Definition. (Page 58) Let $\class{k}$ denote the congruence class to which $k$ belongs $\!\pmod{n}.$

Definition. (Page 59) For $\class{a}\in\Z_n,$ define $\class{a}\oplus\class{b}$ by \[ \class{a}\oplus\class{b}=\class{a+b}. \]

Example 11.1. If $n=5$ then \[ \class{3}\oplus\class{4} =\class{3+4} =\class{7} =\class{2} \] and \[ \class{-29}\oplus\class{7} =\class{-22} =\class{3}. \]

Definition. (Page 59) If $\star$ is an operation on $S$ then it is said to be well-defined.

Lemma 11.1. Addition on integers modulo $n$ is well-defined. That is, in $\Z_n,$ if $\class{a_1}=\class{a_2}$ and $\class{b_1}=\class{b_2}$ then $\class{a_1+b_1}=\class{a_2+b_2}.$

Durbin Note. (Page 59) Problem 11.18 is designed to remove doubts about whether Lemma 11.1 is really necessary.

Problem 11.18. Define an equivalence relation on the set of integers by letting $a\sim b$ mean that either both $a$ and $b$ are negative or both are nonnegative. Attempt to define an operation $\boxplus$ on $\{\class{-1},\class{0}\},$ the complete set of equivalence class representatives of $\sim,$ by \[ \class{a}\boxplus\class{b} =\class{a+b} \] for $a,b\in\Z$ in analogy with the definition of $\oplus$ on $\Z_n.$ Show that $\boxplus$ is not well-defined.

Guevara Notes

Guevara Definition 11.1. (Page 59) Let $*$ be an operation on $S,$ and $\varphi:S\rightarrow T.$ Define an operation $\star$ on $T$ by $\varphi(a)\star\varphi(b) = \varphi(a*b)$ for all $a,b\in S.$ Then $\star$ is well-defined if $\varphi(a_1)=\varphi(a_2)$ and $\varphi(b_1)=\varphi(b_2)$ implies $\varphi(a_1*b_1)=\varphi(a_2*b_2).$

Guevara Note. (Page 59) It is possible to define $\varphi$ and $*$ on $S$ in such a way that $\star$ will not be well-defined. Attempting to define $\star$ in terms of such $\varphi$ and $*$ would fail. Let's see how.

Let $\star$ be defined in the manner given in Guevara Definition 11.1, \begin{equation} \varphi(a)\star\varphi(b) = \varphi(a*b), \text{ for all } a,b\in S \label{eq_1}. \end{equation} By the same definition, suppose $\star$ is not well defined, so that the following statements hold for some $a_1, a_2, b_1, b_2 \in S.$ \begin{align} &\varphi(a_1)=\varphi(a_2)\label{eq_2}\\ &\varphi(b_1)=\varphi(b_2)\label{eq_3}\\ &\varphi(a_1*b_1)\ne\varphi(a_2*b_2)\label{eq_4}\\ \end{align} Consider the set $\Gamma=\{\eqref{eq_2}, \eqref{eq_3}, \eqref{eq_4} \}$ of the previous three statements. Note that $\star$ does not occur in the statements of $\Gamma.$ Therefore, the conditions for $\star$ to be well-defined do not depend on \eqref{eq_1}. Also, \eqref{eq_4} implies $a_1*b_1\ne a_2*b_2,$ which implies $a_1\ne a_2$ or $b_1\ne b_2.$

A model for $\Gamma$ is suggested by Problem 11.18. As given there, let $S=\Z$ and $T=\{\class{-1}, \class{0}\},$ and $\varphi(x)=\class{x}$ for all $x\in S.$ Pick values $(a_1,b_1,a_2,b_2)=(-2,0,-1,1)$ and evaluate the statements of $\Gamma,$ to verify they are all true: \begin{align} &\varphi(a_1)=\class{-2}=\class{-1}=\varphi(a_2)\tag{$2'$}\label{eq_2_p}\\ &\varphi(b_1)=\class{0}=\class{1}=\varphi(b_2)\tag{$3'$}\label{eq_3_p}\\ &\varphi(a_1+b_1)=\class{-2+0}=\class{-1}\ne\class{0}=\class{-1+1}=\varphi(a_2+b_2)\tag{$4'$}\\ \end{align} Therefore, $\Gamma$ is consistent. Note also that $\varphi$ itself is well-defined, in the sense that if $x=y$ then $\varphi(x)=\varphi(y).$

Now, discard the model and suppose the statements of $\Gamma$ hold. That is, assume $\star$ as defined by \eqref{eq_1} is not well-defined. Then \begin{equation} \varphi(a_1)\star\varphi(b_1)=\varphi(a_2)\star\varphi(b_2)\tag{6}\label{eq_6} \end{equation} follows by simple substitution from \eqref{eq_2} and \eqref{eq_3}. Applying \eqref{eq_1} to the left and right sides of \eqref{eq_6} yields \[ \varphi(a_1)\star\varphi(b_1)=\varphi(a_1*b_1) \] and \[ \varphi(a_2)\star\varphi(b_2)=\varphi(a_2*b_2) \] respectively. Combining these results with \eqref{eq_6} gives \[ \varphi(a_1*b_1)=\varphi(a_2*b_2) \] contradicting \eqref{eq_4}. Therefore, $\Gamma\cup\{\eqref{eq_1}\}$ is inconsistent.

Taken together, the consistency of $\Gamma$ and the inconsistency of $\Gamma\cup\{\eqref{eq_1}\}$ shows that it is possible for an operator to be defined according to Guevara Definition 11.1, yet not be well-defined. In particular, if $\star$ is defined in terms of a map $\varphi$ and an operator $*$ on $S,$ as given in Guevara Definition 11.1, then $\varphi$ must satisfy the given conditions in order for $\star$ to be well-defined. The model for $\Gamma$ shows that not all maps and operators $*$ that can be defined on $S$ do satisfy those conditions. Therefore, attempting to define $\star$ in terms of such maps would fail.