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Let $p(x)=ax^2+bx+c, q(x)=dx^2+ex+f$ and $n\in\mathbb{z}$. Okay, I need to define the following opertations on $\mathbb{Z_n}$.

(1) $[r]_n\bigoplus[s]_n=[p(r)+q(s)]_n$

(2) $[r]_n\bigodot[s]_n=[p(r)q(s)]_n$

And determine whether or not $\bigoplus$ and $\bigodot$ are well defined, and prove your answer.

---- To be honest, I have no idea what I am supposed to do here. Do I have to prove like, $[r]_n\equiv r+kn$ or $[r]_n\equiv r\mod {n}$?

If $r=2,s=3$, How do I make these operation work? --$[2]_n\bigoplus[3]_n=[(4a+2b-c)+(9d+3x+f)]_n$? How this operation work?

Wes
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An element $[r]_n \in \mathbb{Z}_n$ is actually an equivalence class. For example $[2] \in \mathbb{Z}_5$ stands for the equivalence class $\{\ldots,-8,-3,2,7,12,\ldots\}$. The operations $\oplus$ and $\odot$ require you to do arithmetic with elements of $\mathbb{Z}_n$. In order to do so we pick a representative from the equivalence class. The operations are called well defined if the outcome does not depend on which representative you pick. For example, we want to have $[2]_5 \oplus [1]_5 = [-8]_5 \oplus [11]_5$, so that the operation $\oplus$ makes sense on the equivalence classes as a whole.

Arthur
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  • THen do I pick arbitrary number for $n,r$ or......oh, please give me some more guide line. How do I deal with $n$? – Wes Oct 11 '13 at 00:37
  • Pick arbitrary representatives $r_1,r_2 \in [r]_n$ and $s_1,s_2 \in [s]_n$ and show that $r_1 \oplus s_1 = r_2 \oplus s_2$. – Arthur Oct 11 '13 at 00:51
  • Then, how do I make connection with $p(r),q(s)$? I am not sure what to do with the $a,b,c,d,e,f$. – Wes Oct 11 '13 at 01:02
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    $a,b,c,d,e,f$ are just random constants. Consider $[r]_k \oplus [s]_k$. Let $r' = nk + r$. We find $[r]_k \oplus [s]_k = ar^2 + br + c + ds^2 + es + f$ and $[r']_k \oplus [s]_k = ar^2 + 2ank + a(nk)^2 + br + bnk + c + ds^2 + es + f$. So $[r]_k \oplus [s]_k$ and $[r']_k \oplus [s]_k$ differ only by a multiple of $k$, which means they are in the same equivalence class of $\mathbb{Z}_k$. So the answer of $\oplus$ does not depend on the representative for the first coordinate. By symmetry it is obvious that it does not depend on the representative for the second coordinate either. So $\oplus$ is.. – Arthur Oct 11 '13 at 01:11
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    ...well defined. Can you do the same for $\odot$ now? – Arthur Oct 11 '13 at 01:11
  • I think I am getting it. But form $[r']\bigoplus[s]$, I am not sure how you said it differs inly by a multiple of $k$. Because, $ar^2+a(nk)^2$ is not the same with a(r+nk)^2$, and $2ank$ doesn't fit to $[r]\bigoplus[s]$. You know what I mean? – Wes Oct 11 '13 at 01:53
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    Sorry, I made a typo. The term $2ank$ should be $2ankr$. What I have done is simply multiply out $a(nk+r)^2 = ar^2 + 2ankr + a(nk)^2$. What I mean by the difference being a multiple of $k$, the difference is equal to $[r'] \oplus [s] - [r] \oplus [s] = 2ankr + a(nk)^2 + bnk = k(2anr + an^2k + bn)$. – Arthur Oct 11 '13 at 02:40