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Proved that a nonempty set containing rational number is a group under multiplication

  1. Aug 29, 2007 #1
    1. The problem statement, all variables and given/known data
    Proved that the set of all rational numbers of the form 3^m *6^n are integers , is a group under multiplication


    2. Relevant equations

    No equations for this particular proof

    3. The attempt at a solution

    Assume that all rational numbers are in the form 3^m *6^n . Therefor 3^m*6^n = p/q where p and q are real numbers.

    Suppose that 3^m*6^n is a group under multiplication. Then this non-empty set must have the following 3 properties: must have an inverse, an identity and the assiociative property

    3^m*6^n is equivalent to (3(sub1)+3(sub2)+3(sub3)+...+3(sub m))*(6(sub 1)+6(sub 2) +6(sub 3)+...+6(sub n))

    On the side note, in an abstract group, 3^3=(3+3+3) and 5^3=(5+5+5).

    To show that 3^m*6^n has an inverse, I must show that 3^m*6^n =6^n*3^m. To be honest I not sure whether I have to show that 3^m*6^n=6^n *3^m or 3^m *6^n=3^-m*6^-n.

    Either way, either expression has to equal e, or einheit.

    I could supposed that b and c are inverses of the expression 3^m*6^n , where b*(3^m*6^n)=e and c*(3^m*6^n)=e => c*(3^m*6^n)= b*(3^m*6^n) => b=c


    To prove that 3^m*6^n has an identity property, I must show that (3^m*6^n)*e=e*(3^m*6^n)=(3^m*6^n)

    I then have to supposed that e and e' are identities of of the Group . Then I can conclude that (3^m*6^n)*e=(3^m*6^n) for all (3^m*6^n) forms in G and e'*(3^m*6^n)=(3^m*6^n) for all (3^m*6^n) forms in G.=> e'e=e' and ee'=e . Thus, e and e' are both equal to e'e and so are equal to each other.

    Not sure how to prove that 3^m*6^n has an assiociative property

    Not sure how to prove that the expression 3^m*6^n as an assiociative property.
     
    Last edited: Aug 30, 2007
  2. jcsd
  3. Aug 29, 2007 #2

    Dick

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    This is really garbled. I suspect that what you actually want to prove is that the set of rational numbers 3^m*6^n, where m and n are both integers is a group under multiplication. Where '*' is ordinary multiplication and '^' is ordinary exponentiation. I'll give you a hint. The identity is m=0, n=0. Because it's 1. Now i) prove closure. ii) what's the inverse of 3^m*6^n? You don't have to prove stuff like associativity because these are real numbers and the reals are already associative. It's a subgroup.
     
  4. Aug 29, 2007 #3
    to prove with closure, wouldn't I have to show that the product of 3^m*6^n is equal to an an integer since the exponents of 3 and 6 are integers. The inver of 3^m*6^n is equal to 3^-m*6^-n
     
  5. Aug 29, 2007 #4
    I think the question is to show that it is a group (not a subgroup). The distinction is rather trivial but they will entail different approaches. The subgroup approach requires less work as you will be using the fact that (Q, *) is a group.

    Benzoate, I would suggest that you go back to your definitions. You seem to be struggling with them. Do come back if you're having anymore trouble.
     
  6. Aug 29, 2007 #5
    I don't understand, I do know my definitions. I understand my definitions I am suppose to know for the proof. For example, inverse of a^n would be a^-n.

    Why do you think I do know understand the definitions?
     
  7. Aug 29, 2007 #6

    Dick

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    To prove closure, you want to show 3^(m1)*6^(n1)*3^(m2)*6^(n2)=3^(m3)*6^(n3), show if m1,n1,m2,n2 are integers then you can take m3 and n3 to be integers also. That's closure. You have the inverse right. But be careful 3^m*6^n does not have to be an integer. As your inverse requires, m or n could be negative.
     
  8. Aug 29, 2007 #7
    You are right. So then 3^m *6^n fails under closure since , for example, 3^-1 *6^-2 =(1/3)(1/36)=1/108 clearing not an integer. If the probably where to ask to prove that 3^m*3^n forms a rational number, then closure would work for this particular form of integer
     
  9. Aug 29, 2007 #8

    Dick

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    Not even a little, I already told you that elements of the group don't have to be integers. -1 and 0 are integers, so 3^(-1)*6^0=1/3 is an element of your group. And it's not an integer. You are barking up the wrong tree. It is closed. Reread my last post and remember laws of how to combine exponents.
     
  10. Aug 30, 2007 #9

    matt grime

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    Not all rational numbers are of this form, so don't assume something false

    don't invoke the real numbers, it ain't going to help you, and you never mention this fact again

    This is clearly wrong. Firstly you only ever write + for abelian group operations, nt the operation in an abstracte group, secondly, you've been told what the group and operation are 3^2=9....


    neither of those is true, and neither as anythin to do with inverses. It is more traditional to find the identity first, since you can't find an inverse if you don't know what the identity is.


    you must find an e with that property.

    It inherits associativity from the fact you're just mutliplying rational numbers. This is the only way you ever prove associativity without drawing out a latin square.

    If I multiply 3^a*6^b by 3^c*6^d what do I get?
     
  11. Aug 30, 2007 #10

    HallsofIvy

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    Perhaps it would be a good idea for you to give the actual problem as written in your textbook.

    Originally you said
    which, even allowing for the "is a group is a group", does not make sense. I think it should be "Prove that the set of all rational numbers of the form 3m6n where m and n are integers is a group under multiplication."

    There is no requirement that the numbers 3m6n must be integers, only that m and n are.
     
  12. Aug 30, 2007 #11
    I acknowledged that 3^m*6^n fails under closure because I gave a specific example that 3^m*6^n , where m and n are integers , but 3^m *6^n is not an integer
     
  13. Aug 30, 2007 #12

    matt grime

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    Please read Halls's post above to realize that you haven't written the question out properly. Most of us filled in the blanks and corrected it ourselves.

    You are undoubtedly supposed to show that the set

    { 3^m*6^n : m,n in Z}

    with the usual multiplication of rational numbers, is a group.
     
    Last edited: Aug 30, 2007
  14. Aug 30, 2007 #13

    learningphysics

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    Have you posted the question exactly as it is written?

    "Proved that the set of all rational numbers of the form 3^m *6^n are integers , is a group under multiplication"

    Are you sure this is exactly what the question says? Because nothing is mentioned here about what values m and n can take...
     
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