Someone explain wheres error

>>8952475
sqrt(ab)!=sqrt(a)sqrt(b) when a or b is neg I guess

[math]\sqrt{(-1)(-1)}=1\neq-1=i\cdot{i}[/math]

Edgar Allen Poe knows dick about math, that's your problem.

https://math.stackexchange.com/questions/49169/why-sqrt-1-times-1-neq-sqrt-12

there's an interesting article on euler's misuse of the product rule in Elements of Algebra, too

http://www.martinezwritings.com/m/Euler.html

>>8952487
But
[math] \sqrt{(-1)(-1)} = \sqrt{-1} \cdot \sqrt{-1} = \i \cdot \i [/math]

>>8952487
No, it's a problem with the formulation of [math]i[/math]. Wildberger was right.

Example:
[math]3=\sqrt{9}=\sqrt{3\cdot3}=\sqrt{3}\cdot\sqrt{3}=3[/math]

>>8952508
First equal sign isn't valid, brainlet.
>[math]1=\sqrt{(-1)(-1)}=\sqrt{-1}\sqrt{-1}=-1[/math]
This is you being a brainlet.

>>8952477
literally just this

nothing more to it

Bro... -1 x -1= 1....

>>8952518
>3=3
Really showed me there. Learn what a principal root is, fagtron.

>>8952528
The OP image is a completely valid decomposition of 1. I just did something analagous and it worked while his example didn't.

>>8952533
>I just did something analagous
Your father should have worn a condom.

Also (2^0)(2^0) = (2^0)^2 which simplifies to 2^0...exponents multiply when they're in brackets like that

>>8952475
that's what happens what you make up something to solve equations with no solutions.
>muh i isn't made up
>muh sqrt-1 isn't gibberish

>>8952475
Both -1 and 1 are equal to Sqrt1, then the equation has 2 possible outcomes, 1 being 2=1+1, which is true, and the second being 2=1-1, which is false

>>8952561
square root of -1 you stupid brainlet

>>8952533
>The OP image is a completely valid decomposition of 1

No it is not. The 4 fourth equals sign is false.

[math] \sqrt{ - 1} \cdot \sqrt{ -1 } = e^{i \frac{\pi}{2}}\cdot e^{i \frac{\pi}{2}} = e^{\pi} = -1 [/math]

while

[math]\sqrt{-1 \cdot -1} = 1 = e^{i \cdot 0} [/math]

It becomes clear when you define the square root function precisely for complex numbers.

For eg, if you define it as Sqrt: re^(it) -> sqrt(r)*e^(it/2), where sqrt() is the usual square root function for non-negative reals and t lies in [0, 2pi), then Sqrt(a*b) = Sqrt(a) * Sqrt(b) only when the sum of their angles is in [0, 2pi)

>>8952578
good job looking stupid

>>8952583
What are you talking about?

>>8952477
That's a funny identity you've got there, bud.

>>8952477
There's a bit more to it but that's the idea

>>8952475
>(((imaginary))) numbers

>>8952585
It's just a brainlet sperging out.

[math]1^2=-1^2[/math]
[math]1=-1[/math]
look everyone I broke math xd

>>8952477
what does factorial has to do with anything?

>>8952648
!= is a notation for not equal sign

The square root of -1 can also be -i

So, take other as i and other as -i and you get -i*i=1.

You have to remember that square roots have always two solutions which can cumulate.

>>8952657
Not for human communication. You use it only when talking to compilers, interpreters and software """engineers""".

>>8952657
that would be =/=, or since we have tex commands here, [math]\neq[/math].

>>8952660
this

>>8952475
>2^0 = -1

Wtf

>>8952834
that i^2 dumbass

>>8952475
Sqrt (-1) actually has two values: i and -i, so, technically, [math]i \neq \sqrt {-1} [/math]

>>8952523
I'm not arguing there isn't a problem, I just understand why it works for positive integers but not negative ones.

[math]2 = \sqrt{4} = \sqrt{2 \cdot 2} = \sqrt{2} \cdot \sqrt{2} = 2[/math]

[math] 2 = \sqrt{4} = \sqrt{-2 \cdot -2} \neq \sqrt{-2} \cdot \sqrt{-2} = -2 [/math]

What I don't understand is why the equals sign isn't valid. An answer to that question from your part would warrant you to call others brainlets. Until then, I will assume that like most people on /sci/, you're just larping.

>>8952840
Yeah I think you got it right. Assuming one of them is -i and the other is i then the expression works out fine. Otherwise the logic is basically: >>8952638

>>8952841
That rule, sqrt(ab)=sqrt(a)sqrt(b) is valid only for positive real a and b.

>>8952897
I imagine it is, but is there a rationale as to why? Otherwise it's just a stipulation, which is not ideal.

>>8952841
>I understand why it works for positive integers but not negative ones
>I don't understand is why the equals sign in [math]\sqrt{(-1)(-1)}=\sqrt{-1}\cdot\sqrt{-1}[/math] isn't valid
Pick one.

>>8952907
typo. I meant 'don't understand' for both.

>>8952902
>>8952910
It's not a stipulation. This identity doesn't hold for all a and b because square root is a multivalued function. That's why it's commonly restricted to principal root.

>>8952839
Where is the I?

>>8952920
Okay thanks!

Brainlets were a mistake.

>>8952475
i.i does not equal sqrt(-1)(-1) if you multiply the negative ones together first before you square root

>>8952902
Yes, because if you allow it with negative numbers and complex ones, you get the contradiction 1 = -1. You are familiar with proof by contradiction right?

>>8953017
Fair point.

>>8952840
Wrong.
The sqrt is defined as a function and there is only one value for each square root.

Also, -i^2 = 1 not -1 so you're double fucked.>>8952888

File: 1425706734001.jpg (44KB, 640x560px) Image search: [Google]

44KB, 640x560px

>>8953126

>>8953126
>[math](-i)^2=1[/math]
>the absolute state of this brainlet

>>8953126
>The sqrt is defined as a function and there is only one value for each square root.
Square root is a multifunction, you brain dead cretin. Stop posting and kys yourself.

>>8952527
Brainlet boy. i =/= -1, i^2 = -1

>>8953180
That's clearly not what anon meant in his post about there being two roots for sqrt -1.
>>8953184
Nope. The sqrt of a number is always the positive root.

What youre thinking of are solutions of the form y = sqrt (x). Clearly, x = +- sqrt (y) but the sqrt refers specifically to |sqrt (y)|.

How does it feel knowing you aren't fit to teach elementary operations?

>>8953126
>The sqrt is defined as a function and there is only one value for each square root.
NO, IT IS NOT
the [math]\sqrt{i} = -1[/math] is the biggest cancer I have ever seen holy fuck

>>8953184
>>8953499
[math]
\sqrt {x^2} \ne \pm x, \quad \sqrt {x^2} = \left | x \right |
[/math]

>>8952475
can just take the square root of individual terms to get rid of the entire square root function