By:

>>14251957
sqrt(a)*sqrt(b) = sqrt(a*b) is only true when a,b >= 0

Anonymous

Anonymous Thu 24 Feb 2022 05:43:43 No.14260531 Report

Quoted By:

>>14251946
$\sqrt{a}\sqrt{b}\neq\sqrt{ab}$ in the general case for complex numbers,
because the logarithm is discontinuous in $\mathbb{C}$ along the negative reals.
In general, you can derive complex numbers from the assumption that
1) there is a number field that contains $\mathbb{R}$ and
2) you only need finitely many nonreal numbers to fully define it.

Anonymous

Anonymous Thu 24 Feb 2022 05:50:49 No.14260537 Report

Quoted By:

>>14251946
>?-1 * ?-1 which would lead to ?1
Nice try but that trick doesn't work when the negatives are under a radical.

>>14251967
Also this

Anonymous

Anonymous Thu 24 Feb 2022 06:17:18 No.14260556 Report

Quoted By: >>14260557 >>14260577

To expand upon this purely with what I assume to be middle school knowledge
and being able to google a few fucking symbols. Suppose we have a set $\mathbb{U}\supset\mathbb{R}$ .
Since the reals are a proper subset we know $\exists j\in\mathbb{U}: j\not\in\mathbb{R}$ .
If it is a number field (meaning essentially you can multiply, divide, add
and subtract as you would with reals, including associativity and distributivity),
then we know that $\forall a,b\in\mathbb{R}: a+bj\in\mathbb{U}$ .
We already got infinitely more nonreal numbers that the single one we were able to assume
should a set such as $\mathbb{U}$ be possible.
There is no reason that we restrict ourselves to multiplying with reals,
proper numbers allow for arbitrary products, hence it must be true that
$\forall a,b,x,y\in\mathbb{R}: (a+bj)(x+iy)=ax+(bx+ay)j+byj^2\in\mathbb{U}.$
Now the question is: what properties must $j$ have for us not to end up at a contradiction?
Well turns out, either $\exists A,B\in\mathbb{R}: j^2 = A+Bj$ ,
meaning the subset $\mathbb{V} = \{a+bj:a,b\in\mathbb{R}\}$ is in itself a self-contained number field,
or no integer power of $j$ exists such that
$j^n = \sum_{l=0}^n c_l j^l,~~ c_0,\dots,c_n\in\mathbb{R}$
which just means we need an infinite set of nonreal numbers to form $\mathbb{U}$ (or any subfield of it that is not the reals).
The latter is beyond the scope of what a non-mathematician needs to know about complex numbers.
By assuming the former you will find that either the field does not exist
or there is a number $i=R+Mj$ with real $R,M$ such that $i^2=-1$ .

Anonymous

Anonymous Thu 24 Feb 2022 06:19:46 No.14260557 Report

Quoted By:

>>14260556
Fucking 4chan
>Since the reals are a proper subset we know*

Anonymous

Anonymous Thu 24 Feb 2022 06:47:45 No.14260577 Report

Quoted By: >>14260645 >>14260993

>>14260556
nobody's going to read that crap, lol

Anonymous

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Anonymous Thu 24 Feb 2022 07:54:59 No.14260645 Report

Quoted By:

>>14260577
>no words, only answer

Anonymous

Anonymous Thu 24 Feb 2022 13:37:10 No.14260993 Report

Quoted By:

>>14260577
the answer is b.)

Anonymous

Anonymous Thu 24 Feb 2022 14:41:10 No.14261071 Report

Quoted By:

>>14251946
You got the wrong definition.

Anonymous

Anonymous Thu 24 Feb 2022 14:42:09 No.14261073 Report

Quoted By:

>>14255425
A square root is the inverse of the exponent with a power of two, which means its result is the same as what would be the input in an exponent of 2 for the value plugged into the root.
The funny thing about this is it means all square roots have two answers because squares are the same for both positive and negative answers, and this is represented by a plus/minus sign
So i is equal to plus or minus the square root of -1

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