Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Chapter on functions

  1. May 7, 2005 #1
    I am going on my own through a chapter on functions and here is something that puzzled me in the definition:

    Each element in the domain is paired with just one element in the range.

    I guess my calculus knowledge interferes, but what about function like f(x) = sqrt(x). It has two roots: + and -. How does set theory account for that? Or is sqrt(x) not a function in set-theoretic terms :confused:
    Although f(x) = x^2 fits the definition of the function.
    Could someone please explain?

    Thanks in advance.
  2. jcsd
  3. May 7, 2005 #2
    [itex]f(x)=\pm\sqrt{x}[/itex] is not a function, since f(x) has more than one value for all x > 0. However, [itex]f(x)=\sqrt{x}[/itex] is a function; [itex]\sqrt{x}[/itex] always refers to the positive square root.
  4. May 7, 2005 #3


    User Avatar
    Gold Member

    Yes, just look at the definition of function. If something doesn't meet the requirements of the definition, it just simply isn't a function.
    But how can you tell whether something is a function until you specify its domain and the superset of its range? [itex]\sqrt{x}[/itex] isn't a function from N to N.
  5. May 7, 2005 #4
    Thanks :smile:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Chapter on functions
  1. Gaussian function (Replies: 1)

  2. Gaussian function (Replies: 1)

  3. Indicator function (Replies: 2)