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B Why negative numbers inside root can't be separated?

  1. Nov 15, 2016 #1
    Why √[(-a).(-b)] can't be written as √(-a).√(-b)
    • Is it only because complex number do not work for this statement.
    • Just like here: √ab = √[(-a).(-b)] = √a√bi^2 = -√ab which is wrong.
    • We can separate √(-4)(9) = √-36 = 6i , √4i.√9 =6i, but why can't we separate for two negative numbers inside the root?
    • Is it only because of example in the 2nd point does not satisfies or there is a different reason.
     
  2. jcsd
  3. Nov 15, 2016 #2

    mfb

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    There are rules for roots that work for positive real numbers only, ##\sqrt{ab} = \sqrt a \sqrt b## is the most important one. You cannot apply them if you have (or get) other values in the roots. The main problem with the more general root in the complex numbers: you have to choose which sign you want, and no matter how you choose that, you ruin some of the relations that work for positive real numbers.
     
  4. Nov 15, 2016 #3

    fresh_42

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  5. Nov 15, 2016 #4
    Basically its a paradox,
     
  6. Nov 15, 2016 #5
    Okk, then why it is applicable here: √(-4)(9) = √-36 = 6i , √4i.√9 =6i, why in this example we divided negative number's root and positve numbered root. Why can we do this operation if only one of them is negative, why can't we do it for both the negative numbers.
    I think answer is because it gives wrong answer.
     
  7. Nov 15, 2016 #6

    fresh_42

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    This boils down to the question why ##\sqrt{-1} = i## which is explained in the insight I quoted in #3.
    To define a complex square root, one has to get exponential, since ##\sqrt{z}=z^\frac{1}{2}##. This is done by the exponential function and logarithm. However neither of them is one-to-one anymore but splits into branches instead each time a full circle in the complex plane is reached. As a consequence of it, the rules of the reals don't apply one-to-one to complex numbers either.
     
  8. Nov 15, 2016 #7

    mfb

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    You don't use any of the problematic rules in those cases.
    (-4)*9 = -36, and sqrt(-36) = 6i (if you define the square root that way).
    ##\sqrt 9=3## and ##\sqrt{4}=2##, therefore ##\sqrt{4} \cdot i\sqrt{9}=2\cdot i \cdot 3 = 6i##
     
  9. Nov 15, 2016 #8
    Wow , thanks, i got the idea, we can't apply all the rules of real numbers to the complex numbers
     
  10. Nov 15, 2016 #9

    Stephen Tashi

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    You need to distinguish between the concept of "a negative number" and the concept of "a variable with a minus sign in front of it". The expression "-a" does not necessarily symbolize a negative number. For example, if a = -3 then "-a" represents 3.

    The fact that a certain algebraic equation only applies in particular cases isn't a "paradox" unless you can prove that it ought to work in all cases. We have no proof that ##\sqrt{a\ b} = \sqrt{a} \sqrt{b} ## is correct for all numbers ##a,b##. You can say that the fact it doesn't work for all cases is a "limitation" of the equation.

    Another equation that doesn't work for all cases is ##(a^b)^c = a^{bc}##.
    For example, try a = -1, b = 2, c = 1/2.
     
  11. Nov 15, 2016 #10

    Mark44

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    No, not at all.
    Of course not. The real numbers are different from the complex numbers. You shouldn't expect that a rule (or theorem) that is applicable to one set will also be applicable in the other set.
     
  12. Nov 15, 2016 #11
    True, food items are just like a group of numbers , we can boil them in microwave to get hot food, but we can't boil a egg which belongs to same group.therefore we use different way to boil eggs :biggrin::biggrin:o0)
     
  13. Nov 15, 2016 #12

    Mark44

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    I'm not sure this is a good analogy. If you're trying to boil an egg by putting it in a microwave by itself, that wouldn't be boiling an egg. To boil something, you put it in water. I think if you put an egg in a non-metallic container, with water, you could boil it in a microwave. I haven't tried this, so I don't know whether it works
     
  14. Nov 15, 2016 #13

    fresh_42

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    A better example might be the decomposition of integers into primes: ##14=2\cdot 7## in ##\mathbb{Z}## but in ##\mathbb{R}## there are no primes. Or subtraction in ##\mathbb{N}##.
     
  15. Nov 15, 2016 #14

    mfb

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    I wouldn't put full eggs with shell into the microwave due to the risk of an explosion, but in principle you can boil eggs in a microwave without water. I don't understand how the analogy works even if you could not do so.
     
  16. Nov 30, 2016 #15
    You can understand with this easy way also. As it helps me a lot.

    When problems with negatives under a square root first appeared, mathematicians thought that a solution did not exist. They saw equations such as x2 + 1 = 0, and wondered what the solution RadNeg3.gif really meant. Assume imaginary number.

    The imaginary number "i" is the square root of negative one.
    RadNegi.gif

    An imaginary number possesses the unique property that when squared, the result is negative.

    irad1.gif

    Consider: irad2a.gif
     
  17. Nov 30, 2016 #16

    Svein

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    Yes, but that is not the whole story. First: [itex] -1=e^{i\pi}[/itex]. But due to periodicity we also have [itex]-1=e^{3i\pi} [/itex]. Now square roots in the complex plane using polar coordinates is given by the square root of the modulus (which is 1) and half the angle. Therefore we have both [itex]\sqrt{-1}=e^{i\frac{\pi}{2}}=i [/itex] and [itex] \sqrt{-1}=e^{i\frac{3\pi}{2}}=-i[/itex].
     
  18. Dec 2, 2016 #17

    haruspex

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    I think you have missed the point of Parshyaa's question. mfb answered it best in post #2.
     
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