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Posers and Puzzles

Posers and Puzzles

  1. Standard member TheMaster37
    Kupikupopo!
    26 Feb '04 13:39
    Ok, the complex number i is defined as sqrt(-1), but that definition is not really a good definition:

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

    Can someone tell me EXACTLY where the problem is in this chain of equalities (aside the fact that -1 isn't equal to 1)? All i've used above is the definition of || and the definitin of i. I've asked many people about this, but none can give me a good answer...i hope someone here can.

    Ton
  2. Standard member noodlepdx
    Kibitzing Regicides
    26 Feb '04 14:33
    Why do you take the absolute value in the second to last step? That is the same as taking the absolute value for any expression in the string of equations, which would make the equations balance.

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

    In other words, all that hand-waving to get to sqrt(-1)^2 does not allow you to enclose the next expression in || unless you do it for every expression down the line. Taking the absolute value of a negative is an operation like any other -- you need to balance your equation.
  3. Standard member TheMaster37
    Kupikupopo!
    27 Feb '04 12:41
    Originally posted by noodlepdx
    Why do you take the absolute value in the second to last step? That is the same as taking the absolute value for any expression in the string of equations, which would make the equations balance.

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

    In other words, all that hand-waving to get to sqrt(-1)^2 does not allow you to enc ...[text shortened]... bsolute value of a negative is an operation like any other -- you need to balance your equation.
    You were correct for how i typed it up, what i meant to type was:

    -1 = i * i = sqrt(-1) * sqrt(-1) = sqrt(-1 * -1) = |-1| = 1

    The || is a length measure, and in a 1D space the length of A is defined as:

    |A| := sqrt( A^2 ). And |A| being the familiar absolute value of A.

    So my question is again, where's the error?
  4. Standard member Fiathahel
    Artist in Drawing
    27 Feb '04 19:44
    Originally posted by TheMaster37
    You were correct for how i typed it up, what i meant to type was:

    -1 = i * i = sqrt(-1) * sqrt(-1) = sqrt(-1 * -1) = |-1| = 1

    The || is a length measure, and in a 1D space the length of A is defined as:

    |A| := sqrt( A^2 ). And |A| being the familiar absolute value of A.

    So my question is again, where's the error?
    Oh Ton, oh Ton, oh Ton. How many times do I have to say this:

    sqrt(1) can also be -1!
    sqrt(-1) can also be -i!

    It is about tha same as saying:
    (-1)^2=1, 1^2=1 then -1 = 1

    Oh, I realise now, after all this thinking, is that what I told you over and over again before was incorrect and that this is the truth
  5. Standard member TheMaster37
    Kupikupopo!
    28 Feb '04 14:41
    Lol, i'm inclined to belive you Fiathahel. But i am bugged by it still.

    sqrt(-1) can also be -i yes, but it's defined to be i...

    And sqrt(X^2) is defined to be |X|...
  6. Standard member royalchicken
    CHAOS GHOST!!!
    29 Feb '04 02:57
    Originally posted by TheMaster37
    Lol, i'm inclined to belive you Fiathahel. But i am bugged by it still.

    sqrt(-1) can also be -i yes, but it's defined to be i...

    And sqrt(X^2) is defined to be |X|...
    No. sqrt(-1) is NOT defined to be i. Complex numbers are defined first, with addition and multiplication, and then the equality i^2 = -1 is deduced from this definition.
  7. Standard member TheMaster37
    Kupikupopo!
    29 Feb '04 12:37
    Originally posted by royalchicken
    No. sqrt(-1) is NOT defined to be i. Complex numbers are defined first, with addition and multiplication, and then the equality i^2 = -1 is deduced from this definition.
    Oh how are they defined then? I was under the assumption tht they first defined i, and with that defined complex numbers to be a+bi, with a and b reals.
  8. Standard member Fiathahel
    Artist in Drawing
    29 Feb '04 13:22
    Originally posted by TheMaster37
    You were correct for how i typed it up, what i meant to type was:

    -1 = i * i = sqrt(-1) * sqrt(-1) = sqrt(-1 * -1) = |-1| = 1

    The || is a length measure, and in a 1D space the length of A is defined as:

    |A| := sqrt( A^2 ). And |A| being the familiar absolute value of A.

    So my question is again, where's the error?
    Let's spell it out this time:
    (a V b = a or b)

    S := sqrt(-1 * -1) = sqrt(1) = 1 V -1

    SS := sqrt(-1) * sqrt(-1) = (i V -i) * (i V -i)
    = (i V -i) * i V (i V -i) * -i
    = (-1 V 1) V (1 V -1)
    = -1 V 1 V 1 V -1
    = 1 V -1

    The outcomes are not depending of each other so therefor S and SS aren't equal.

    If you analyse your statement from the left to the right thinking about what I just said you see:
    -1 = i * i = sqrt(-1) * sqrt(-1) = sqrt(-1 * -1) = 1 V -1,
    which of course is true

    and also from the right to the left you see:
    1 = sqrt(-1 * -1) = sqrt(-1) * sqrt(-1) = 1 V -1
    which of course is also true.

    a last thing I want to say is that I've never seen the sqrt-function when working with complex numbers, though I believe the definition of the sqrt would be sqrt(z) = x where x^2 = z and Re(x) >0


  9. Donation Acolyte
    Now With Added BA
    29 Feb '04 23:08
    Originally posted by royalchicken
    No. sqrt(-1) is NOT defined to be i. Complex numbers are defined first, with addition and multiplication, and then the equality i^2 = -1 is deduced from this definition.
    If you don't specify whether sqrt(-1) is i or -i, sqrt isn't a well-defined function on C. However, there isn't a standard definition of sqrt(-1); the usual approach to defining sqrt on C is to cut out the negative real axis, set sqrt(1) = 1 and then choose the other values so that sqrt is continuous on the cut plane D, which gives a unique value of sqrt everywhere on D. You can define sqrt(-1) to be i, but there's no particular reason to do so.
  10. Standard member royalchicken
    CHAOS GHOST!!!
    29 Feb '04 23:26
    Originally posted by Acolyte
    If you don't specify whether sqrt(-1) is i or -i, sqrt isn't a well-defined function on C. However, there isn't a standard definition of sqrt(-1); the usual approach to defining sqrt on C is to cut out the negative real axis, set sqrt(1) = 1 and then choose the other values so that sqrt is continuous on the cut plane D, which gives a unique value of sqrt everywhere on D. You can define sqrt(-1) to be i, but there's no particular reason to do so.
    Right.
  11. 01 Mar '04 18:39
    Originally posted by royalchicken
    Right.
    I seem to remember reading somewhere that complex maths also holds together pretty well if we assume that i, j and k=ij are 3 distinct roots of -1, which is where quaternions come from.
  12. 03 Mar '04 22:46
    I don't know--maybe I'm wrong--, but the last time I checked sqrt(a) is the positive square root of a, not the positive and negative root of a.
    In other words, sqrt(4)=2, not 2 and -2.
    Though technically (2)^2 and (-2)^2 both equal 4, the only way that sqrt(4) equals both is if it is specified as +-sqrt(4) (stated as "plus or minus the square root of 4&quot.
    The best reason I believe I can give for why this is true is that the function f(x)=sqrt(x) would not be a function if the negative values for the square root were accepted. Therefore, algibraically the root function only derives the positive values, unless otherwise specified, or if it is used as the inverse of the squared function (but in that case, the "+-" should still be specified and not assumed).
    I'm sorry I could not give a more clear reason, but I hope this makes sense. But, as I said, I might be wrong. At lease I can say I tried.
  13. Standard member Fiathahel
    Artist in Drawing
    04 Mar '04 13:09
    Originally posted by econundrum
    I don't know--maybe I'm wrong--, but the last time I checked sqrt(a) is the positive square root of a, not the positive and negative root of a.
    In other words, sqrt(4)=2, not 2 and -2.
    Though technically (2)^2 and (-2)^2 both equal 4, the only way that sqrt(4) equals both is if it is specified as +-sqrt(4) (stated as "plus or minus the square root of ...[text shortened]... on, but I hope this makes sense. But, as I said, I might be wrong. At lease I can say I tried.
    yes of course the sqrt(4) =2 and not -2, but the sqrt used here is not the real sqrt but the inverse of x^2. Perhaps it was better to say x^2 = 4 so x = sqrt(4) v -sqrt(4) in stead of sqrt(x) = 2 v -2.
  14. 07 Mar '04 05:05
    Originally posted by Fiathahel
    yes of course the sqrt(4) =2 and not -2, but the sqrt used here is not the real sqrt but the inverse of x^2. Perhaps it was better to say x^2 = 4 so x = sqrt(4) v -sqrt(4) in stead of sqrt(x) = 2 v -2.
    I agree with you, but the reason is that you have basically stated what I have said, but in fewer words. I admit that I'm a bit verbose; nonetheless, sqrt(-1) is not specified as the inverse of x^2=-1, if it did, then I would agree with the questions and arguments stated in this thread.
    Thanks for indirectly proving my own statement.
  15. Standard member TheMaster37
    Kupikupopo!
    07 Mar '04 12:26
    Ok, i'm going to pay attention to teachers in the future. They should introduce i with the remark that i is A number with the property i^2 = -1, not THE number.