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

[srfriggen]

Homework Statement

Prove that the set of all rational numbers of the form 3ⁿ6^m, m,n$\in$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.

(3^m6ⁿ)(3^k6^l) = 3^m+k6^n+l.

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

(3^m6ⁿ$\ast$3^k6^l)$\ast$3^p6^q = 3^m6ⁿ$\ast$(3^k6^l$\ast$3^p6^q)


4. Associativity:

 

[LCKurtz]

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 3^m6ⁿ are closed under multiplication, which is what you showed. If it is to hand in I would show a bit more detail.

 

[srfriggen]

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 3^m6ⁿ 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!

 

[SammyS]

...

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

(3^m6ⁿ)(3^k6^l) = 3^m+k6^n+l.

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

(3^m6ⁿ$\ast$3^k6^l)$\ast$3^p6^q = 3^m6ⁿ$\ast$(3^k6^l$\ast$3^p6^q)

...

The following are in addition to the comments of LCKurtz.

For Closure:

Of course it's true that (3^m6ⁿ)(3^k6^l) = 3^m+k6^n+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 3^r6^s, where r,s∈ℤ.

Why is it that 3^m+k6^n+l is of the desired form? It's because the integers are closed under addition.​

 

[srfriggen]

The following are in addition to the comments of LCKurtz.

For Closure:

Of course it's true that (3^m6ⁿ)(3^k6^l) = 3^m+k6^n+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 3^r6^s, where r,s∈ℤ.

Why is it that 3^m+k6^n+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.

 

[SammyS]

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 ..