/sci/ - Science & Math » Thread #10760396

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The imaginary numbers are WRONG

Anonymous Thu 27 Jun 11:33:55 2019 No.10760396 View Reply Original Report

Quoted By: >>10760404 >>10760405 >>10760418 >>10760431 >>10760457 >>10760491 >>10760600 >>10760605 >>10760733 >>10760756 >>10760934 >>10760938 >>10761048 >>10761707 >>10761717 >>10761723 >>10761750

Just admit it. You can't just invent numbers that don't exist. SQRT (-1) is an illegal number so it is a huge mistake to use it. It amounts to mathematical fraud. It is in fact a shameful hoax on humanity. Those that that have such egos to think they are above mathematical law will be condemned to mediocrity. No truths can be found and no are problems are solved if one pursues the route of the imaginary, complex theoretical pseudo mathematician. PERIOD.

Anonymous

Anonymous Thu 27 Jun 2019 11:40:14 No.10760404 Report

Quoted By: >>10760415 >>10760719

>>10760396
>SQRT (-1) is an illegal number so it is a huge mistake to use it.
Prove it

>It amounts to mathematical fraud.
Prove it

>Those that that have such egos to think they are above mathematical law will be condemned to mediocrity
What mathematical laws say that sqrt(-1) can't be done?

>No truths can be found and no are problems are solved if one pursues the route of the imaginary, complex theoretical pseudo mathematician
Electrical engineering, aeroelastic flutter, and oscillation damping disagree with you

Anonymous

Anonymous Thu 27 Jun 2019 11:40:55 No.10760405 Report

Quoted By: >>10760719

>>10760396
Stupid troll, shut up moron

Anonymous

Anonymous Thu 27 Jun 2019 11:48:40 No.10760415 Report

Quoted By: >>10760427 >>10760437 >>10760904

>>10760404
>Prove it
Prove that it isn't

>Prove it
Prove that it doesn't

>What mathematical laws say that sqrt(-1) can't be done?
Give me a number, such that multiplied by itself gives as -1. Protip, you can't

>Electrical engineering, aeroelastic flutter, and oscillation damping disagree with you
These are all empirical branches of science. Where all discovery has been made by experimental approaches. The inclussion of SQRT (-1) in this branches is a ploy by pseudo-intellectual mathematicians to gain recognition.

Anonymous

Anonymous Thu 27 Jun 2019 11:49:51 No.10760418 Report

Quoted By: >>10760719

>>10760396
Confirmed for failing calculus II.

Anonymous

Anonymous Thu 27 Jun 2019 11:55:25 No.10760427 Report

Quoted By:

>>10760415
Shut up moron, your iq is low and you are stupid.

Anonymous

Anonymous Thu 27 Jun 2019 11:56:38 No.10760431 Report

Quoted By: >>10760719

>>10760396
>You can't just invent numbers that don't exist.
Boy, you're in for a surprise

Anonymous

Anonymous Thu 27 Jun 2019 11:59:02 No.10760437 Report

Quoted By:

>>10760415
>Give me a number, such that multiplied by itself gives as -1. Protip, you can't
i

Shut up you piece of click clack without a clock

Anonymous

Anonymous Thu 27 Jun 2019 12:00:06 No.10760441 Report

Quoted By: >>10760473 >>10760493 >>10761051 >>10762759

To everyone saying that i'm wrong, try to explain this.

We know that:
$i^{2} = -1$

But then we have:
$i^{2} = i*i = \sqrt{-1} *\sqrt{-1} = \sqrt{-1 * -1} =\sqrt{1} = 1$

So you are implying that 1 = -1. Pathetic.

Anonymous

Anonymous Thu 27 Jun 2019 12:07:19 No.10760457 Report

Quoted By: >>10760719

>>10760396
$\mathbb{C} = \mathbb{R}[x] / \langle x^2+1 \rangle \\ i = x+\langle x^2+1 \rangle$

Anonymous

Anonymous Thu 27 Jun 2019 12:18:59 No.10760473 Report

Quoted By: >>10760488

>>10760441
i isn't exactly a sqrt(-1), because sqrt(-1) can be both i and -i. Think of it this way:
Imagine ordered pairs of real numbers (a,b) with operations + and • which are defined as (a1,b1)+(a2,b2)=(a1+a2,b1+b2) and (a1,b1)•(a2,b2)=(a1*a2-b1*b2,b1*a2+b2*a1) where + and * are addition and multiplication of real numbers. Observe that (0,1)•(0,1)=(-1,0) and that operations on pairs (x,0) closely resemble that of real numbers. Now, we call (a,b) a complex number and write it as a+bi. (0,1) = i. Even though it is entirely possible to use complex numbers with only ordered pairs of reals, a+bi notation is way more convenient to use. It is commonly used in AC electronics, signal processing etc.

Anonymous

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Anonymous Thu 27 Jun 2019 12:31:33 No.10760488 Report

Quoted By: >>10760582

>>10760473
Damn when you put it like that it actually makes sense. Thanks i guess anon.

Anonymous

Anonymous Thu 27 Jun 2019 12:33:09 No.10760491 Report

Quoted By: >>10760719

>>10760396
lol coping undergrad failing math 1

Anonymous

Anonymous Thu 27 Jun 2019 12:34:45 No.10760493 Report

Quoted By: >>10760528

>>10760441
Can you prove that $\sqrt{a} \cdot \sqrt{b}= \sqrt{a\cdot b}$ for both a,b<0?

Anonymous

Anonymous Thu 27 Jun 2019 13:13:25 No.10760528 Report

Quoted By: >>10760551 >>10761051

>>10760493
can you prove that it isn't?

Anonymous

Anonymous Thu 27 Jun 2019 13:27:36 No.10760551 Report

Quoted By:

>>10760528
>can you prove that it isn't?
You prove that $\sqrt{a} \cdot \sqrt{b}= \sqrt{a\cdot b}$ for positive numbers via $(\sqrt{a} \cdot \sqrt{b})^2= a\cdot b$ , after that you have to prove that $\sqrt{a} \cdot \sqrt{b}$ is positive. If you drop this last part you can have $\sqrt{a} \cdot \sqrt{b}= -\sqrt{a\cdot b}$ , which is the relation we have to use for $i^2$ .

Anonymous

Anonymous Thu 27 Jun 2019 13:56:15 No.10760582 Report

Quoted By:

>>10760488
>when you put it like that
That's literally the standard definition. Do people seriously post math threads without knowing shit about math?

Anonymous

Anonymous Thu 27 Jun 2019 14:07:32 No.10760600 Report

Quoted By: >>10760719

>>10760396
You clearly don't have any clue what the formal notion is. If we have a unital commutative ring, like R[x] (the ring of polynomials with coefficients in R) we want to be able to find the roots of elements. But we can't find a root of x^2 + 1. So we can throw in that root into R to get a larger field. This is equivalent to taking the quotient R[x]/<x^2 + 1>, i.e. modding by the principal ideal of "polynomials with a factor of x^2 + 1". It is a basic argument to prove that such a ring is in fact a field since x^2 + 1 is irreducible and thus the ideal is maximal, and the isomorphism with your current notion of C is very clear. Merely send 1 to 1 and x to i. This is how one formally defines the complex numbers.
You can see that x=i is in fact a square root of -1, since x^2 + 1 = 0 under the quotient.

Anonymous

Anonymous Thu 27 Jun 2019 14:11:25 No.10760605 Report

Quoted By: >>10760719

>>10760396
It's just a tool, and it turns out that it was invented in a way that it's pretty useful ^^

Anonymous

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Anonymous Thu 27 Jun 2019 15:16:23 No.10760719 Report

Quoted By:

>>10760404
>>10760405
>>10760418
>>10760431
>>10760457
>>10760491
>>10760600
>>10760605
mfw all this people respond to obvious bait

Anonymous

Anonymous Thu 27 Jun 2019 15:24:28 No.10760733 Report

Quoted By:

>>10760396
the absolute state of /sci/

Anonymous

Anonymous Thu 27 Jun 2019 15:36:06 No.10760756 Report

Quoted By: >>10760824

>>10760396
>You can't just invent numbers that don't exist.
does 2 exist ? does -2 exist ? does 1/2 exist ? does pi exist ?

Anonymous

Anonymous Thu 27 Jun 2019 16:03:26 No.10760824 Report

Quoted By: >>10760930

>>10760756
>does 2 exist ?
yes
>does -2 exist ?
yes
>does 1/2 exist ?
yes
>does pi exist ?
NO

Anonymous

Anonymous Thu 27 Jun 2019 16:35:44 No.10760904 Report

Quoted By: >>10760920

>>10760415
> electrical engineering is empirical
yeah, no. electrical engineering is not the same as being an electrician, retard.

Anonymous

Anonymous Thu 27 Jun 2019 16:41:35 No.10760920 Report

Quoted By:

>>10760904
Engineering in general is empirical, you fuckin moron. Hell, the entire physical SCIENCE of electrodynamics is empirical.

Anonymous

Anonymous Thu 27 Jun 2019 16:49:16 No.10760930 Report

Quoted By: >>10760936

>>10760824
Okay you fucking retard, give me a 2. Don't give two OF something, give me a "pure" two.

Anonymous

Anonymous Thu 27 Jun 2019 16:50:22 No.10760934 Report

Quoted By:

>>10760396
I will not "admit" to something that is obviously not true.

Anonymous

Anonymous Thu 27 Jun 2019 16:50:58 No.10760936 Report

Quoted By:

>>10760930
*reaches into platonic realm*
here you are

Anonymous

Anonymous Thu 27 Jun 2019 16:51:37 No.10760938 Report

Quoted By: >>10760957 >>10761037

>>10760396
The only reason i [spoiler:lit]and -1[/spoiler:lit] exist is because of squareroots. The definition of a squareroot is flawed and incomplete, and the "root" of your problems.
While imaginary numbers like i and negatives have undeniably proven their worth, the exact meaning of what an "squareroot" is still needs to be reevaluated, especially given its lack of commutativity and production of multiple equivalent values.
>inb4 "why does the squareroot need to be commutitative?"

Anonymous

Anonymous Thu 27 Jun 2019 16:58:10 No.10760957 Report

Quoted By:

>>10760938
>unary operator
>commutative
besides that, why the fuck would it need to be commutative? Whatever the fuck that means here.

Anonymous

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Anonymous Thu 27 Jun 2019 17:32:24 No.10761019 Report

Quoted By: >>10761043 >>10762464

$\sqrt{-1} = -i$
$\sqrt{-1}^2 = (-i)^2$
$-1 = -1$
therefor $\sqrt{-1} = \pm i$ or in short, it's all a fucking meme

Anonymous

Anonymous Thu 27 Jun 2019 17:43:28 No.10761037 Report

Quoted By:

>>10760938
>the exact meaning of what an "squareroot" is still needs to be reevaluated
z -> z^2 is a two-sheeted covering C* -> C* of Lie groups. a squareroot is the fibre. to get "the" squareroot amounts to picking a section, this is not possible globally as the covering is not trivial. nothing to be "reevaluated", this is all there is to it.

Anonymous

Anonymous Thu 27 Jun 2019 17:47:56 No.10761043 Report

Quoted By: >>10761060

>>10761019
That's true but $\sqrt{}$ denotes the absolute value of square root, so root(-1) only equals i

Anonymous

Anonymous Thu 27 Jun 2019 17:53:58 No.10761048 Report

Quoted By:

>>10760396
>he believes in sqrt()
brainlet. i exists as an extension of the fundamental theorem of algebra. That is all that is needed. """Square Root""" is a convenience, it doesn't really exist.

I bet you believe division exists too.

Anonymous

Anonymous Thu 27 Jun 2019 17:54:14 No.10761051 Report

Quoted By:

>>10760528
Yes, of course. In fact, you essentially proved it yourself with the argument in >>10760441. If it was true, then we'd have a contradiction. Therefore it cannot be true.

Anonymous

Anonymous Thu 27 Jun 2019 18:02:49 No.10761060 Report

Quoted By: >>10761674

>>10761043
>That's true but ...
Not really; the post you're replying to is misleading at best. Their logic goes the wrong way. They assumed a conclusion and deduced a truth. That's not a proof unless all the steps can be done in reverse (i.e. all statements are "if and only if"). In this case, the middle statement does NOT imply the first statement, so it's not a proof at all.

Anonymous

Anonymous Thu 27 Jun 2019 21:29:20 No.10761674 Report

Quoted By:

>>10761060
It's valid logic for the square root function. (But not the principal root function of course)

Anonymous

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Anonymous Thu 27 Jun 2019 21:41:29 No.10761707 Report

Quoted By: >>10761737 >>10762765

>>10760396
>SQRT (-1) is an illegal number
can we invent a new set of numbers larger than the imaginary ones called ILLEGAL numbers pretty please?
I need this in my life

Anonymous

Anonymous Thu 27 Jun 2019 21:49:02 No.10761717 Report

Quoted By:

>>10760396
>You can't just invent numbers that don't exist
Then how can you have math at all nerd?

Anonymous

Anonymous Thu 27 Jun 2019 21:51:52 No.10761723 Report

Quoted By: >>10761728 >>10761746

>>10760396
This guy gets it.

I stopped caring about math when I was introduced to the concept of imaginary numbers. What a crock of shit. If your equation can only be solved by inventing numbers that can't exist, like some kind of math deity , then you are fucking wrong and the math is flawed. Same for algebra solutions that basically say "the correct answer is whatever the correct answer is". Thats what the math said transcribed to words but god forbid if i wrote in down in english instead of the ancient math runes the teacher word mark me wrong.

Math is logical and numbers never lie my ass. Math is just as flawed as any other human construct.

Anonymous

Anonymous Thu 27 Jun 2019 21:54:28 No.10761727 Report

Quoted By:

Niggers, all of you

Anonymous

Anonymous Thu 27 Jun 2019 21:54:49 No.10761728 Report

Quoted By:

>>10761723
brainlet niggers, you and op

Anonymous

Anonymous Thu 27 Jun 2019 21:59:04 No.10761737 Report

Quoted By:

>>10761707
$\pi + i_\text{fish}$

One illegal number.

Anonymous

Anonymous Thu 27 Jun 2019 22:02:20 No.10761746 Report

Quoted By: >>10761775

>>10761723
>"imaginary" numbers
*yields extreme utility in everyday applications of science and engineering*
Huh, looks like they're just as real as "real" numbers afterall

Anonymous

Anonymous Thu 27 Jun 2019 22:04:02 No.10761750 Report

Quoted By: >>10761759

>>10760396
>wave and oscillation don't exist
weakest bait I've seen in a while

Anonymous

Anonymous Thu 27 Jun 2019 22:09:15 No.10761759 Report

Quoted By: >>10761767

>>10761750
You know you can describe these with sines and cosines, right?
>inb4 sine and cosine are complex exponentials

Anonymous

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Anonymous Thu 27 Jun 2019 22:10:36 No.10761764 Report

Quoted By: >>10761797

Moron's like OP are why mathematicians now admit that calling $\sqrt{-1}$ "imaginary" was probably a mistake.

And oh, OP, about numbers... they're ALL imaginary!

Anonymous

Anonymous Thu 27 Jun 2019 22:12:51 No.10761767 Report

Quoted By:

>>10761759
Have fun trying to write the frequency response of an LC circuit down without complex exponentials

Anonymous

Anonymous Thu 27 Jun 2019 22:14:34 No.10761775 Report

Quoted By: >>10762472 >>10762549

>>10761746
>"God" doesn't exist
>*yields extreme utility in everyday applications of life and human existence*
Huh, looks like He's just as real as anything else afterall

Anonymous

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Anonymous Thu 27 Jun 2019 22:15:45 No.10761778 Report

Quoted By:

Lmao faggots $i=\sqrt{-1}$ is literally a wrong definition, because roots must have one solution. $i^2=-1$ is correct.

Anonymous

Anonymous Thu 27 Jun 2019 22:21:34 No.10761797 Report

Quoted By:

>>10761764
>Moron's like OP are why mathematicians now admit that calling ?1???? "imaginary" was probably a mistake.
Exactly like this. If only people listened to Gauss and called them lateral numbers. OP and his followers are brainlets who cannot grasp that "imaginary" numbers are as areal and natural as any other.

Anonymous

Anonymous Fri 28 Jun 2019 04:46:55 No.10762464 Report

Quoted By:

>>10761019
You can write down the same thing for square roots of positive numbers as well. What's your point? The squaring is not an injective function.

Anonymous

Anonymous Fri 28 Jun 2019 04:49:26 No.10762472 Report

Quoted By:

>>10761775
Is your crux with imaginary numbers literally just that they're called imaginary?

Anonymous

Anonymous Fri 28 Jun 2019 05:33:57 No.10762549 Report

Quoted By:

>>10761775
God does exist, and I'm sure glad He's not the pathetic, degenerate false one the chr*stians call "father"

Anonymous

Anonymous Fri 28 Jun 2019 07:13:41 No.10762661 Report

Quoted By: >>10762861

Why do retards have problems with "imaginary" numbers but not negative or irrational numbers? They are as imaginary.

Anonymous

Anonymous Fri 28 Jun 2019 08:52:41 No.10762759 Report

Quoted By:

>>10760441
$ \sqrt{-1} \cdot \sqrt{-1} \neq \sqrt{-1 * -1} $

Anonymous

Anonymous Fri 28 Jun 2019 08:55:30 No.10762765 Report

Quoted By:

>>10761707
$\displaystyle \boxed{ \mathbb{O} \; \boxed{ \mathbb{H} \; \boxed{ \mathbb{C} \; \boxed{ \mathbb{R} \; \boxed{ \mathbb{Q} \; \boxed{ \mathbb{Z} \; \boxed{ \mathbb{N}}}}}}}} $

Anonymous

Anonymous Fri 28 Jun 2019 10:40:01 No.10762861 Report

Quoted By:

>>10762661
they have imaginary in the name, that's literally the only reason

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