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The square root

  1. Oct 5, 2013 #1
    Hello everyone,

    What is the square root of a square of a negative number equal to? For example: [itex]\sqrt{-1}^{2}[/itex]

    It seems there are two possible ways of doing this, the problem is that I am getting two different answers using these two approaches i.e; We can first take the square of -1 and then take square root of the square of -1. This gives 1 as the answer. On the other hand, if we replace the radical sign by 1/2 as the exponent, we should then just be left with -1 as the answer. The correct answer, I think, is +1, but I can't figure out why there's an apparent contradiction...

    Please help me understand where and why I am wrong.

    Thank you!
     
  2. jcsd
  3. Oct 5, 2013 #2

    pwsnafu

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    As written, most mathematicians would (probably*) parse that as ##(\sqrt{-1})^2##. The square root evaluates to either i or -i depending on where your branch cut is, but they both square to -1.

    *: It's ambiguous though.

    With complex numbers your assumptions of exponentiation derived from the reals breaks down. In this case you are assuming ##(a^{b})^c = (a^c)^b##, and this is simply not true. Further ##(a^b)^c \neq a^{bc}## in general.
     
    Last edited: Oct 5, 2013
  4. Oct 5, 2013 #3

    arildno

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    I find your notation unclear.
    Do you mean: [itex]a: \sqrt{(-1)^{2}} b: (\sqrt{-1})^{2}[/itex]
     
  5. Oct 5, 2013 #4

    Curious3141

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    Your notation is ambiguous, as others pointed out, but your verbal explanation of the problem is clear. You want to evaluate ##\sqrt{(-1)^2}## or ##{((-1)^2)}^\frac{1}{2}##.

    In essence, they're the same thing. Both exponentiation operations are of equal precedence, so you work from inside out. You're left with ##\sqrt{1}## or ##{1}^\frac{1}{2}##. Those are just different ways of writing the same thing. The usual implication of exponentiation to a fractional power is the extraction of the principal root, which is the unique positive value, if it exists. In this case, that's simply ##1##.

    Doing it the second way, you might be tempted to use the law of exponents ##(a^b)^c = a^{bc}## to cancel the ##2## and the ##\frac{1}{2}##, but this would be wrong. That law only applies (without qualification) when ##a## is non-negative.
     
  6. Oct 5, 2013 #5

    arildno

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    remember that:
    [tex]\sqrt{x^{2}}=|x|[/tex]
    not x.
     
  7. Oct 5, 2013 #6

    Mark44

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    Based on what VC described rather than the notation used, this is the problem: ##\sqrt{(-1)^2}##. This expression simplifies to 1.
     
  8. Oct 5, 2013 #7

    Mark44

    Staff: Mentor

    Actually, the notation used was clear and unambiguous, but did not agree with his description.
     
  9. Oct 5, 2013 #8
    This is what I meant: ##\sqrt{(-1)^2}##.

    I'm sorry for the ambiguity guys. I tried putting the parentheses around -1 before, but the code broke down for some reason, so I removed it.

    I think the greatest gap in my understanding was that I didn't know that the rule (a[itex]^{b}[/itex])[itex]^{c}[/itex] = a[itex]^{bc}[/itex] couldn't be applied for a<0 (still a bit confused about this point. Isn't (-2[itex]^{3}[/itex])[itex]^{2}[/itex] = (-2)[itex]^{6}[/itex]?

    I knew that [itex]\sqrt{x}^{2}[/itex]=[itex]\left|x\right|[/itex], but it felt like this was inconsistent with the former rule (law of exponents), thanks to my ignorance. :redface:

    Thank you so much, all of you! I seriously appreciate your help more than you probably know. :smile:
     
    Last edited: Oct 5, 2013
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