Results 1 to 4 of 4

Math Help - Induction question..

  1. #1
    Newbie
    Joined
    Nov 2007
    Posts
    17

    Induction question..

    Hello,

    Q - Ler R be a symmetric relation defined on a set C. Prove or disprove thatr R^5 is a symmetric relation.

    My question is.. do I have to prove that by Induction ??? or is there any other way to solve it ??

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,963
    Thanks
    1784
    Awards
    1
    If R is symmetric then the standard proof that so is R \circ R follows:
    \begin{gathered}<br />
  (a,b) \in R \circ R \Rightarrow \;(a,c) \in R \wedge (c,b) \in R \hfill \\<br />
   \Rightarrow \;(c,a) \in R \wedge (b,c) \in R \hfill \\<br />
   \Rightarrow (b,a) \in R \circ R \hfill \\ <br />
\end{gathered}.

    You should be able to extend that
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2007
    Posts
    17
    Quote Originally Posted by Plato View Post
    If R is symmetric then the standard proof that so is R \circ R follows:
    \begin{gathered}<br />
(a,b) \in R \circ R \Rightarrow \;(a,c) \in R \wedge (c,b) \in R \hfill \\<br />
\Rightarrow \;(c,a) \in R \wedge (b,c) \in R \hfill \\<br />
\Rightarrow (b,a) \in R \circ R \hfill \\ <br />
\end{gathered}.

    You should be able to extend that
    Yeah.. I do get what you mean.
    because we have 5 as a power.. 5 is an odd number. So we will be missing a pair making the relation not symmetric..
    So the answer should be No.. I guess..??

    Anyway, thanks for the help man. Really appreciate it.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,963
    Thanks
    1784
    Awards
    1
    Quote Originally Posted by MMM88 View Post
    Yeah.. I do get what you mean.
    because we have 5 as a power.. 5 is an odd number. So we will be missing a pair making the relation not symmetric..
    So the answer should be No.. I guess..??
    You have been mislead by my posting!
    We have shown that R \circ R is symmetric.
    Now suppose that (x,y) \in R \circ R \circ R then for some z, (x,z) \in R \circ R\quad  \wedge \quad (z,y) \in R.
    But we already have shown that (x,z) \in R \circ R\quad  \Rightarrow \quad (z,x) \in R \circ R.

    This is the reason that you may have been lead to think that it is a proof by induction.
    In fact, R^5  = R \circ R \circ R \circ R \circ R is symmetric.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Induction question
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: December 7th 2009, 01:44 PM
  2. Induction question
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: May 2nd 2009, 02:32 AM
  3. induction question..
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 6th 2009, 07:23 AM
  4. Question on induction
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: July 18th 2008, 09:49 PM
  5. Induction Question!!!
    Posted in the Algebra Forum
    Replies: 11
    Last Post: June 15th 2007, 10:51 PM

Search Tags


/mathhelpforum @mathhelpforum