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!

Proving something is a Group

  1. Aug 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that the set of all rational numbers of the form 3n6m, m,n[itex]\in[/itex]Z, is a group under multiplication.

    2. Relevant equations



    3. The attempt at a solution

    For this problem I attempted to show that the given set has 1. an Identity element, 2. each element has an inverse, 3. Closure under multiplication, and 4. Associativity.

    1. The identity element is 1

    2. The inverse is 3-n6-m

    3. Closure: the rationals are closed under multiplication, so closure holds, i.e.

    (3m6n)(3k6l) = 3m+k6n+l.

    4. Associativity: This is a property of the rationals and holds, i.e.

    (3m6n[itex]\ast[/itex]3k6l)[itex]\ast[/itex]3p6q = 3m6n[itex]\ast[/itex](3k6l[itex]\ast[/itex]3p6q)


    4. Associativity:
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 29, 2012 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I don't see a question here. It looks like you have the idea. But I would write the identity as 3060. And closure is not because the rationals are closed under multiplication, it is because rationals of the form 3m6n are closed under multiplication, which is what you showed. If it is to hand in I would show a bit more detail.
     
  4. Aug 29, 2012 #3
    I guess my question would be, "Is this correct?"

    I'm glad I was on the right path. I will certainly clean it up before handing it in. Thank you!
     
  5. Aug 29, 2012 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    The following are in addition to the comments of LCKurtz.

    For Closure:
    Of course it's true that (3m6n)(3k6l) = 3m+k6n+l. This is true due to properties of rational numbers under multiplication, particularly the commutative and associative properties.

    But, for this equation to demonstrate closure, the quantity on the right side of the equation needs to be of the form 3r6s, where r,s∈ℤ.

    Why is it that 3m+k6n+l is of the desired form? It's because the integers are closed under addition.​
     
  6. Aug 30, 2012 #5
    Can I just write, "m+k, n+l are integers, since the integers are closed under addition"? Or should I write m+k=r, where r is an integer since the integers are closed under addition.
     
  7. Aug 30, 2012 #6

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    It should be good enough to write, "m+k, n+l are integers, since the integers are closed under addition" .

    I was merely trying to write a complete statement in my earlier post, with the r & s ..
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proving something is a Group
Loading...