The square of the square root of -1????

Discussion in 'Recreational Math' started by Zanziba, Mar 4, 2004.

  1. Zanziba

    Zanziba Guest

    A friend set me a paradox puzzle that I cannot fault his logic on. HELP please

    You take the square root of -1 and square ir, the answer must be -1, however....

    He says, write it as root -1 multiply root -1

    Then place them all under the same root (Root (-1 times -1))

    Now multiply -1 by -1 to get +1

    Hence root -1 squared = root 1 = 1

    Which is obviously pish posh, but i can't fault the algebra logic...

    Zanziba, Mar 4, 2004
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  2. Zanziba

    Ken Oliver Guest

    The property sqrt(a)*sqrt(b) = sqrt(a*b) holds for non-negative numbers
    only. Thus the following line of demonstration is invalid.
    Ken Oliver, Mar 4, 2004
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  3. Zanziba

    philippe 92 Guest


    Also for sqrt(a)/sqrt(b).
    Both a and b should be non negative real numbers, otherwise :
    i=sqrt(-1) = sqrt(1/-1) = 1/sqrt(-1) = 1/i
    so i^2 = 1 and 1 = -1

    which was the variant of the joke I knew.
    philippe 92, Mar 4, 2004
  4. Zanziba

    Dave Guest

    You're both only telling half the story. Square rooting creates TWO
    solutions, not one, so square root of -1 is not i, but +/-i:

    -i*-i=(minuses cancel)i*i=-1. (constitutes proof at least in the form of
    a counterexample to the hypothesis "sqrt(-1)!=-i")

    Not sure what Ken's going on about ("sqrt(a)*sqrt(b) = sqrt(a*b) holds
    for non-negative numbers only"). That's probably true if you assume
    positive numbers only have positive square roots:

    a=-9; b=-16
    =-1*3*4 OR -1*-3*4 OR -1*3*-4 OR -1*-3*-4
    =-12 or +12 or +12 or -12.

    a=9; b=16
    =3*4 or -3*4 or 3*-4 or -3*-4
    =+12 or -12 or -12 or +12


    You could create more "paradoxes" using the fact that the nth root of
    something produces n solutions. What are the three cube roots of -1?
    (Hint: two of them are complex, and IIRC the angles between the three
    vectors to each root are equal (if so that's .5+/-i.sin(pi/3) but I
    could be miles off; however (.5+i*.866)^3 is fairly close...))

    Dave, Mar 4, 2004
  5. Zanziba

    philippe 92 Guest

    "square root of" may be considered either as a multivalued function
    (what you do), or a convention as "principal root of", which is usually
    that is square root of 4 is +2, so we write the quadratic solutions as
    (-b +/- sqrt(delta) )/2a, with the +/- inside the formula.
    With a multivalued function we could write just +sqrt(delta) which would
    include the two values.
    Here some paradox which may be related with cube roots of 1 :
    Let x solution of x^2 + x + 1 = 0
    x is not 0, so dividing by x : x + 1 + 1/x = 0
    But from original equation : x + 1 = -x^2, Then -x^2 + 1/x = 0
    That is x^2 = 1/x or x^3 = 1 and x = 1.
    And from original equation : 3 = 0 !
    (of course the bug is that x^3 = 1 as _three_ solutions, x=1, j and j^2.
    j = -0.5 + i*sqrt(3)/2, j^2 = -0.5 - i*sqrt(3)/2.
    j and j^2 are the solutions of the original equation, x = 1 has been
    added surreptitiously).
    philippe 92, Mar 4, 2004
  6. Zanziba

    Dave Guest

    Ok, so what's the principal root of -1?

    If an assuption X (in this case "'square root of' means only 'the
    principal root of'") leads to a paradox, that generally constitutes
    proof that X is false (reductio ad absurdum, IIRC). The OP's paradox
    ("sqrt(-1).sqrt(-1)=-1 BUT sqrt(-1).sqrt(-1)=sqrt(-1.-1)=sqrt(1)=1")
    stems from the use of principal roots, and the solution to the paradox
    is to point out that principal roots are not the only roots, and that
    there is in fact no paradox.

    With false assumptions you can create paradoxes to your heart's content,

    assume 1+1!=2
    Then 1+(1+1)!=3 Paradox!

    (IAACPPP (C++ programmer), hence != for "doesn't equal". =/= always
    throws me, and there's no exception handler big enough for that :)

    Dave, Mar 4, 2004
  7. Zanziba

    philippe 92 Guest

    IMHO +i, by definition of i, which is _one_ number in |C (not a couple
    of numbers)
    if X (principal root) _and_ Y (sqrt(a)*sqrt(b)=sqrt(a*b) for any a,b)
    result in a contradiction, so _either_ X or Y is false...
    Use of principal root (X) => Y false :
    sqrt(a)*sqrt(b) != sqrt(a*b) if a or b <0 or not real

    If we want to keep the usefull property sqrt(a)*sqrt(b)=sqrt(a*b), I
    agree with you, it's impossible to use "principal roots" and we have to
    list all cases as you did to solve the "paradox".

    It's exactly what I said :
    Either you use "principal roots" and you forbide a multiplicative sqrt()
    _or_ you list all roots as you did (multivalued function).
    In both cases the paradox is solved.
    Of course...
    The false assumptions may be explicit as in your example
    or implicit when you apply badly some rules like dividing by 0 or
    extracting only one square root or extending a function's property
    defined for positive real numbers to |C and s.o.
    philippe 92, Mar 4, 2004
  8. Zanziba

    Virgil Guest

    There is no such thing as "the" square root of -1, since there are two
    such square roots, not just one.

    Also, sqrt(a)*sqrt(b) = sqrt(a*b) only holds when a and b are both
    non-negative and "sqrt" means the principle square root of a
    non-negative real number.

    One has to be very careful when doing root extractions on negatives or
    Virgil, Mar 5, 2004
  9. Zanziba

    Zanziba Guest

    Here some paradox which may be related with cube roots of 1 :
    Now that's one I can "Get him back with"

    Thanks guys. I understand most of what you have said. ANyone know of
    an interesting bookob complex numbers that I may like to read? Or
    interesting websites for beginners to the subject?

    Zanziba, Mar 5, 2004
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