/sci/ - Science & Math » Thread #11587372

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Anonymous Tue 21 Apr 08:50:27 2020 No.11587372 View Reply Original Report

Quoted By: >>11587442 >>11587456 >>11587537 >>11587541 >>11587543 >>11587650 >>11587713 >>11587718 >>11588707 >>11588733 >>11589143 >>11589208 >>11591026 >>11591627

I don't understand, what is the significance of this number? Helps a brainlet out.

Anonymous

Anonymous Tue 21 Apr 2020 08:54:09 No.11587375 Report

Quoted By:

It's unsolvable so it's just a completely new number. A "dark number", if you will.

Anonymous

Anonymous Tue 21 Apr 2020 08:55:08 No.11587378 Report

Quoted By:

*hits bong*
dudeee its like a normal number but imaginary, lets just invent this out of thin air lmaaaao

Anonymous

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Anonymous Tue 21 Apr 2020 09:31:26 No.11587442 Report

Quoted By: >>11589064

>>11587372
There's this common misconception that 'i' is just mathematicians making up a nonsensical rule and taking it to its extreme. That's because people are taught about how 'i' works before they actually see how it's used.

Imaginary numbers are a useful mathematical tool because they can describe rotation off of the number line. The easiest way to show this is to look at Euler's identity, which makes a lot more sense when you look at it geometrically.

$e^{i \pi}+1 = 0$

https://www.youtube.com/watch?v=v0YEaeIClKY

You can describe things that oscillate or track a circle, like electromagnetic waves. For instance, the Gaussian beam approximation (pic related) uses 'i', not because light is a magical property of 'imaginary math', but because it allows you to write a vector that oscillates up and down in a clean mathematical way without trigonometric functions or matrices.

Anonymous

Anonymous Tue 21 Apr 2020 09:35:59 No.11587456 Report

Quoted By: >>11587464 >>11587537 >>11588480 >>11589102

>>11587372
It's wrong. i^2 = -1 does not imply i = sqrt(-1).
Roots of negative numbers aren't defined.

Anonymous

Anonymous Tue 21 Apr 2020 09:39:29 No.11587464 Report

Quoted By: >>11587531 >>11588480

>>11587456
but 'i' wasn't defined as i^2 = -1 to begin with

why would anyone inventing imaginary numbers describe it like that?

Anonymous

Anonymous Tue 21 Apr 2020 10:32:15 No.11587531 Report

Quoted By: >>11587540 >>11587650 >>11588480

>>11587464
It literally was. Look up "algebraically closed". They were looking for a solution to the polynomial x^2 = -1, so they defined i as a number that solves this polynomial for x = +-i and started to investigate its properties. Again, it does not mean that i = sqrt(-1).

Anonymous

Anonymous Tue 21 Apr 2020 10:35:29 No.11587537 Report

Quoted By:

>>11587372
This image hurts to look at
>>11587456
This. Thank you.

Anonymous

Anonymous Tue 21 Apr 2020 10:37:18 No.11587540 Report

Quoted By: >>11587558

>>11587531
In what situations would the true value of 'i' by itself actually matter though? The fact that it rotates things on the number line comes from the properties of i^2.

I won't dispute the original invention of 'i' because I'm not a math historian and what you've written sounds plausible. But I also don't really see how it matters besides technical pedantry.

Anonymous

Anonymous Tue 21 Apr 2020 10:37:31 No.11587541 Report

Quoted By: >>11587544

>>11587372
it started in the 17th century

x^3-5x+1=0 has three REAL solutions
but you have to use i to solve it
https://www.google.com/search?q=y=x^3-5x%2B1
https://youtu.be/_qvp9a1x2UM?t=3m

Anonymous

Anonymous Tue 21 Apr 2020 10:39:01 No.11587543 Report

Quoted By: >>11587657 >>11587742

>>11587372
it started in the 16th century

x^3-5x+1=0 has three REAL solutions
but you have to use i to solve it
https://www.google.com/search?q=y=x^3-5x%2B1
https://youtu.be/_qvp9a1x2UM?t=3m

Anonymous

Anonymous Tue 21 Apr 2020 10:40:40 No.11587544 Report

Quoted By:

>>11587541
This is also a great example of what I talked about with 'i' being useful for rotating off the real number line. One of my Youtube favorites.

Anonymous

Anonymous Tue 21 Apr 2020 10:56:35 No.11587558 Report

Quoted By: >>11587568 >>11588132

>>11587540
It doesn't, you're absolutely right.
What does matter though is that the square root of negative numbers isn't a thing and assuming i to be defined like that leading to various misconceptions and errors.
But yeah, everything you said is true.

Anonymous

Anonymous Tue 21 Apr 2020 11:03:23 No.11587568 Report

Quoted By:

>>11587558
>assuming i to be defined like that leading to various misconceptions and errors.

What kinds of misconceptions can you derive from not defining it based on its square?

Anonymous

Anonymous Tue 21 Apr 2020 11:57:58 No.11587650 Report

Quoted By: >>11587657

>>11587372
complex numbers were invented to deal with polynomial equations which do not have solutions in the real numbers, see >>11587531. "i" satisfies

i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1

and then the pattern repeats, i^5 = i etc. this "periodic" behavior is typical for complex numbers and it's something which real numbers don't have at all. in the end it turns out that complex numbers are extremely convenient for describing any oscillatory phenomena, from circular motion to signal processing. this is what makes them important.

Anonymous

Anonymous Tue 21 Apr 2020 12:01:21 No.11587657 Report

Quoted By: >>11587673 >>11587742

>>11587650
>to deal with polynomial equations which do not have solutions in the real numbers
lol no, the exact opposite, see >>11587543

Anonymous

Anonymous Tue 21 Apr 2020 12:12:51 No.11587673 Report

Quoted By:

>>11587657
you're right, my bad

Anonymous

Anonymous Tue 21 Apr 2020 12:41:25 No.11587713 Report

Quoted By:

>>11587372
Consider the field F that has elements {0,1,2} with addition and multiplication modulo 3. But oh no, the polynomial x^2+1 has no roots in this field. Bad bad field. We repair. Let i be the name of some kind of thing that is a root of x^2+1. Let F(i) be the expressions of the form a+bi with a and b in F. Then using i^2=-1 you can prove that F(i) is a field with 9 elements.

Anonymous

Anonymous Tue 21 Apr 2020 12:43:52 No.11587718 Report

Quoted By: >>11587739

>>11587372
What's there not to understand? It's the number i defined such that i*i = -1

Anonymous

Anonymous Tue 21 Apr 2020 12:51:26 No.11587734 Report

Quoted By: >>11588147 >>11591241

What triplet of complex numbers $x,y,z\neq 1$ satisfies $x^2+y^2+z^2+xyz-4=0$ ?

Anonymous

Anonymous Tue 21 Apr 2020 12:53:19 No.11587739 Report

Quoted By: >>11587964

>>11587718
For one thing why you gloss over the fact that there are two numbers i for which i^2=-1.

Anonymous

Anonymous Tue 21 Apr 2020 12:55:44 No.11587742 Report

Quoted By: >>11588407

>>11587657
>>11587543
how would it be solved? with quadratic formula?

Anonymous

Anonymous Tue 21 Apr 2020 14:30:47 No.11587964 Report

Quoted By:

>>11587739
i is not two numbers, it's one number.
What you meant to say is that there are two roots of x^2 + 1 = 0 and that has nothing to do with his post.

Anonymous

Anonymous Tue 21 Apr 2020 15:16:56 No.11588132 Report

Quoted By:

>>11587558
The square root is defined as the principal value of the inverse square. If i^2=-1, then sqrt(-1) MUST equal either i or -i, depending on convention.

Anonymous

Anonymous Tue 21 Apr 2020 15:23:27 No.11588147 Report

Quoted By: >>11588164

>>11587734
https://en.wikipedia.org/wiki/Quaternion
https://en.wikipedia.org/wiki/Split-complex_number
So the split-quaternions?
How do split-complex numbers even work? they seem like half-baked quackery (like wheel theory)

Anonymous

Anonymous Tue 21 Apr 2020 15:29:16 No.11588164 Report

Quoted By:

>>11588147
study clifford algebras. this is the framework where "split-complex numbers" and shit like that naturally appears.

Anonymous

Anonymous Tue 21 Apr 2020 15:35:08 No.11588177 Report

Quoted By: >>11588387

Every EE knows i is for losers. Real men use j instead.

Anonymous

Anonymous Tue 21 Apr 2020 16:59:37 No.11588387 Report

Quoted By:

>>11588177
>i is for losers
you use it a LOT

Anonymous

Anonymous Tue 21 Apr 2020 17:04:13 No.11588407 Report

Quoted By:

>>11587742
https://i.stack.imgur.com/H9211.png

Anonymous

Anonymous Tue 21 Apr 2020 17:32:20 No.11588480 Report

Quoted By: >>11589232

>>11587456
>>11587464
>>11587531
I mean yes, i is adjoined to the reals by modding out by ideal (x^2 + 1), but this construction implicitly gives a name to to such an x, which we call i. So it’s not wrong to say i = sqrt(-1) as long as you understand it’s done formally by a construction that extends you to the complex plane.

Anonymous

Anonymous Tue 21 Apr 2020 18:31:57 No.11588707 Report

Quoted By: >>11588851

>>11587372
Google the fundamental theorem of algebra.

i is also used a lot in EE for AC circuits, signals and control systems

Anonymous

Anonymous Tue 21 Apr 2020 18:36:39 No.11588733 Report

Quoted By:

>>11587372
i^2 = j^2 = k^2 = ijk = -1

Anonymous

Anonymous Tue 21 Apr 2020 19:20:17 No.11588851 Report

Quoted By: >>11588885

>>11588707
that's not exactly what makes i significant mathematically. Being able to introduce geometric intuition by construction the complex plane using adjoining elements is mathematically significant and a fairly eloquent idea.

Anonymous

Anonymous Tue 21 Apr 2020 19:30:21 No.11588885 Report

Quoted By:

>>11588851
>NOOOO YOUR APPLICATION ISN'T SIGNIFICANT BECAUSE I LIKE MY APPLICATION BETTER

Anonymous

Anonymous Tue 21 Apr 2020 19:38:24 No.11588912 Report

Quoted By:

it allows you to deal with certain properties such as current and voltage with relation to capacitors and inductors

Anonymous

Anonymous Tue 21 Apr 2020 20:19:22 No.11589064 Report

Quoted By: >>11589088 >>11589307

>>11587442

>can describe rotation off of the number line

Seems pretty arbitrary desu. Like you can have a y-axis number line in base pi (instead of base i) and it would be exactly as useful. "Multiplying by pi results in a rotation off the number line" makes about as much sense as "multiplying by i results in a rotation off the number line".

Show me where I'm wrong.

Anonymous

Anonymous Tue 21 Apr 2020 20:26:09 No.11589088 Report

Quoted By:

>>11589064
No you cannot, because pi is a real number, so multiplication by pi on the real number line will always result in a scaling of the real number line, not any rotation.
i is not a real number, so you can define it in such a way that multiplication by i does indeed correspond to a rotation of the real number line.
Also, yes it is arbitrary. Every definition in math is arbitrary; that is unimportant though, because these things that seem to be arbitrarily defined end up having amazing applications, and can be used to simplify things immensely.

Anonymous

Anonymous Tue 21 Apr 2020 20:28:58 No.11589102 Report

Quoted By: >>11589232

>>11587456
you forgot a+bi

Anonymous

Anonymous Tue 21 Apr 2020 20:37:41 No.11589143 Report

Quoted By:

>>11587372
Let p(x) = x2+1. Then i = x + ?p(x)? in the field R[x]/?x2+1?. This gives you
p(i) = p(x)+?p(x)? = 0+?p(x)?, thus providing a field that contains an embedding of R by the mapping af(x)=a, while also containing a root for the polynomial p(x)=x2+1. Furthermore, R[x]/?x2+1? is the smallest field with this property, up to isomorphism.

What does all of this mean? It means git gud lmaomao

Anonymous

Anonymous Tue 21 Apr 2020 20:53:23 No.11589208 Report

Quoted By: >>11589228

>>11587372
Let $p(x) = x^{2}+1$ . Then $i = x + \langle p(x)\rangle$ in the field $\mathbb{R}[x] / \langle x^{2}+1\rangle$ . This gives you
$p(i) = p(x)+ \langle p(x) \rangle = 0+ \langle p(x) \rangle$ , thus providing a field that contains an embedding of $\mathbb{R}$ by the mapping $a \mapsto f(x)=a$ , while also containing a root for the polynomial $p(x)=x^{2}+1$ . Furthermore, $\mathbb{R}[x]/\langle x^{2}+1 \rangle$ is the smallest field with this property, up to isomorphism.

What does all of this mean? It means git gud lmaomao

Anonymous

Anonymous Tue 21 Apr 2020 20:59:51 No.11589228 Report

Quoted By:

>>11589208
Shid, I meant $a \mapsto a+\langle x^{2}+1 \rangle$ , not $a \mapsto f(x)=a$ . I need to git gud too. For real though, just google what I wrote if you're actually interested.

Anonymous

Anonymous Tue 21 Apr 2020 21:00:17 No.11589232 Report

Quoted By: >>11589295

>>11588480
>it’s not wrong to say i = sqrt(-1)
Roots are only defined on the cut complex plane. Otherwise they're not functions and, e.g. cannot be applied to equations. Every rotation of 2pi would be a solution.
Sure, you can also cut elsewhere, but convention is negative real numbers.

>>11589102
?

Anonymous

Anonymous Tue 21 Apr 2020 21:18:29 No.11589295 Report

Quoted By: >>11591346

>>11589232
That's just where the discontinuity occurs. There's no problem defining the principal value of a square root for the negatives, or defining a multivalue function. Unless you're being a pedant and explaining to middle-schoolers, there's no problem with saying sqrt(-1)=i, if it's clear you and your audience know what you're saying.

Anonymous

Anonymous Tue 21 Apr 2020 21:21:35 No.11589307 Report

Quoted By:

>>11589064
>Like you can have a y-axis number line in base pi (instead of base i) and it would be exactly as useful

Multiplying by a real number by 'i' gives you something that isn't on the real number. That's a 90 degree rotation on the complex plane.

Multiplying by pi just scales your original number by pi.

But hey, if you can find some notational trick that makes it work and describe things like light and solutions to differential equations, then power to you.

Anonymous

Anonymous Wed 22 Apr 2020 10:54:11 No.11591014 Report

Quoted By:

>Be maths lad
>Try and do maths with the root of negative numbers for funsies
>Realise that you can use it make sense of a bunch of stuff
>Physics lads show that these results also work for physical, measurable things

That's basically it. Maths is all about trying stuff, and seeing if it makes sense.

Anonymous

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Anonymous Wed 22 Apr 2020 11:00:21 No.11591026 Report

Quoted By:

>>11587372
you can create nice graphs with the number

Anonymous

Anonymous Wed 22 Apr 2020 13:35:10 No.11591241 Report

Quoted By:

>>11587734
x=0
y=0
z=2
But really you can plot a surface of solutions in R&3.

Anonymous

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Anonymous Wed 22 Apr 2020 13:48:10 No.11591269 Report

Quoted By:

Let's say that you are doing some complicated algebra, shuffling around equations and the like.

At the beginning, you start with a real number.
At the end, your result is a real number.
But the middle, you have imaginary numbers.

It's sort of like you are storing a hypothetical construct that you eliminate from your equation later.
You can treat this thing algebraically as if its a real thing, knowing that in the end it will be flushed out.

Imaginary numbers also have rules associated with how they work. You can exploit these rules to accomplish things that you desire.
So for example, you might set up an equation where the real part represents something, and the imaginary part represents something else (like voltage, or energy, or whatever). But you set it up in advance knowing that the way imaginary numbers work will do the algebra for you in order to get the result out you want at the end.

I hope that makes some sense. In general it's a tool you can use to produce a result.

Anonymous

Anonymous Wed 22 Apr 2020 14:17:51 No.11591346 Report

Quoted By: >>11591587

>>11589295
>multivalue function
Not a function. With all the detriments.

Anonymous

Anonymous Wed 22 Apr 2020 14:37:00 No.11591377 Report

Quoted By:

The "real" in real numbers is a misnomer. All numbers are just logical fictions anyway, tools to solve problems.

El Arcón

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El Arcón Wed 22 Apr 2020 15:01:17 No.11591417 Report

Quoted By:

Anonymous

Anonymous Wed 22 Apr 2020 16:24:51 No.11591587 Report

Quoted By:

>>11591346
Of course it's not a function, it's a multi-valued function. You have to introduce principal values and branches and such. But it's incorrect to say that the square root of negative numbers is not defined.

Anonymous

Anonymous Wed 22 Apr 2020 16:38:06 No.11591627 Report

Quoted By:

>>11587372
The wave function is a complex number

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