Why isn't there an answer to sqrt(x)=-1

[math]\sqrt{-1} [/math]

>>8300936
sqrt(1)=+/-1
kys

Whatever the solution of
[math] \sqrt{x} = -1 [/math]
was, if would fulfill
[math] (\sqrt{x})^2 = (-1)^2 = 1 [/math]
and as long as we stick to [math] (\sqrt{x})^2 = x [/math],
this means the task was to find an object with
[math] x^2 = 1 [/math]

If we read 1 as the two-dimensional unit matrix
I = ((1,0), (0,1)),
then
i = ((0, -1), (1, 0))
does the trick.
In the field of matrices for the form
a·I+b·i
the equation
[math] x^2 = I [/math]
has a solution, namely
[math] x = i [/math]

Now to your question.
If we stick to + meaning a commutative operation and we want to extend the real numbers (a vector space over R), then a solution to
[math] \sqrt{x} = -1 [/math]
implies a solution to
[math] \sqrt{1+y} = -1 [/math]
for some y
but then
[math] 1+y = (-1)^2 = 1 [/math]
and
[math] y = 0 [/math]
IF there is a solution then it's
x = 1+y = 1+0 = 1

But if the square root ought to be a function (not related to values poping out of continuations alla Riemann), then we take [math] \sqrt{1} = +1 [/math]

>>8300968
Thank very much

>>8300948
Retard spotted.

>>8300968
>square root ought to be a function
[math] \displaystyle
\sqrt {x^2} \ne \pm x, \quad \sqrt {x^2} = \left | x \right |
\\
|x| =
\begin{cases}
\;\;\; x & ,x \geq 0 \\
-x & ,x < 0
\end{cases}
[/math]

I'm not sure if this is the right place to ask but it seemed the most correct

This is like grade 10 algebra.
For 2(2a * b)

Would it be 4a * 2b

Or would it be 4a * b.

My friend says it would be the latter and I insist it is the former. Anything outside the brackets acts as a coefficient, right? It multiplies everything inside.. Doesnt it?

If it is infact the latter, could you explain why?
I'm not the smartest of people but that is what i was taught through high school.

>>8300936
because square root by definition is always positive

>>8301048

Not sure if this is bait but yes your friend is literally retarded and I highly suggest he should consider suicide.

>>8301960
Uhmm I dont know if youre baiting but his friend is right as 2(2a * b) = 2*(2*a*b) = 2*2*a*b = 4a*b

>>8300936
sqrt(x) = -1 technically is true for x = 1, as (-1)^2 = 1, so -1 = sqrt(1), however, the conventional definition of the sqrt function only maps to the positive solutions, so sqrt(x) = -1 would have no solution. If we use an atypical definition of the sqrt function, where the function maps to the negative solution instead of the positive solution, we could prove that sqrt(x)=-1 has a solution, but sqrt(x)=1 would no longer have a solution.

>>8301040
sqrt(x) is undefined for values smaller than 0

>>8301048
what >>8302092 said