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Square roots

  1. Mar 22, 2010 #1

    BruceW

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    I understand the basics of raising numbers to powers of whatever, but i've been wondering:

    When you square a number then square root it, you end up with plus or minus the original number. (i.e. 5 times 5 gets 25, then root gets 5 or -5). But using the rules of raising numbers to powers of other numbers:
    (5^2)^0.5=5^1
    And i would assume 5^1=just 5, not -5. But the left-hand-side of the equation gives 5 or -5.

    Another example: 2^2=2^(4/2)=(2^4)^0.5=16^0.5= 4 or -4
    But surely 2^2=4 and not -4

    So there seems to be a contradiction. Is there some rule that defines when there is an uncertainy in the sign of the number?
     
  2. jcsd
  3. Mar 22, 2010 #2
    [tex](2)^2=4[/tex] and [tex](-2)^2=(-2)(-2)=4[/tex]

    [tex]\therefore[/tex] if [tex]x^2=4[/tex], x can be [tex]\pm 2[/tex]
     
  4. Mar 22, 2010 #3
    No you don't: you end up with the modulus of the number:

    [tex]\sqrt{x^2}=\left|x\right|[/tex]

    By the way, these operations don't commute:

    [tex]\left(\sqrt{x}\right)^2=x[/tex]

    You are confusing the square roots of a number with the function [itex]f\left(x\right)=\sqrt{x}=x^{0.5}[/itex], so [itex]16^{0.5}=4[/itex], not 4 or -4.
     
  5. Mar 23, 2010 #4

    BruceW

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    @JSuarez: So the function f(x) equals the positive root, so why don't the operations x and (x^0.5)^2 commute? They both have all the same answers for any given input. And (x^2)^0.5 would also commute, since we are saying the function equals the positive root only.
    ps thanks for your help :)
     
  6. Mar 23, 2010 #5

    Mentallic

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    x and [tex](x^\frac{1}{2})^2[/tex] are the same. Take some positive x, you'll end up back at x after sqrt root then squaring. For some negative, you'll have a complex number then square it to get the negative back again.

    However, when you square x first, you lose information as to whether x was positive or negative before squaring, so when you take the root it's not exactly x, but rather |x|.

    When you square root x first, for x positive, you have some result positive (by the way, this is how square rooting has been defined, only as the positive square root, not plus/minus), but for x negative you have a complex number. You don't lose information in this case.

    It's somewhat like taking the derivative of a function. Doing the inverse (integrating) means you lose the info as to what the constant was.
     
  7. Mar 23, 2010 #6
    When I said "commute", I meant that, given functions f(x) e g(x):

    [tex]f\left(x\right)=\sqrt{x}[/tex]

    [tex]g\left(x\right)=x^2[/tex]

    Then [itex]f\circ g\left(x\right)=\left|x\right| \neq g\circ f\left(x\right)=x[/itex].
     
  8. Mar 24, 2010 #7

    Mark44

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    No, they are not. As a counterexample, [itex]-2 \neq ((-2)^{1/2})^2[/itex]
     
  9. Mar 24, 2010 #8

    Mentallic

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    [tex](-2)^{1/2}=\sqrt{2}i[/tex]

    [tex](\sqrt{2}i)^2=-2[/tex]

    Please explain why you're claiming they're not equivalent.
     
  10. Mar 24, 2010 #9

    HallsofIvy

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    Mark44 would not have made such a mistake- it must be a typo.

    I suspect he meant to say that [itex]((-2)^2)^{1/2}\ne -2[/itex].
     
  11. Mar 24, 2010 #10

    Mark44

    Staff: Mentor

    It wasn't a typo. I was thinking in terms of the square root function defined on the nonnegative reals, but I didn't add this qualification. If you attempt to take the square root of a negative number on a calculator, the operation is undefined -- that's the sense of the square root I had in mind. Sorry I didn't make myself clearer.
     
    Last edited: Mar 24, 2010
  12. Mar 24, 2010 #11
    Yes there's a way to solve this uncertainty, but using what's called "Complex numbers", where there's nothing called "not possible", like taking the square-root of a negative number.

    If you would like to learn how this contradiction is solved, you have to learn the De Moivre's theorem, you can find details about it on wikipedia:

    http://en.wikipedia.org/wiki/De_Moivre's_formula

    In this theorem, every square-root gives you 2 outcomes, and every cube-root gives 3 outcomes, and every quad-rood gives 4 outcomes, and so on! So the number of outcomes depends on the power of the root you're applying. Those outcomes are not always nice and "symmetric" as they are for square-roots, positive and negative. They are sloppy somewhat and need graphical representation to make sense.

    So the square-root of 1 is (1,-1). But the bad news is, that the cube-root of 1 is not only 1, but there are another two complex numbers, namely: cos(120) + i sin(120), and cos(240) + i sin(240). And "1", which is the first outcome we mentioned may be calculated with the same rule: cos(0) + i sin(0) = 1.

    i is the kernel of the of the complex numbers, basically it's defined as the square-root of -1.

    Sorry for complicating things, but this is why mathematics is nice, everything is symmetric, and everything has an answer, even if it doesn't make sense in the first sight :D

    Hope this helps, any questions are welcome!

    Ciao!
     
  13. Mar 25, 2010 #12

    Mentallic

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    I also gave a mention to square roots of negatives, mainly:
    Calculators are not a nice way of disclaiming a fact.

    For example, if you want to calculate the cotangent of a number that is equal to zero, most calculators don't have a cot function ready, so you need to take the reciprocal of the tangent function, but this leads to 1 divided by (undefined), but we know this is equal to 0:
    [tex]\frac{1}{tan\left(\frac{\pi}{2}\right)}=0[/tex]

    The calculator unsurprisingly gives us a math error since it tries to calculate [itex]tan(\pi/2)[/itex].

    Similarly for the square root of a negative, just because the calculator gives us an undefined answer, doesn't mean it is undefined in the real sense.
     
  14. Mar 25, 2010 #13

    BruceW

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    @ JSuarez:
    Cool, that does make sense to me. So basically i just have to take care in the order in which i take those two functions.
    Thanks, that's pretty much the answer i was looking for :)
     
  15. Mar 28, 2010 #14

    jgens

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    Mentallic, you're correct if you're working in the field of complex numbers, but if you're working strictly in the field of real numbers, then [itex](\sqrt{x})^2 \neq x[/itex] since negative reals have no square roots. I believe that this is Mark44's point when he is talking about calculators because most calculators work exclusively with real numbers.
     
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