# Quotient set of Z

• December 6th 2009, 08:26 AM
oldguynewstudent
Quotient set of Z
The following is from a handout. We have not gone over this material in class yet and I cannot find adequate explanations in the text. As finals are 9 days away, I could use some help understanding this material.

6. Theorem: Let n be a fixed positive integer. The relation $\equiv$n is denoted on Z by the rule x $\equiv$n y iff

n | (x-y) (In many books, you will see the notation x $\equiv$y (modulo n)).
a) $\equiv$n is an equivalence relation on Z.
b) [a] $_\equiv$n will be denoted by [a] for brevity, and

[a] = {a + kn |k $\epsilon$Z}

c) The quotient set of Z by $\equiv$n is denoted Zn and it has n elements

Zn = {[0]; [1]; [2]; ... ; [n -1]}.

When we define functions or operations where the domain is a quotient set, we often come across the problem of deciding if the rule given is well-defined.

7. Consider the function'
h : Z5 $\longrightarrow$ Z given by the rule h([a]) = $a^2$for a $\epsilon$Z. Is h well-defined?

8. Consider the function' l : Z3 $\longrightarrow$ Z given by the rules l([a]) = a + 1 for a = 0; 1; 2. Is l well-defined?

9. Consider the `function' f : Z5 $\longrightarrow$Z7 given by the rule f([a]) = [a] for a $\epsilon$Z. Is f well-defined?

For number 7) I came up with h([0]) = $0^2$ = 0
h([1]) = $1^2$ = 1
h([2]) = $2^2$ = 4
h([3]) = $3^2$ = 9
h([4]) = $4^2$ = 16

Should I have written instead h([1]) = { 1 + k*5, k $\epsilon$ Z} ? and the same for the other equivalence classes?

I said that h was well-defined.

For number 8) I came up with l([0]) = 0 + 1 = 1
l([1]) = 1 + 1 = 2 and l([2]) = 2 + 1 = 3, since only the domain is limitted and not the range, all equivalence classes will still be in the range so l is well-defined. ( Should this also be of the form a + k*3?

I'm a little confused by number 9. Does this mean I come up with each equivalence class for mod 5, then these should be equal to an equivalence class for mod 7?

Thanks

Sung to the tune of the Alvin and the Chipmunks song:

Final, final exams are here, time for panic and time for fear!
All my logic does loop de loops, I still want a universal proof.
I can hardly stay awake, just shoot me, erase my fate!
• December 6th 2009, 09:12 AM
Plato
That is even too much to quote. Moreover, I consider it is badly written.
That relation works this way: $x \equiv _n y$ if and only if $x~\&~ y$ have the same remainder when divided by $n$.

So $15 \equiv _7 22$ or $-3 \equiv _5 2$.
That second one takes some thought: $2-(-3)=5$.

The equivalence classes of $\equiv _5$ are $[0],~ [1],~[2],~[3],~[4]$.
Where $[2]=\{\cdots -8,-3,2,7\cdots\}$. Notice those are five units apart.

I hope this helps.