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Math Help - Determining if this subset is a subring

  1. #1
    Newbie driegert's Avatar
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    Determining if this subset is a subring

    Hi there. Firstly, sorry if the title wasn't overly descriptive.

    I'm working on an assignment question dealing with group/ring theory that is:

    Let R=\{f : [0, 1]  \to \mathbb{R}\} be the ring of all real-valued functions defined on the closed interval [0, 1]. Decide if the folowing subset of R is a subring of R:

    S = \{f \in R | f(x) = 0 has an infinite number of solutions in [0, 1] \}

    I know since it is already a subset of R that I need to check if:

    for g, h \in S if

     g - h \in S and gh \in S

    So basically whether

    g - h = 0 and gh = 0

    will also have an infinite number of solutions.

    Another question was similar but the subset consisted of the functions of R that had only a finite number of solutions in [0,1]. I was able to determine by contradiction using an example that it indeed was not, so my intuition here is that this subset is a subring. I'm just not sure how to start it.

    Thanks for any help,
    --
    Dave
    Last edited by driegert; October 10th 2010 at 06:57 PM. Reason: Additional Information
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  2. #2
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    Quote Originally Posted by driegert View Post
    Hi there. Firstly, sorry if the title wasn't overly descriptive.

    I'm working on an assignment question dealing with group/ring theory that is:

    Let R=\{f : [0, 1]  \to \mathbb{R}\} be the ring of all real-valued functions defined on the closed interval [0, 1]. Decide if the folowing subset of R is a subring of R:

    S = \{f \in R | f(x) = 0 has an infinite number of solutions in [0, 1] \}

    I know since it is already a subset of R that I need to check if:

    for g, h \in S if

     g - h \in S and gh \in S

    Another question was similar but the subset consisted of the functions of R that had only a finite number of solutions in [0,1]. I was able to determine by contradiction using an example that it indeed was not, so my intuition here is that this subset is a subring. I'm just not sure how to start it.

    Thanks for any help,
    --
    Dave


    Take g(x):=\left\{\begin{array}{ll}0&if\,0\leq x<\frac{1}{2}\\1&if\,\frac{1}{2}\leq x\leq 1\end{array}\right.\,,\,\,h(x):=\left\{\begin{arra  y}{ll}1&if\,1\leq x<\frac{1}{2}\\0&if\,\frac{1}{2}\leq x\leq 1\end{array}\right. to show S is not a subring.

    Obviously you people don't require a subset to contain the ring's unit in other to be considered a candidate

    to be a subring, otherwise S is automatically not a subring.

    Tonio
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  3. #3
    Newbie driegert's Avatar
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    In order to be a ring doesn't it just have to be associative under addition and multiplication, commutative under addition, 0 and negatives exist, and obey the distributive law? A ring with unity (identity element) is in addition to a basic ring I thought.

    I could be completely wrong however and any clarification would help my understanding!

    Thank you as well for your help Tonio. Really appreciate it.
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