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Help me thru

  1. Apr 30, 2005 #1
    i was trying to determine the sqrt of -1,and here is wat i have done.
    sqrt of -1=-1^2/4,this further gives [[-1]^2]^1/4, which further gives 1^1/4
    therefore the sqrt of -1 is 1^1/4 which is 1
    have u any objections,let me know
  2. jcsd
  3. Apr 30, 2005 #2

    matt grime

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    square 1, what is the answer? is it -1?
  4. Apr 30, 2005 #3
    but at least u saw my proof ,is there any mistake
  5. Apr 30, 2005 #4

    matt grime

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    of course there is. you said that the square root of -1 is 1, if so squaring 1 should give you -1, it doesn't therefore there is a mistake, and a rather obvious one: you're reasoning is wrong, simple as that.

    If you square a number, then square root it do you necessarily get what you started with? No, because you need to choose a branch of the sqaure root, or the 4'th root in this case, taking 4th powers is a 4-1 function, so taking 4th roots doesn't uno taking 4th powers.
  6. Apr 30, 2005 #5
    I cant find where his proof breaks any math rules.

    Aside from the obvious distinction matt put out, I still dont see a reason why his proof is incorrect (although I know that [itex] i != 1 [/itex]

    Is it the 4th root?
  7. Apr 30, 2005 #6

    is only valid in general for non-negative, real a, where roots are defined to always be non-negative real numbers. In other words, you simply can't prove that this equality holds when [itex]a[/itex] is negative (and you define roots appropriately). Notice that [itex]\pm i[/itex] are fourth roots of [itex]1[/itex] - but that doesn't mean that every fourth root of 1 is a square root of -1.

    All abia ubong has done is give an example to show precisely what I said - that you can't (necessarily) express powers in that way - it gives you a wrong answer!

    You really shouldn't think about it as "the square root of -1," anyway. The square root operation can be defined for negative numbers - but this is a definition. For example, you can define [itex]\sqrt{a} = ib[/itex] where [itex]\mathbb{R} \ni b > 0, \ b^2 = -a[/itex] when [itex]a<0[/itex]. You restrict its range appropriately so that it gives you exactly one answer. This does not guarantee that your new operation keeps all the nice properties that it had when it was only defined for positive numbers.
    Last edited: Apr 30, 2005
  8. Apr 30, 2005 #7
    Ok, I didnt know that. thanks.
  9. May 1, 2005 #8
    this is a easy way to think about it

    (-a)^1/2 * (-b)^1/2 does not equal (ab)^(1/2)

    unless a=-1 and b is positive

    make sense (:
  10. May 1, 2005 #9
    pls take a look at this i/i=sqrt of (-1/-1)which eventually gives 1
    also i/ican be expressed as 1/i *i now 1/i equals sqrt of (1/-1)which is sqrt of -1
    that is i .now 1/i *i =i *iwhich is -1
    what i have done now means that i/i=1=-1. i hope this answers matt question that 1=-1 which makes sqrt of -1 to be 1
  11. May 1, 2005 #10


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    So you are telling us that you do believe that 1= -1?

    For those of us who do not believe that 1= -1, (1)2= 1, not -1 and so 1 is not the square root of -1!
  12. May 1, 2005 #11
    This thread addresses this issue. Essentially, as matt explained, you seem to be confused about the fact that you can't mix and match different branches of the square root function and expect things to work out nicely.
  13. May 1, 2005 #12
    abia you really need to think about what your saying.
    according to your proof the roots of x^2+1 are -1 and 1.
    such function has no REAL ROOTS, but does have imaginary roots i,-i.
    make sense?
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