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Homework Help: Defining the root of a number

  1. Apr 9, 2008 #1
    1. The problem statement, all variables and given/known data

    I have been doing elementary algebra for a couple of weeks now, and gotten quite used to the idea that the square root of n gives +- n1.

    Then moving on to arithmetic and n-roots / powers, I was given the current definition of the square root:

    The square root of a number N is the positive number M which gives M^2 = N.

    So, what is going on here? Why are we suddenly only interested in the positive number?

    Is this related to the fact that the square root of N can be written as N^(1/2) ?

  2. jcsd
  3. Apr 9, 2008 #2
    The reason you're only interested in positive numbers, is that you can only ever root a positive number.

    Consider that two positives are a positive, and two negatives are a positive.

    With the exclusion of imaginary numbers (i), there are no numbers which can square to equal a negative number.

    so when you take your number N and you put a square root symbol in front of it you get M.
    If M was a negative number the negative sign would be inside the root symbol, which is impossible.
    It is, however, possible to put the negative on the outside of the root symbol, as that is after the calculation.

    First time response, so let me know if i'm just writing crap :/
  4. Apr 9, 2008 #3
    That would depend on the degree of the root, you can take the cube root of a negative number.

    The cube root of -8 would be -2, since (-2)^3 = -8.

    Actually, you can take the N-th root of a negative number as long as N is odd.

    And I am sorry, but I don't see how this relates to the "problem"?
  5. Apr 9, 2008 #4


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    Hi kenewbie! :smile:

    Sort of.

    The point is that we mustn't define anything that's ambiguous.

    If we want to write N^(1/2), or √N, and there are two possible values for it, then we must specify in advance which value it is to be. :smile:

    Same for things like arcsin and log-of-a-complex-number!
    Hi Lore! :smile:

    No … it's a thoroughly good answer … though not exactly to the right question ! :rolleyes:
  6. Apr 9, 2008 #5
    More practice needed methinks..\pm

    Thanks for that Tim, but could you do me a favour kenewbie and clarify what the problem actually was?
    I get that I missed the point, but I still can't find the point :redface:

    And that the powers cancel to 1, yes. Am I getting warmer?

    Or are you asking about the [tex]\pm[/tex] as Tim was referring to?

    Or maybe both?

    Sorry if this has already been answered, but i'm confuddled..
  7. Apr 9, 2008 #6
    But.. but.. why? Algebra is perfectly happy with getting two results and then just branching in two directions for the rest of the calculation. Why not when we work with actual numbers?

    I gather that something, somewhere, relies on the fact that the result of a root is positive? But then if that is true, why allow something like the cubic root of a negative number, the result is negative!

    I want to see a contradiction or something (preferably something that can be read with my level of math). Some reason which explains why using both results would end up being a bad thing.

    Let me try to give an example of what I mean by a contradiction:

    The square root of 4 is 2 and -2. But, 2^2 is not equal to (-2)^2 and to avoid this problem, we define the root of N to be the positive number M which gives M^2 = N.

    The thing is, I cant find such a contradiction? (2^2 is obviously equal to (-2)^2)

    I sort of get the feeling that I'm staring something obvious in the face here and calling it a UFO :/


  8. Apr 9, 2008 #7


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    ok … saying " let f(a) be a solution of x^2 - 2x + a = 0" is fine. :smile:

    But saying " let f(a) be the solution of x^2 - 2x + a = 0" is not fine! :frown:

    "The square root of 4 is 2 and -2" is an abuse of the word "is".
    Mathematical proofs are linear … they don't operate "in parallel". We can't use "both" results at the same time. :cry:
    No … but a lot of things rely on it being uniquely defined!
  9. Apr 9, 2008 #8
    Aha. Ok, I think I finally see your point. The difference between "a solution" and "the solution" is quite substantial.

    The question then becomes, can you show me a simple problem that would occur if the result of the square root was NOT uniquely defined? :)

    I gather that the whole thing would come crashing down, but there might be some small easily shown paradox that would occur? I'm not talking about "well, the root of a square with an area of 4 certainly cant be -2", but a fault which would arise in the structure of the math that ties in to roots.. if that made sense.

    Thanks a lot for your patience tiny-tim, I really appreciate it.
  10. Apr 9, 2008 #9


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    Actually, some things do rely on the square root being specifically positive.

    We want to be able to define the function f(x)= ax for all x. Also we want to be able to do all the "algebra" things we do with other functions. In particular we want to be able to use such properties as (ax)y= axy. If we defined x1/2= [itex]\sqrt{x}[/itex] as the negative number y such that y2= x, we would run into trouble with (x1/2)1/2.
  11. Apr 9, 2008 #10


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    Hi kenewbie! :smile:

    Well … for example, any integration over a "triangular" region defined by x > y, where y = √(something). :smile:
  12. Apr 10, 2008 #11
    Aha! Now that is the sort of thing I was looking for, thanks :)

  13. Apr 10, 2008 #12
    Ok, I'll file that under things to check at a later point :) I'm not quite at integration yet.

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