Results 1 to 7 of 7

Math Help - stumped on this equivalence relation

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    5

    stumped on this equivalence relation

    define ~ on Z by a~b iff 3a+b is a mulitple of 4.

    so far i only have reflexive...

    reflexive: 3a+a=4a

    how can i prove that its symmetric and transitive?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by glopez09 View Post
    define ~ on Z by a~b iff 3a+b is a mulitple of 4.

    so far i only have reflexive...

    reflexive: 3a+a=4a

    how can i prove that its symmetric and transitive?
    Symmetric:

    a\sim b\implies 3a+b=4n,

    But 3a+b=4n\implies 9a+3b=4(3n)\implies3b+a=4(3n)-8a \implies 3b+a=4(3n-2a)\implies b\sim a

    Transitive:

    I'll start this, and then I'll leave it for you to finish.

    a\sim b\implies 3a+b=4n and b\sim c\implies {\color{red}3b+c}=4k

    Now, 3a+b=4n\implies 3a+(b+3b)+c-({\color{red}3b+c})=4n\implies\dots

    Can you finish this?
    Last edited by Chris L T521; September 26th 2009 at 10:08 PM. Reason: fixed error.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2009
    Posts
    5
    i am not understanding either of those...?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello glopez09
    Quote Originally Posted by glopez09 View Post
    i am not understanding either of those...?
    You need to work out what it is you need to prove, in order to show that \sim is symmetric and transitive.

    First it's symmetric if a\sim b \Rightarrow b\sim a.

    Now a\sim b means that 3a+b is a multiple of 4. And if we're going to show that b\sim a, we shall need to show that this will mean that 3b+a is also multiple of 4. So, read carefully through Chris L T521's answer. He has clearly shown that if 3a+b is a multiple of 4 then 3b+a is also a multiple of 4.

    Then, what does it mean to prove that \sim is transitive? It is this: if a\sim b and b\sim c, then we must prove that a\sim c. Translating this into multiples of 4, this is: if 3a+b is a multiple of 4, and 3b+c is a multiple of 4, then we must prove that 3a+c is also a multiple of 4. Chris L T521's answer has shown you how to build up an expression for 3a+c starting with 3a+b and 3b+c. Study it again, and make sure that you can see how this shows that 3a+c is a multiple of 4.

    Grandad
    Last edited by Grandad; September 27th 2009 at 10:07 PM. Reason: Corrected typos
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2009
    Posts
    5
    OK. i got some time to work on it. and i now understand the symmetry. but that -(3b+c) is throwing me through a loop?

    this was my attempt without that.

    3a+b+3b+c = 4n+4k

    3a+4b+c = 4n+4k

    3a+c = 4n+4k-4b

    3a+c = 4(n+k-b)

    would that work?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by glopez09 View Post
    OK. i got some time to work on it. and i now understand the symmetry. but that -(3b+c) is throwing me through a loop?

    this was my attempt without that.

    3a+b+3b+c = 4n+4k

    3a+4b+c = 4n+4k

    3a+c = 4n+4k-4b

    3a+c = 4(n+k-b)

    would that work?
    Perfect!

    In the transitive case, I had to add zero:

    3a+b=4n\implies 3a+b+{\color{blue}0}=4n\implies3a+b+(3b+c){\color{  red}-(3b+c)}=4n \implies 3a+4b+c-4k=4n...

    Then your result follows. Does this clarify things more?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Sep 2009
    Posts
    5
    ahhh ok thanks so much.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: April 7th 2011, 12:46 AM
  2. Equivalence Relation
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: March 4th 2010, 06:43 AM
  3. equivalence relation and equivalence classes
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: January 7th 2010, 07:36 PM
  4. Equivalence relation and order of each equivalence class
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 30th 2009, 10:03 AM
  5. Equivalence relation and Equivalence classes?
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: January 7th 2009, 04:39 AM

Search Tags


/mathhelpforum @mathhelpforum