/sci/ - Science & Math

[Anonymous No.10688456]

>inb4: undefined.

[Anonymous No.10688463]

X = -1^2

easy

[Anonymous No.10688466]

x=1

[Anonymous No.10688468]

>>10688463
-1^2 = 1

[Anonymous No.10688473]

>>10688466
Obviously doesn't work. sqrt(1) = 1 and 1 != -1

[Anonymous No.10688474]

z=-1

[Anonymous No.10688479]

>>10688468
(-1)^2/2=(-1)^1=-1

[Anonymous No.10688480]

[eqn]e^{2i\pi}[/eqn]

[Anonymous No.10688487]

>>10688480
>>10688473

[Anonymous No.10688488]

>>10688473
square root is multivalued, im sorry you had to find out this way.

[Anonymous No.10688490]

>>10688456
It's the negative branch of the square root function so x=1. In reality sqrt(1)=(-1,1).

[Anonymous No.10688495]

>>10688487
[eqn]\sqrt{e^{2i\pi}}=(e^{2i\pi})^{1\over2}=e^{2i\pi\cdot{1\over2}}=e^{i\pi}=-1[/eqn]

[Anonymous No.10688506]

>>10688488
No it's not, it has range of principal values. If it's multi-valued then it's not a function by definition of a function. If you accept that it has two solutions then you actually broke the whole math, because you quickly find out that almost all functions are actually multivalued and we can't do algebra with it. The whole point is - range of principal values is a big flow of modern mathematics. You ether have to redefine squire root or -1*-1 to be not equal to 1, or come up with whole sets of new fundamental axioms.

[Anonymous No.10688514]

>>10688495
Wrong.

[Anonymous No.10688607]

>10688456
x=1. since the squareroot divides the unit circle in half, it makes sense that the sqrt(1) = -1.

If you wanted sqrt(x) = 1, you would need x=1*1; that is, two unit circles would have to be explicitly defined in order to get a positive root.

[Nigerai No.10689175]

Pee pee

[Anonymous No.10689178]

Defined at [math]\sqrt{-1} \eq 1[/math]

[Anonymous No.10689202]

If [math]y\eq \sqrt{x}[/math] then
[math]x=y^{2}[/math]
[math]y\eq -1[/math]
[math]x=-1^{2}[/math]
[math]x=1[/math]

[Anonymous No.10689207]

>>10688514
Explain because that says it's right.

[Anonymous No.10689223]

>>10689207
What part of [math]\frac{1}{2} \notin \mathbb{Z}[/math] should I explain

[Anonymous No.10689233]

>>10689223
1/2 is in Z

[Anonymous No.10689307]

>>10689233
I guess it's just trolling, but there is possibility that you are really confused. So try this
https://en.wikipedia.org/wiki/Integer

[Anonymous No.10689430]

>>10689307
Oh, I actually thought Z meant complex numbers

[Anonymous No.10689445]

>>10689233
[math] \displaystyle
\boxed{ \mathbb{O} \;
\boxed{ \mathbb{H} \;
\boxed{ \mathbb{C} \;
\boxed{ \mathbb{R} \;
\boxed{ \mathbb{Q} \;
\boxed{ \mathbb{Z} \;
\boxed{ \mathbb{N}}}}}}}}
[/math]

[Anonymous No.10689456]

>>10688456
[math] \displaystyle
a+ib \leftrightarrow
\begin{bmatrix}
a & -b \\
b & a
\end{bmatrix}
\\
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
=
\begin{bmatrix}
-1 & 0 \\
0 & -1
\end{bmatrix}
[/math]

[Anonymous No.10689515]

>>10689456
elegant

[Anonymous No.10689519]

>>10689175
poo poo

[Anonymous No.10689562]

>>10688456

If the square root of -1 = i, then -1 = the square root of x when x is i.

Is that right, or am I just imagining it?

Pun intended.

[Anonymous No.10689570]

>>10688506
So how come this doesn't cause any actual issues?

[Anonymous No.10689588]

>>10688456
Curious.. has anyone model this shit or is it just theoretical BS

[Anonymous No.10689590]

[math] \displaystyle \sqrt{1}= \pm 1[/math]
...so x = 1 ... EZPZ fagitts

[Anonymous No.10689610]

>>10689570

It does. Any mathematician who doesn't recognize that the existence of quadrant III in the Cartesian plane system is a rogue free-radical in terms of what might exist numerically is just ignoring an obvious issue.

Sure, the square root of -1 is imaginary. That's great. But then what? So there's a thing that can't consistently exist in the system of mathematics by which we have mapped the entire universe.

That just means there's a part of the universe we haven't rationally determined yet.

So... like... it's pretty funny.

[Anonymous No.10689614]

>>10689223
That simple says an exponent of that form is periodic over some integer variable. It's doesn't say you can *only* raise to a power of an integer.

[Anonymous No.10689647]

>>10689614
>an exponent of that form is periodic over some integer variable

Do you not see that attempting to split the "Real Numbers" in half creates a paradox that is equivalent to the Law of the Excluded Middle?

Real numbers are both odd and even.

Even numbers can be split into multiples without fracturing, but odd numbers cannot.

Thirteen is a six and a seven, unless you break that law and imagine one of them as a half.

Three is a one and two, unless you do the same thing.

What is half of a real number, though?

Half of one is not real, because no matter how you halve it, there is part of it that only exists in the imaginary.

In terms of integers, one is equivalent to negative one, in terms of absolute value - yet, there is no square root of negative one that doesn't belong to the imaginary quadrant.

Bro, do you even Pythagoras, bro?

[Anonymous No.10689654]

>>10689647
https://youtu.be/-rAwdOdhUrA

[Anonymous No.10689684]

>>10689590
Correct. The sqrt-operation is equivalent to the question: what number, multiplied by itself, produces the number under the squareroot?

(-1) * (-1) = 1 is one true answer.

>>10689610
Quadrant III has nothing to do with OPs question whatsoever. The real issue is that f(x) = x2 (pic related) is not surjective over the whole R -> R domain, which makes its reverse function f^(-1) be multivalued and thus not a function in the strict sense, maybe a distribution.

Imo x = +1 for a root function of R -> R+
and x = -1 for a fuction of R -> R-

Pic related: two branches of x^2,a rotation by 90 degrees should remain both branches.

[Anonymous No.10689685]

>>10689647
And this kids is why drugs are bad.

[Anonymous No.10689700]

>>10688480
>>10688495
Incorrect due to the fact that [math]e^{2i\pi} = 1[/math] which fails to solve the equation.

There is no answer but yet again /sci/ takes the bait.

[Anonymous No.10689703]

>>10689684
>Quadrant III has nothing to do with OPs question whatsoever. The real issue is that f(x) = x2 (pic related) is not surjective over the whole R -> R domain, which makes its reverse function f^(-1) be multivalued and thus not a function in the strict sense, maybe a distribution.

If f(x) = x^2 is not subjective over the whole R -> R domain, then that means there is a division between both solutions that divides it into parts that are irresolvable without a third category that eschews the Law of the Excluded Middle that allows for an integer that both exists as a square root of negative one and also retains its status as something not divided.

That's what makes it imaginary.

Bivalence is the shell of the egg from which we hatch, I'm hoping.

[Anonymous No.10689708]

>>10689685

That's not really a valid critique, as much as you think you're cool for saying it. I'm sure you're cool, though, so go ahead and gloat about it. Like, wow, you're sober, good job.

[Anonymous No.10689710]

>>10688514
I think that shouldn't matter because [math]\sqrt{a}=b[/math] such that [math]a^2=b[/math] and for [math]e^{2\pi i}[/math] we have [math]e^{\pi i}\cdot e^{\pi i}=e^{2\pi i}[\math] and hence [math]\sqrt{e^{2\pi i}}=e^{\pi i}=-1[\math]

[Anonymous No.10689712]

>>10688456
1

because the only way to test if two square roots are equal is to square them and see if the results are equal.

Check mate faggots.

[Anonymous No.10689716]

>>10689703
>That's what makes it imaginary.

Not imaginary. Imaginary would be asking what value of x^2 equals to -1: the answer number is imaginary as no value in R does so - see pic related.

[Anonymous No.10689726]

>>10689710

Euler's Identity is not proof that the Law of the Excluded Middle is resolved, though.

It is proof that the result of ignoring the ambiguity of squares in terms of negative numbers can result in equations that surpass the ability of human understanding, though.

What do you do after that, though?

It's like finding a precious artifact in the attic.

Yes, it's amazing, and it's as valuable as everyone thinks it is... but what good does it actually do outside of what it can be traded for?

If you want to define value, you'll have to find another system.

[Anonymous No.10689728]

>>10689703
>>10689716

Instead, OP asks what number in roots equals -1, and the answers lie in the R2-Domain, but are, if you wish, bivalent.

Pic related, A and B.

[Anonymous No.10689731]

>>10689728
>what number in roots equals -1

Wait - is this a trick question?

There are no two roots who, when multiplied, might equal -1 in the set of real numbers.

That's literally why the the square root of negative one is called "imaginary."

[Anonymous No.10689743]

>>10689731
It doesn't matter tho. You allowed to use complex numbers, quaternions, octonions, split-complex number or any rigorous definition of a number. including [math]\sqrt{\theta } = -1[/math]. The actual trick here that it eventually lead to contradiction no matter what, like [math]\theta = 1 and \theta \neq 1[/math]. But there is no good reason why it cannot be defined. What essentially means that roots are broken in general.

[Anonymous No.10689759]

>>10689743

That sounds like the Law of the Excluded Middle being broken.

Maybe they're the same thing. Maybe they're different.

But 1=0 in a both/and system. For the same reason that θ=1 and θ≠1.

Every time it doesn't, there's a space where it does.

[Anonymous No.10689784]

>>10689731

(-1) * (-1) = 1 is the solution to the sqrt(x) = -1,
because sqrt asks what number multiplied by itself equals the number under the squareroot.

Simple example:

sqrt(49) = 7 because 7 * 7 = 49.
sqrt(25) = 5 because 5 * 5 = 25
sqrt(1) = -1 because (-1) * (-1) = 1

But, as >>10688488 pointed out, sqrt is multivalued, eg.

sqrt(49) = -7 because -7 * -7 = 49.
sqrt(25) = -5 because -5 * -5 = 25
sqrt(1) = 1 because (1) * (1) = 1

So the full solution is sqrt(x) = ±y, if you solve for y that gives y1 = +sqrt(x) and y2 = -sqrt(x).

In the function domain, this corresponds both to >>10689728 aswell as to pic related (the X-coordinate of A is 1), but >>10689716 is a whole different problem.

Hope I was not confusing you.

[Anonymous No.10689812]

>>10689784
>because sqrt asks what number multiplied by itself equals the number under the squareroot.
No. Square root cannot produce negative number. If it does it's not a square root anymore and not a function either. There is a reason why |x| defined as sqrt(x^2). In your world length of the objects are negative and positive at the same time.

[Anonymous No.10689822]

>>10689784
>So the full solution is sqrt(x) = ±y, if you solve for y that gives y1 = +sqrt(x) and y2 = -sqrt(x).

I think the confusion lies in this. The squared-function is not surjective and thus its inverse function (squareroot) is multivaluable.

Last pic, function x^2 and it's inverse.

[Anonymous No.10689824]

>>10689812
>Square root cannot produce negative number

How do you define the square root function?

> In your world length of the objects are negative and positive at the same time.

Mathematics is not physics, and length is defined as |x|, but x may be negative - think of vectors.

[Anonymous No.10689832]

>>10689824
There is no vectors with negative length.

[Anonymous No.10689837]

>>10689812
>There is a reason why |x| defined as sqrt(x^2).

x^2 cannot be negative in R unless you extend upon C with x = i, but this is not what OP asked (problem in pic >>10689716), however this doesn't imply that a squareroot cannot produce a negative value, which it must, imo.

If it doesn't, (-1)*(-1) cannot have a solution too, and we would limit ourself to the first quadrant.

[Anonymous No.10689840]

>>10689832
but there are vectors with negative values.

[Anonymous No.10689844]

>>10689784
>(-1) * (-1) = 1 is the solution to the sqrt(x) = -1

No, it's not. Unless there's no difference between 1 and -1, then that's not actually any solution at all.

That's why we can't accept it within a rational universe.

There's some awesome shit that might happen if we abandon a rational universe, but it's also really scary, but I'm willing to risk it because trying to remain completely bound by a rational universe is super fucking boring.

I want to sing in ecstasy. I want to live dreams.

I am a raw nerve boiling in the fabric of a universe that is asking to be built.

Feel me.

[Anonymous No.10689851]

2+2=5
>inb4: wrong

[Anonymous No.10689854]

>>10689840
Values and length is a quite different thing.

[Anonymous No.10689860]

>>10689844
OK then, how do you calculate a square root?

[Anonymous No.10689865]

>>10689854
Yes.

[Anonymous No.10689874]

>>10689851
By the same logic, you can say that i^2 = -1 is invalid question.

[Anonymous No.10689880]

>>10689874
It is. There isn't a proper answer. It's like division by zero.

[Anonymous No.10689898]

>>10689880
Exactly. But division by zero is actually even easier to define. If you accept that functions might be multivaluted and the result is always a set of numbers, not a particular number, then you can easily say that 1 / 0 = R

[Anonymous No.10689912]

>>10689898
>if you accept that
>if you delude yourself that
Life doesn't work that way

[Anonymous No.10689917]

>>10689898

Division by 0 is equivalent to a green point on the y-axis of pic related - call it however you wish, undefined, complex infinite, singularity, but it sure is not R.

OPs question is equivalent to x-coordinate of A in said pic >>10689784

[Anonymous No.10689931]

>>10689912
This is another way of saying that current math is not totally consistent and should be deeply rethought.

>>10689917
You just don't get me.

[Anonymous No.10689990]

>>10689931
>This is another way of saying that current math is not totally consistent and should be deeply rethought.

Your brain is not totally consistent, math is perfectly fine, but in order to understand and apply math a calculator and wikipedia is not enough.

This https://www.youtube.com/watch?v=1o0Nuq71gI4
is more consistent with realty and is entirely based on math and physics.

[Anonymous No.10690009]

>>10689990
>math is perfectly fine
Yeah, that why we have infinities all over the place in physics. Everything is fine..

[Anonymous No.10690017]

>>10690009

One drink plus half a drink plus a quarter of a drink plus one eight of a drink plus ... = two drinks.

[Anonymous No.10690057]

>>10688456
/thread

https://www.wolframalpha.com/input/?i=root(x)+%2B+1+%3D+0

>Solutions:
>(no solutions exist)

[Anonymous No.10690090]

>>10690057
Genius

[Anonymous No.10690141]

>>10688456
-1/12

[Anonymous No.10690463]

>>10690017
Name one instance in nature where this happens.

[Anonymous No.10690493]

>>10688456
Define what you mean by [math] \sqrt{\quad} [/math]. Is it a function from [math] \mathbb{R} \to \mathbb{R} [/math]? Then we need to restrict ourselves to either [math] \mathbb{R}_0^+ \to \mathbb{R}_0^+ [/math] or [math] \mathbb{R}_0^+ \to \mathbb{R}_0^{-} [/math]. Clearly if you choose the first option (which is standard practice), there is no solution. The second option has a trivial solution, [math]x = 1[/math].

Do you want to include all of [math] \mathbb{R} [/math] in the domain? Then we need to define a new function whose target is now some subset of [math] \mathbb{C} [/math]. Now you have the problem of branch cuts, which is widely studied in the field of complex analysis. (In this context of "widely studied", you may as well take the function defined on [math] \mathbb{C} \to \mathbb{C} [/math] and all its properties, and then restrict its domain to \mathbb{R} if you really need to.) So which branch cut do you want? Clearly, taking the one that goes "above" the complex plane cannot yield a value that has negative real part, so you need the branch cut that goes "below". So again it depends entirely on how you define the [math] \sqrt{\quad} [/math] mapping.

[Anonymous No.10690512]

[math] \displaystyle
\sqrt {x^2} \ne \pm x, \quad \sqrt {x^2} = \left | x \right |
\\
[/math]

[Anonymous No.10690648]

>>10690493
You are absolutely correct. I know that this is the only right answer to the question in the mathematical context that exists today. But than we can as ourselves is it context right? Look the choice between R+ and R- are arbitrary, but then we use this exact math with ARBITRARY choices in physics and hope that world is working the way as we choose it to work.