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A strange inconsistency with square roots

  1. Dec 10, 2014 #1
    Hi,
    I have a question that came into my mind while solving some problems. If I have a constant times an expression in a square root like ##4\sqrt{16}## I can square the constant and push it into the square root: ##4\sqrt{16}=\sqrt{4^2 16} = 16##. But what if the constant outside of the square root is negative? Then I will get a contradiction: ##-4\sqrt{16} = \sqrt{(-4)^2 16} = \sqrt{4^2 16} = 16 ≠ -16##. Why is this happening? Clearly I cannot square negative numbers the same way as I do with positive ones, but why?
    Any comments will be appreciated!
     
  2. jcsd
  3. Dec 10, 2014 #2

    jedishrfu

    Staff: Mentor

    You have to remember that when you take the square root of a positive number there are always two answers one positive and one negative.
     
  4. Dec 10, 2014 #3

    Mark44

    Staff: Mentor

    For starters, let's limit the discussion to the square roots of nonnegative numbers.

    A property of square roots that you're using is this: ##\sqrt{ab} = \sqrt{a}\sqrt{b}##. This is valid only when a and b are nonnegative.

    The problem your work is that ##\sqrt{(-4)^2} \neq -4 ##, so the first equality in your equation above isn't valid.
     
  5. Dec 10, 2014 #4

    Mark44

    Staff: Mentor

    I disagree. When you take the square root of a positive number, you get a single (positive) number. For example, ##\sqrt{9} = 3##. It's not ##\pm 3##. The square root symbol represents the principal (i.e., positive) square root of its argument.
     
  6. Dec 10, 2014 #5

    jedishrfu

    Staff: Mentor

    You're right, I was thinking of square root of x^2 where we don't know whether x is positive or negative.
     
  7. Dec 10, 2014 #6

    Mark44

    Staff: Mentor

    And it's the same here, meaning that you get one answer:

    ##\sqrt{x^2} = |x|##

    If the value of x is not known, you can't say the following
    ##\sqrt{x^2} = x##

    Also, the following isn't true
    ##\sqrt{x^2} = \pm x##
     
  8. Dec 10, 2014 #7

    jedishrfu

    Staff: Mentor

    Okay got it. I keep forgetting the principal square root is the convention.

    http://mathworld.wolfram.com/PrincipalSquareRoot.html

    This must be one of the reasons I majored in Physics and not Math.
     
  9. Dec 10, 2014 #8
    if you define square root x1/2 as the inverse function of square x2, then yes, there will be two answers, and one will be the negative of the other.
    However, if you remember your algebra teacher telling you about the "vertical line test"(if somewhere on the graph you have 2 outputs=y for one input=x), then square root is not a function at all, because it has 2 outputs for each input.


    In case your wondering, a justification for why a negative times a negative is a positive:

    suppose the money in your bank account is increasing by x dollars per year.

    then x*n=amount of money you made in n years.

    if x was negative, then it would mean you are losing money
    then x*n would be negative

    Heres where this comes into play

    If n=number of years was negative
    and x=money per year was positve,
    then x*n would be negative because it would show how much money you would lose if you went back in time(when you are actually gaining money as time moves forward).

    so, if both x and n were negative?
    then because x is negative, you are losing money,
    but because n is negative, you are also going backward in time
    therefore you are gaining money as you go backwards in time.

    Its kinda like how if you play country music backwards the guy gets his truck fixed, gets married, goes sober, and his dog comes back to life. :p
     
    Last edited: Dec 10, 2014
  10. Dec 10, 2014 #9

    Mark44

    Staff: Mentor

    No, this isn't right, either. The squaring function is not one-to-one, so doesn't have an inverse function.
     
    Last edited: Dec 10, 2014
  11. Dec 10, 2014 #10
    Thanks very much for the replies!
    So if I understand correctly, I was unconsciously missing the step ##4\sqrt{16}=\sqrt{4^2}\sqrt{16}## even with the positive number and just focused on pushing it into the square root, and thus the flawed step with the negative case was ##-4\sqrt{16}≠\sqrt{(-4)^2}\sqrt{16}##?
     
  12. Dec 10, 2014 #11

    Mark44

    Staff: Mentor

    Right.

    We get questions about square roots like this fairly often, and there are a lot of misconceptions around this concept, that seem to lead to contradictions. One of these is this one:
    ##1 = \sqrt{1 * 1} = \sqrt{(-1)(-1)} = \sqrt{-1} \sqrt{-1} = i * i = -1 ##
    which appears to show that 1 and -1 are the same number. This breaks down at the step ##\sqrt{(-1)(-1)} = \sqrt{-1} \sqrt{-1}## for reasons I already gave.

    Another misconception that shows up a lot is that, for example, ##\sqrt{4} = \pm 2##. The root of this confusion, I believe, is that we're taught that every positive real number has two square roots. As already mentioned, though, by definition and common usage, the √ symbol means the principal, or positive square root.
     
  13. Dec 10, 2014 #12
    im pretty sure if you put the numbers into polar coordinates
    z=re
    you dont need to be as careful with these steps.
    so if
    a=raea
    and
    b=rbeb
    and i want to find
    sqrt(a)*sqrt(b)
    then i have
    (raea)1/2(rbeb)1/2
    then i can still use the property sqrt(a)sqrt(a)=sqrt(ab) to get:

    (rarbeaeb)1/2

    because e^(x)e^(y)=e^(x+y) i can change this to:

    ((rarb)ei(θab))1/2
    then because (e^x)^y=e^(xy), i can say:

    (rarb)1/2ei(θab)/2


    now z is positive if θ=0, and negative if θ=pi
    so, if both are negative, then (θab)/2=pi
    so we have eipi=-1
    then because ra is just |a| and rb=|b|
    we have -(|a||b|)1/2, which is correct for sqrt(a)*sqrt(b) where a and b are negative, even though i used sqrt(a)sqrt(a)=sqrt(ab)

    to be honest, its alot more work than is needed :p
     
    Last edited: Dec 10, 2014
  14. Dec 10, 2014 #13

    pwsnafu

    User Avatar
    Science Advisor

    This is not true for arbitrary complex x or y. See wikipedia.
     
  15. Dec 10, 2014 #14
    also i forgot to take the root of unity lol

    i think the whole axbx=(ab)x thing can be reduced to the idea that if f(x)=a and b, it doesnt necessarily mean a has to equal b, but from what i read on the page it seems to be implying that the inconsistencies with
    (e^x)^y=e^(xy)
    can be explained and the real answer is
    2f8b3e67ccc1d2ce6c6dfc4d84f0cdd8.png

    while the inconsistencies with
    ln(b^x)=x*ln(b)

    cant be explained with a multivalued interperetation. pretty strange
     
  16. Dec 11, 2014 #15
    That's a pretty interesting "intuitive" explanation, but the fact does follow immediately from the distributive property.
     
  17. Dec 16, 2014 #16
    Square root always returns a non-negative number. It's also only defined on non-negative numbers. Not every step is reversible. When you square a number and then take its square root, you lose information about the sign of the number, since this will always return a non-negative number.
     
  18. Dec 17, 2014 #17

    statdad

    User Avatar
    Homework Helper

    Mark44: Edited by adding quote tags for the statements that statdad is citing.
    If you are referring to the principal square root (commonly written [itex] \sqrt x [/itex]), but even in the realm of real numbers something like 25 has two square roots: 5 and -5.

    If you are working in the realm of reals for domain and range, yes.

    Yes (again, in the reals), but this is more clearly seen as a feature of the fact that [itex] y = x^2 [/itex] is not a one-to-one function. (The others are also, in sense).

    But the point of all this is?
     
    Last edited by a moderator: Dec 17, 2014
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