Proving the Rational Numbers of the Form 3n6m is a Group under Multiplication

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In summary, the set of all rational numbers of the form 3n6m, m,n\inZ, is a group under multiplication. The identity element is 1, each element has an inverse 3-n6-m, and closure under multiplication holds. Associativity is also a property of the rationals, and holds so that (3m6n)(3k6l) = 3m+k6n+l.
  • #1
srfriggen
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Homework Statement



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

Homework Equations





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:
 
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  • #2
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.
 
  • #3
LCKurtz said:
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.

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!
 
  • #4
srfriggen said:
...

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)

...
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.​
 
  • #5
SammyS said:
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.​

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.
 
  • #6
srfriggen said:
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.
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 ..
 

1. What are the four criteria for proving something is a group?

The four criteria for proving something is a group are closure, associativity, identity, and inverses.

2. How do you show closure in a group?

Closure in a group is shown by demonstrating that for any two elements in the group, their combination or operation also results in an element within the group.

3. Can a group have more than one identity element?

No, a group can only have one identity element. This element is unique and must satisfy the criteria that when combined with any other element in the group, it remains unchanged.

4. What is the significance of inverses in a group?

Inverses in a group are necessary to ensure that every element in the group has a corresponding element that results in the identity element when combined. This allows for the existence of a neutral element and the ability to "undo" operations within the group.

5. Are all groups commutative?

No, not all groups are commutative or abelian. Commutativity refers to the property that the order of elements in an operation does not affect the result. Some groups, such as the symmetric group, are not commutative.

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