1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Undefined values

  1. Oct 25, 2008 #1
    Let us say that 1/x + 1/y = 1/z; z is a function of R, so that z = 1/R
    For R = 0, z does not exist in set C, in which C is the most general set for this case.
    However, is it possible to say that 1/z = 1/(1/0) = 0/1 = 0 for R = 0?...is it valid?
     
  2. jcsd
  3. Oct 25, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, it would not be. Since z is not defined when R= 0, neither is 1/z. It would be valid to say that the limit of 1/z, as R goes to 0, is 0.
     
  4. Oct 25, 2008 #3
    Does that imply that there is some sort of discrepancy when we look at it as 1/z = 1/(1/R) = R?
     
  5. Nov 15, 2008 #4
    ...o_o...
     
  6. Nov 15, 2008 #5

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    In a manner of speaking you can think of it that way. This is basically a question of defining a function carefully.

    A function is defined as a map from a set A to a set B (notation: f:A -> B) where each element a in A is assigned a unique element in B (written f(a)). Note not all elements in B must be mapped to, and elements in B can have more than one element in A mapping to them. A is called the domain, B is called the co-domain

    If you have f:A->B a function and g:B->C a function (g could also map D->C for D a subset of B) , then g(f(x)) is a function g(f):A->C (as any value in A is carried to a value in C). So if we take f(x)=1/x, the domain and codomain of this function is going to be R-{0} (R being the set of real numbers.... alternatively it could be the set of complex numbers) Then f(f(x)) is a function that maps R-{0} to R-{0} where every element x is mapped to itself. But the function isn't defined at 0 itself, as 0 isn't in the domain (so f(f(0)) is as meaningful as f(f(apple)) or f(f(red)).

    It's possible to add additional values (such as an infinity symbol) to the codomain in order to allow 0 to be in the domain, and in such cases you'll find f(f(0)) ends up being 0, but these are special cases
     
  7. Nov 15, 2008 #6
    What is infinity (I've come up with a vague definition, but I don't think it suffices in more modern script)?
     
  8. Nov 16, 2008 #7
    There are many different types of infinities used in mathematics. The infinity that Office_Shredder is referring to is the infinity of the Projectively Extended Real Numbers, which is defined in the link. The extended real number sets unfortunately do not have a field algebra, so you will not see them used much in everyday arithmetic.
     
  9. Nov 16, 2008 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, there is a discrepancy.
    If z= 1/R, then 1/z= R for all R except R= 0.

    A variation often seen in beginning Calculus is this: [tex]\frac{x^2- 4}{x- 2}= x+2[/tex] for all x except x= 2. Any good textbook will make that point.
     
  10. Nov 16, 2008 #9
    Oh...I see...
     
  11. Nov 18, 2008 #10
    I thought that was just called a removable discontinuity?
     
  12. Nov 18, 2008 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes it is a "removable discontinuity". What is your point?
     
  13. Nov 18, 2008 #12
    So why is there a discrepency? And why the original question from the other guy?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Undefined values
  1. Undefined Points (Replies: 6)

  2. Is this undefined? (Replies: 4)

  3. Undefinable numbers (Replies: 31)

Loading...