# Proving something is a Group

srfriggen

## Homework Statement

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

## 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$\ast$3k6l)$\ast$3p6q = 3m6n$\ast$(3k6l$\ast$3p6q)

4. Associativity:

## The Attempt at a Solution

Homework Helper
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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.

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 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!

Staff Emeritus
Homework Helper
Gold Member
...

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$\ast$3k6l)$\ast$3p6q = 3m6n$\ast$(3k6l$\ast$3p6q)

...

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.​

srfriggen

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.

Staff Emeritus