MHB Prove Z7 is a Ring Under + and x Operations

[Zoey93]

Hey there,

I need some help with this assignment:

Use the definition for a ring to prove that Z7 is a ring under the operations + and x as defined as follows: [a]7+7 = [a+b]7 and [a]7 x 7 = [a x b]7

1. state each step of your proof
2. provide written justification for each step.But First I must use the six definitions of a ring to prove that z7 satisfies them.
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+7+[c]7 Given
~[a]7+7+[c]7=[a]7+(7+[c]7) Addition of Integers is Associative
~[a]7+(7+[c]7)=[a]7+[b+c]7 Given

2. Additive Commutativity Property
[a]7+7=7+[a]7
~[a]7+7=[a+b]7 Given
~[a+b]7=[b+a]7 Addition of Integers is Commutative
~[b+a]7=7+[a]7 Given

 

[Deveno]

I find your proof of associativity hard to follow. To me, it should run like this:

$[a]_7 + (_7 + [c]_7) = [a]_7 + [b+c]_7$ (by definition of addition in $\Bbb Z_7$)

$= [a + (b + c)]_7$ (by definition of addition in $\Bbb Z_7$, again)

$= [(a+b) + c]_7$ (by the associativity of addition in the integers)

$= [a+b]_7 + [c]_7$ (by definition of addition in $\Bbb Z_7$)

$= ([a]_7 + _7) + [c]_7$ (by definition of addition in $\Bbb Z_7$), QED.

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

Your proof of commutativity is better, line 3 is the critical step.

 

[Zoey93]

Hello Deveno,

Thank you for your help. So I finished the assignment but there are a few things I need to fix. I need help correcting number 4, my professor said that my statement is incorrect. Also, I need help correcting part B the zero divisor, my professor said that it was incorrect. Here is the link to my assignment.

https://www.dropbox.com/s/9x2mows1fb5l7a7/Task 2 AA zk.docx?dl=0

 

[Deveno]

For number 4, you need to show that for any $[a]_7 \in \Bbb Z_7$, there exists $_7 \in \Bbb Z_7$ such that:

$[a]_7 + _7 = _7 + [a]_7 = [0]_7$, and that such a $_7$ is unique (that is, $b$ is unique up to an integer multiple of $7$).

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

So it then falls upon you to find a suitable $b \in \Bbb Z$ that works.

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

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:

$7 - 0 = 7 = 1\cdot 7 \in 7\Bbb Z$), and $a \neq 0$, in which case $b \in \{1,2,3,4,5,6\}$).

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 $a \neq 0$, it follows that $[a]_7 + [7-a]_7 = [a+(7-a)]_7 = [a+(-a+7)]_7 = [(a+-a)+7]_7 = [0+7]_7 = [7]_7 = [0]_7$.

Thus for any $a \in \{0,1,2,3,4,5,6\}$, there is at least one $b \in \{0,1,2,3,4,5,6\}$, such that $[a]_7 + _7 = [0]_7$.

The equation $_7 + [a]_7 = [0]_7$ is a direct result of $+$ being commutative in $\Bbb Z_7$. However, we're not quite done yet. We still have to show that if $b \in \{0,1,2,3,4,5,6\}$ is such that $[a]_7 + _7 = [0]_7$, that no other $c \neq b \in \{0,1,2,3,4,5,6\}$ has this property.

Assume, for the sake of showing a contradiction, that such a $c \neq b$ exists.

We have $[a+b]_7 = [a]_7 + _7 = [a]_7 + [c]_7 = [a+c]_7$, that is:

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

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

The set $\{0,1,2,3,4,5,6\}$ is called a set of representatives for $\Bbb Z_7$, 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 $a \in \{0,1,2,3,4,5,6\}$.

I'm not sure of your professors' issue for number 4, but I suspect it is that you argued using $[-a]_7$, instead of finding $-([a]_7)$.

***********************

With regards to part B, you have made a mistake in arguing:

since $7$ is prime, neither $a$ nor $b$ divide $7$. This is not true, as either $a$ or $b$ might be $1$, which *does* divide $7$.

It is easier to make the reverse argument, that if $[a]_7\cdot_7 = [0]_7$, then either $[a]_7 = [0]_7$ or $_7 =[0]_7$ (or both). To make this argument, assume that $[a]_7 \neq [0]_7$, and show that we *must* have $_7 = [0]_7$.