What is the squre root of -1

December

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  • Aug 24, 2009
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    Kaduwela
    Can somebody explain this...

    I will use "sqrt" to denote square root...

    In mathematics we say

    sqrt(-1) = i This is a definition

    Now i^2 = -1 (e.g. sqrt(4) = 2 ; therefore 2^2 = 4 )

    i^2 = sqrt(-1)*sqrt(-1)

    This can be written as

    i^2 = sqrt((-1)*(-1)) (e.g. sqrt(2)*sqrt(2) = sqrt(2*2) )

    so i^2 = sqrt(1) (as (-1)*(-1) = 1 )

    i^2 = 1 (as sqrt(1) = 1 )

    taking the square root of both sides

    i = 1

    But at the beginning i = sqrt(-1)

    So sqrt(-1) = 1

    :P
     
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    nagaya

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    Mar 18, 2007
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    this is wrong bro ;)


    for the problem in the thread;
    [sqrt(-1)]^2=-1 not +1,so it can't be done like this
     
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    Tom Riddle

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    I think the mistake you are making is when you are saying

    sqrt(-1)*sqrt(-1) = sqrt( (-1)*(-1) )

    That rule must only apply to positive numbers (I haven't seen that it only applies to positive numbers in a book, but ALL the other steps that you have taken are correct)

    I'll try to get a confirmation on the above.
     
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    koondeGoda

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    Nov 28, 2007
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    ₪ Hyper_Cube ₪
    Can somebody explain this...

    I will use "sqrt" to denote square root...

    In mathematics we say

    sqrt(-1) = i This is a definition

    Now i^2 = -1 (e.g. sqrt(4) = 2 ; therefore 2^2 = 4 )

    i^2 = sqrt(-1)*sqrt(-1)

    This can be written as

    i^2 = sqrt((-1)*(-1)) (e.g. sqrt(2)*sqrt(2) = sqrt(2*2) )

    so i^2 = sqrt(1) (as (-1)*(-1) = 1 )

    i^2 = 1 (as sqrt(1) = 1 )

    taking the square root of both sides

    i = 1

    But at the beginning i = sqrt(-1)

    So sqrt(-1) = 1

    :P

    (-1) ^ 1/2 * (-1 )^ 1/2 = -1(1/2+1/2)= -1
    hence
    i^2 = -1
    i = sqrt(-1)
    back to the def
     
    Last edited:

    reshadie

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  • Oct 9, 2008
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    Can somebody explain this...

    I will use "sqrt" to denote square root...

    In mathematics we say

    sqrt(-1) = i This is a definition

    Now i^2 = -1 (e.g. sqrt(4) = 2 ; therefore 2^2 = 4 )

    i^2 = sqrt(-1)*sqrt(-1)

    This can be written as

    i^2 = sqrt((-1)*(-1)) (e.g. sqrt(2)*sqrt(2) = sqrt(2*2) )

    so i^2 = sqrt(1) (as (-1)*(-1) = 1 )

    i^2 = 1 (as sqrt(1) = 1 )

    taking the square root of both sides

    i = 1

    But at the beginning i = sqrt(-1)

    So sqrt(-1) = 1

    :P
    this is wrong bro:eek:
     

    kasuncs

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  • May 21, 2007
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    Definition is
    i^2=-1

    It is not sqrt(-1)=i

    Square root function only defined for positive numbers.




    Edit: Error corrected.
     
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    kasuncs

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  • May 21, 2007
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    :no::no: Both are correct

    A complex number is a number comprising a real and imaginary part. It can be written in the form a + bi, where a and b are real numbers, and i is the standard imaginary unit with the property i 2 = −1.[1] The complex numbers contain the ordinary real numbers, but extend them by adding in extra numbers and correspondingly expanding the understanding of addition and multiplication.
    Code:
    http://en.wikipedia.org/wiki/Complex_numbers
     
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    Tom Riddle

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    A complex number is a number comprising a real and imaginary part. It can be written in the form a + bi, where a and b are real numbers, and i is the standard imaginary unit with the property i 2 = −1.[1] The complex numbers contain the ordinary real numbers, but extend them by adding in extra numbers and correspondingly expanding the understanding of addition and multiplication.
    Code:
    http://en.wikipedia.org/wiki/Complex_numbers

    Can you please read the whole article?

    "This difficulty eventually led to the convention of using the special symbol i in place of sqrt(-1) to guard against this mistake" - check it out.

    And check out other sites, maths books etc. It's all over the place.

    We don't need another definition for -1. We need a definition for sqrt(-1). That incidentally leads to i^2 = -1
     

    December

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  • Aug 24, 2009
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    Hmmm Hmmmm

    The rule is for

    sqrt(a)*sqrt(b) to be equal to sqrt(a*b) at least one of a and b should be positive...

    This solve the mistake...

    although sqrt(2)*sqrt(2) = sqrt(2*2)
    sqrt(-1)*sqrt(-1) != sqrt ((-1)*(-1))