Minus one is an integer and √(-1) is and imaginary number. í means iota.
í = √(-1)
í * í = √(-1) * √(-1)
í ² = √(-1) * √(-1)
í ² = [ √(-1) ]²
í ² = -1
Roots and powers have a higher priority then multiplication and division. This must be taken into consideration with you calculations and you will get -1, not + 1.
í ² = √(-1) * √(-1) ╪ √(-1 * -1). This is where the error is.
2006-08-10 20:30:29
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answer #1
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answered by Brenmore 5
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Because of the discontinuous nature of the square root function in the complex plane, the law is in general not true. Wrongly assuming this law underlies several faulty "proofs", for instance the following one showing that -1 = 1:
-1=i*i=sqrt(-1)*sqrt(-1)=sqrt (-1*-1) = sqrt(1)=1
This is an invalid proof
2006-08-10 20:01:43
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answer #2
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answered by starflower 3
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The definition of "*" from i*i is not the one which you now for R, Q, N numbers (real, fractions, natural). It is another operation defined specialy for "Complex" numbers and it has the property described by you in the first place (i*i=-1). But this * cannot apply to the sqrt... the sqrt "*" is only defined for the "R, Q, N"... Hope you understand what I mean...:D
2006-08-10 19:58:16
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answer #3
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answered by None A 3
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It's fine until you try to combine them as sqrt(-1 * -1), when it becomes false. This only works for positive radicands.
2006-08-10 19:55:17
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answer #4
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answered by Charles G 4
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here is how it is done
sqrt(-1) * sqrt(-1) doesn't equal sqrt(-1 * -1) = sqrt(1) = 1
sqrt(-1) * sqrt(-1) = (sqrt(-1))^2
where as sqrt and ^2 cancel out, leaving you with
-1 as your answer
2006-08-11 03:23:00
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answer #5
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answered by Sherman81 6
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simplimently it's as a circle i*i= -1 then i*i*i = 1 then i*i*i*i= i^2*i^2 = -1 and that's to make -1 usable under the racine i mean the squar root
2006-08-10 20:08:21
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answer #6
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answered by karim b 1
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i is an imaginary number. i=sqrt of -1
i^2=-1
sqrt(-1)xsqrt(-1)=sqrt(-1)^2
2006-08-10 19:53:36
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answer #7
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answered by Anonymous
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I always thought a complex number was one greater than 10.
P.S. There isn't anything wrong with the equation, "i" just helps our primitive maths system to work correctly.
2006-08-11 01:23:19
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answer #8
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answered by David R 3
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-1= -1
i= sqrt[-1]
if you square i you get -1 so -1 = -1
i*i=sqrt[-1]*sqrt[-1]= -1
i cant find anything wrong with it, should there be? x
2006-08-11 03:38:55
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answer #9
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answered by Lisa D 2
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sqrt (-1)* sqrt(-1)=-1
Its perfectly alright......coz u cant combine both sqrt :)
2006-08-10 19:55:11
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answer #10
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answered by Anonymous
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