Associative, Commutative, and Distributive Laws for Congruence Classes

[bologna121121]

Homework Statement

We are supposed to prove that the addition and multiplication of congruence classes for integers are associative, commutative, and distributive.

Homework Equations

None

The Attempt at a Solution

To me, it seems like this would just be trivial as all these properties follow from the fact that we are working with integers, and integers are associative, commutative, and distributive (with both addition and multiplication). Am I on the right track here? Thanks.

 

[jbunniii]

To me, it seems like this would just be trivial as all these properties follow from the fact that we are working with integers, and integers are associative, commutative, and distributive (with both addition and multiplication). Am I on the right track here? Thanks.

It is fairly trivial, but you still need to show it. A congruence class of integers is not the same thing as an integer. What do you have so far?

 

[bologna121121]

I think I have the idea of what to do, but just to be sure I'll try the commutative addition one.

The addition is commutative if $\overline{a}$ + $\overline{b}$ = $\overline{b}$ + $\overline{a}$

This is the same condition as x + y = y + x for all x $\in$ $\overline{a}$ and all y $\in$ $\overline{b}$ where a,b,x,y are all integers.

Thus, because for any integers x and y, x + y = y + x (commutativity of integers under addition) we would have $\overline{a}$ + $\overline{b}$ = $\overline{b}$ + $\overline{a}$

Is this more or less the correct approach to take here?

 

[jbunniii]

That's more or less right. If you want to be more formal, you might start from the definitions:
$$\overline{a} + \overline{b} = \{x + y : x \in \overline{a}, y \in \overline{b}\}$$
$$\overline{b} + \overline{a} = \{y + x : x \in \overline{a}, y \in \overline{b}\}$$
To show equality of two sets, we show that each contains the other, i.e. $\overline{a} + \overline{b} \subseteq \overline{b} + \overline{a}$ and $\overline{b} + \overline{a} \subseteq \overline{a} + \overline{b}$. To show the first inclusion, take an arbitrary element of $\overline{a} + \overline{b}$. By definition, it is of the form $x + y$ with $x \in \overline{a}$ and $y \in \overline{b}$. By commutativity of integer addition, $x + y = y + x$, and the right hand side is clearly an element of $\overline{b} + \overline{a}$. The reverse inclusion is similar.