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Whats the Flaw in this?

  1. Jan 31, 2008 #1
    sqrt(-1) = (-1)^(1/2) = = (-1)^(2/4) = ((-1)^2)^(1/4)) = 1^(1/4) = 1

    Can someone explain the flaw to me?
  2. jcsd
  3. Jan 31, 2008 #2
    Well i think that the flaw here is that the operatios with square roots are defined only for positive numbers, so basically when you went from sqrt(-1) = (-1)^(1/2) to (-1)^(2/4), i think is not correct.
  4. Jan 31, 2008 #3
    You wrote 1^(1/4) = 1, but this is not well-defined because there are four numbers whose fourth power is 1, namely {1, -1, i, -i}. This is like saying (-2)^2 = 4 and sqrt(4) = 2, hence 2=-2.
  5. Jan 31, 2008 #4


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    Homework Helper

    No, you can't do that, when writing: [tex]\alpha ^ {\frac{b}{c}}[/tex]

    Then b, and c must be relatively prime, i.e their GCD must be 1.

    When you square something, it's always non-negative, and when you square (cube, or forth,...) root it, it's still non-negative. This is where the error lies.

    It's the same as:

    [tex]\sqrt[3]{-1} = (-1) ^ {\frac{1}{3}} = (-1) ^ {\frac{2}{6}} = \sqrt[6]{{\color{red}(-1) ^ 2}} = 1[/tex], which is clearly false.
  6. Jan 31, 2008 #5
    looks like there are multiple reasons. Thanks everyone.
  7. Feb 1, 2008 #6
    looks like masnevets made the most sense out of it.
  8. Feb 1, 2008 #7


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    In fact, defining i as "[itex]\sqrt{-1}[/itex]", while a convenient mnemonic, is logically incorrect- every number (except 0) has two square roots and this doesn't specify which square root of -1 i is. In the real number system, we don't have to worry about that since we can specify the square root as being the positive root. Since the complex numbers is not an ordered field we can't do that. That's even more obvious where you see i "defined" by "i2= -1". Such an equation has two roots. Which one is i?

    The most logical way of handling that is to define the complex numbers as ordered pairs of real numbers: (a, b) and definining (a, b)+ (c, d)= (a+ c,We can, then, say that by (a, b) is "represented" That way, we have immediately that (0, 1)2= (0, 1)(0, 1)= (-1, 0). It is also true that (0, -1)2= (0, -1)(0, -1)= (-1, 0) but now we can distinguish between those two numbers. We can, then represent (a, b) as a(1, 0)+ b(0,1)= a+ bi by representing (1, 0) by 1 and (0, 1) by i.

    With those definitions "paradoxes" like the one given here do not occur.
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