1 = 1^(1/2) = (-1 * -1)^(1/2) = (-1)^(1/2) * (-1)^(1/2)

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1+1=2

>>9152505
Noice

the definition of i is not that it is equal to sqrt(-1) or (-1)^(1/2), no

the definition of i is that it fulfills the equation i^2=-1

therefore, you substitute (-1)^(1/2) with i even though you are not allowed to

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

>>9152534
kek

>>9152534
Sqrt a^2 = |a|

>>9152534
These wojak images are being overused for the stupidest comments. That is actually funny in it's own unique autistic way.

>>9152505
>1^(1/2)
=±1
>(-1 * -1)^(1/2)
=±1
>(-1)^(1/2) * (-1)^(1/2)
=±1
>i * i
=-1
>i^2
=-1

Exponentiation z^w is multivalued for non-integer w.
Your first mistake was 1 = 1^(1/2)
Because -1 is also 1^(1/2)

>>9152642
Wrong. See >>9152600

>>9152650
I see no square roots in OP's post.
you fucking retard
sqrt(x) =/= x^(1/2)

>>9152652
no difference, if an exponent is to be defined as a function then it must be single valued

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>babby's first multivalued function
(-1)^(1/2) * (-1)^(1/2) = i * i
this step is not correct

>>9152533
Fucking retard who doesn't know what he's talking about

>>9152505
>(-1 * -1)^(1/2) = (-1)^(1/2) * (-1)^(1/2)
Wrong
(x*y)^z=(x^z)*(y^z) is not true if x,y are negative.

>>9152652
you != not retarded

>>9153973
>(x*y)^z=(x^z)*(y^z) is not true if x,y are negative.
Why wouldn't it

>>9154007
OP just proved that it wouldn't

>>9154007
Exponentiation is a function - you need to prove that this function has certain properties before you can just use it.

https://math.stackexchange.com/questions/2079710/law-of-exponent-for-complex-numbers

ln(xy)=ln(x)+ln(y) <==> (xy)^z=x^z+y^z

Suppose we have two negative numbers, ln(xy) is a valid evaluation while the right side is invalid.