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

  • Thread starter Benzoate
  • Start date
  • #1
420
0

Homework Statement


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


Homework Equations



No equations for this particular proof

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:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,260
619
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.
 
  • #3
420
0
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.
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
 
  • #4
370
0
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.
 
  • #5
420
0
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.
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?
 
  • #6
Dick
Science Advisor
Homework Helper
26,260
619
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.
 
  • #7
420
0
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.
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
 
  • #8
Dick
Science Advisor
Homework Helper
26,260
619
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.
 
  • #9
matt grime
Science Advisor
Homework Helper
9,395
3
Assume that all rational numbers are in the form 3^m *6^n
Not all rational numbers are of this form, so don't assume something false

Therefor 3^m*6^n = p/q where p and q are real numbers.
don't invoke the real numbers, it ain't going to help you, and you never mention this fact again

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).
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....


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.
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.


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)
you must find an e with that property.

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.
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?
 
  • #10
HallsofIvy
Science Advisor
Homework Helper
41,833
956
Perhaps it would be a good idea for you to give the actual problem as written in your textbook.

Originally you said
Benzoate said:
Proved that the set of all rational numbers of the form 3^m *6^n are integers , is a group is a group under multiplication.
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.
 
  • #11
420
0
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.
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
 
  • #12
matt grime
Science Advisor
Homework Helper
9,395
3
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:
  • #13
learningphysics
Homework Helper
4,099
6
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
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...
 

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

Replies
5
Views
1K
Replies
6
Views
4K
Replies
3
Views
2K
Replies
3
Views
13K
Replies
8
Views
1K
Replies
7
Views
5K
  • Last Post
Replies
1
Views
1K
Replies
1
Views
1K
Top