can 1=(-1)??? check this out

[monu2010]

1=1
1=sqrt[1]
1=sqrt[(-1)*(-1)]
1=sqrt[-1]*sqrt[-1]
1=iota*iota
1=iota^2
and iota raise to the power 2 is -1
therefore
1= -1

where i made a mistake?????
help me out

[Dross]

1=sqrt[-1]*sqrt[-1]
1=iota*iota

the square root of -1 is not just iota, just as the square root of 4 is not just 2 - it is ⁺/_-2, since (-2)² = 4 as well.
Similarly, (-iota)² = -1, so you cannot say that sqrt[-1] = iota, you have to say that sqrt[-1] = ⁺/_-iota ("plus or minus" iota).

[monu2010]

the square root of -1 is not just iota, just as the square root of 4 is not just 2 - it is ⁺/_-2, since (-2)² = 4 as well.
Similarly, (-iota)² = -1, so you cannot say that sqrt[-1] = iota, you have to say that sqrt[-1] = ⁺/_-iota ("plus or minus" iota).

my dear friend u just said that
sqrt[anything]=+/-(some value)

but this time i consider only one value which is not affecting equation
u are right for this type of situation
sqrt[4]=sqrt[4]
-/+2=+/-2
then
-2=+2

then we have to cosider sign either +ve or -ve "on both the sider

my question is totally different in this case i can consider only one sign at a time.

[timeshifter]

What you're doing is simply a violation of the quadratic equation. As previously mentioned, that was an artillery shell that just went off, but that's beside the point.

As previously mentioned, sqrt(4)=⁺/_-2, and sqrt(1)=⁺/_-1, but just because -1²=1 doesn't mean that -1=1. Comprende?

[Dross]

but this time i consider only one value which is not affecting equation

Are you saying you get to choose which of +i and -i you take? Quite simply not the case.

[zaza]

Or, to express timeshifter's point fully, (-x)^2 = (x)^2 for all x. This does not mean that x = -x. it just means that sqrt(x^2) = +/- x.

[capsulecorpjx]

1=1
1=sqrt[1]
1=sqrt[(-1)*(-1)]
1=sqrt[-1]*sqrt[-1]
1=iota*iota
1=iota^2
and iota raise to the power 2 is -1
therefore
1= -1

where i made a mistake?????
help me out

Second line isn't true.
1 <> sqrt[1]

+/- 1 = sqrt[1]