Someone explain wheres error

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

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

>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

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

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

square root of -1 you stupid brainlet

>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)

good job looking stupid

What are you talking about?

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

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

>(((imaginary))) numbers

It's just a brainlet sperging out.

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

what does factorial has to do with anything?

!= 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.

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

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

this

>2^0 = -1

Wtf

that i^2 dumbass

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

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 Veeky Forums, you're just larping.

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:

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

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

>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.

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

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.

Where is the I?

Okay thanks!

Brainlets were a mistake.

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

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?

Fair point.

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.

...

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

>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.

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

That's clearly not what user meant in his post about there being two roots for sqrt -1.
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?

>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

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

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