Use the definition for a ring to prove that Zy is a ring und

Use the definition for a ring to prove that Z_y is a ring under the operations + and x defined as follows [a]_T + [b]_T = [a + b] and [a]_J times [b] = [a times b] state each step of your proof Provide written justification for each step of your proof.

Solution

A ring is a set R equipped with two binary operations,1 here denoted by + and *, that have the following properties.

1. (a + b) + c = a + (b + c) for all a, b and c in R (addition is associative).
2. a + b = b + a for all a and b in R (addition is commutative).
3. There is an element 0 R such that 0 + x = x + 0 = x for all x R.
4. For each element x R, there is a unique element y R such that x + y = y + x = 0. (We denote y by x.)
5. (a * b) * c = a * (b * c) for all a, b, and c in R (multiplication is associative).
6. (The distributive law) a * (b + c) = a * b + a * c and (b + c)* a = b * a + c * a for all a, b, and c in R.

This is what I have so far:

1. Additive Associativity Property
[a+b]7+[c]7=[a]7+[b+c]7
~[a+b]7+[c]7=[a]7+[b]7+[c]7 Given
~[a]7+[b]7+[c]7=[a]7+([b]7+[c]7) Addition of Integers is Associative
~[a]7+([b]7+[c]7)=[a]7+[b+c]7 Given

2. Additive Commutativity Property
[a]7+[b]7=[b]7+[a]7
~[a]7+[b]7=[a+b]7 Given
~[a+b]7=[b+a]7 Addition of Integers is Commutative
~[b+a]7=[b]7+[a]7 Given a]7+([b]7+[c]7)=[a]7+[b+c]7 (by definition of addition in Z7)

=[a+(b+c)]7 (by definition of addition in Z7, again)

=[(a+b)+c]7 (by the associativity of addition in the integers)

=[a+b]7+[c]7 (by definition of addition in Z7)

=([a]7+[b]7)+[c]7 (by definition of addition in Z7), QED.

That is, at some point all of a,b,c should be \"inside the brackets\".

For number 4,we need to show that for any [a]7Z7, there exists [b]7Z7 such that:

[a]7+[b]7=[b]7+[a]7=[0]7, and that such a [b]7 is unique (that is, b is unique up to an integer multiple of 7).

While not strictly necessary, it is usual to consider only a{0,1,2,3,4,5,6}, since any other integer a is equivalent modulo 7 to one of these.

So it then falls upon we have to find a suitable bZ that works.

we are correct that b=a would work, but the usual choice is b=7a which guarantees that b{1,2,3,4,5,6,7}.

This actually breaks down into \"two cases\" a=0 (in which case we prefer [0]7 rather than [7]7, although these are equal since:

70=7=177Z), and a0, in which case b{1,2,3,4,5,6}).

It follows from the identity property of [0]7 that [0]7+[0]7=[0+0]7=[0]7, and that [0]7 is the unique congruence class that has this property for [0]7, that is:

[0]7=[0]7.

If a0, it follows that [a]7+[7a]7=[a+(7a)]7=[a+(a+7)]7=[(a+a)+7]7=[0+7]7=[7]7=[0]7.

Thus for any a{0,1,2,3,4,5,6}, there is at least one b{0,1,2,3,4,5,6}, such that [a]7+[b]7=[0]7.

The equation [b]7+[a]7=[0]7 is a direct result of + being commutative in Z7. However, we\'re not quite done yet. We still have to show that if b{0,1,2,3,4,5,6} is such that [a]7+[b]7=[0]7, that no other cb{0,1,2,3,4,5,6} has this property.

Assume, for the sake of showing a contradiction, that such a cb exists.

We have [a+b]7=[a]7+[b]7=[a]7+[c]7=[a+c]7, that is:

a+b(a+c)=7k, for some integer k. Thus:

bc=7k. Since both b,c{0,1,2,3,4,5,6}, we have |bc|<7. Hence k=0, contradiction.

The set {0,1,2,3,4,5,6} is called a set of representatives for Z7, and is handy for showing uniqueness as we have done here. That is:

[a]7 does not uniquely determine a, but it *does* if we require a{0,1,2,3,4,5,6}.