Abelian Group , proving etc.

ℤ/3ℤ

Show that for all classes:

_ _

x,y ∈ ℤ/3ℤ

The following is true:

_ _ ______ ______

∀x1, x2 ∈ x, y1,y2 ∈ y : x1 + y1 = x2 + y2

I dont now how to make lines directly above the letters :(

Okay so I dont know if I need that, but what does the ℤ/3ℤ say? Am I creating classes depending on Euclidean division?

divisible by 3: 0:= [0] = {...,-9,-6,-3,0,3,6,9,...}

remainder 1 1:= [1] = {...,-8,-5,-2,1,4,7,10,13,...}

remainder 2 2:= [2] = {...,-10,-7,-4,-1,2,5,8,11,14,...}

How can I combine it with the statement I have to prove?

Additionally I need to show that (ℤ/3ℤ , +) is an Abelian group.

In this case I I would draw a matrix and analyze it:

+ 0 1 2

0 0 1 2

1 1 2 0

2 2 0 1

Okay so 0 doesnt change anything. This means that the neutral element is 0

The invers is 0, because it is a part of every row.

It is also symmetrical on the diagonal line.

This means it is an Abelian group.

Is this proof enough?

Re: Abelian Group , proving etc.

$\displaystyle x_1 = 3q_1 + r_1, x_2 = 3q_2 + r_1, y_1 = 3q_3+r_2, y_2 = 3q_4 + r_2$ for some integers $\displaystyle q_1,q_2,q_3,q_4,r_1,r_2$ with $\displaystyle 0\le r_1 < 3, 0\le r_2 < 3$.

Re: Abelian Group , proving etc.

Thanks for your answer, but I would like to understand the solution instead of getting the solution itself.

Re: Abelian Group , proving etc.

I did not give you the solution. I used the division algorithm for each variable. That's all I did. I did not tell you what to do with the variables. I did not prove any aspect of the problem. I just gave you a start.

Re: Abelian Group , proving etc.

Oh okay, it just looks so complicated to me, that I thought it is an actual solution. Sadly, my point stands still. I dont know what to do with it. What are the qs?

Sorry for being such a dumbo. But I seriously dont know how to approach this problem.

Re: Abelian Group , proving etc.

As I said, I used the division algorithm. So, $\displaystyle x_1 = 3q_1 + r_1$ means $\displaystyle q_1$ is the quotient when $\displaystyle x_1$ is divided by 3 and $\displaystyle r_1$ is the remainder. Just go back to the Division Algorithm, and it should all make sense.

Now, to show that $\displaystyle \overline{x_1+y_1} = \overline{x_2+y_2}$, you need to show that $\displaystyle (x_2+y_2) - (x_1+y_1)$ is divisible by 3.