7.5: Las funciones deben estar bien definidas
- Page ID
- 116585
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)La discusión de la aritmética modular ignoró un punto muy importante: las operaciones de suma, resta y multiplicación necesitan estar bien definidas. Es decir, si\(\overline{a_{1}}=\overline{a_{2}}\) y\(\overline{b_{1}}=\overline{b_{2}}\), entonces necesitamos saber que
- \(\overline{a_{1}}+{ }_{n} \overline{b_{1}}=\overline{a_{2}}+_{n} \overline{b_{2}}\),
- \(\overline{a_{1}}-_{n} \overline{b_{1}}=\overline{a_{2}}-{ }_{n} \overline{b_{2}}\), y
- \(\overline{a_{1}} \times_{n} \overline{b_{1}}=\overline{a_{2}} \times_{n} \overline{b_{2}}\).
Afortunadamente, todas estas afirmaciones son ciertas. En efecto, siguen fácilmente desde Ejercicio\(5.1.19\):
- Desde\(\overline{a_{1}}=\overline{a_{2}}\) y\(\overline{b_{1}}=\overline{b_{2}}\), tenemos\(a_{1} \equiv a_{2}(\bmod n)\) y\(b_{1} \equiv b_{2}(\bmod n)\), entonces Ejercicio nos\(5.1.19(1)\) dice eso\(a_{1} + b_{1} \equiv a_{2} + b_{2} (\bmod n)\). Por lo tanto\(\overline{a_{1}+b_{1}}=\overline{a_{2}+b_{2}}\), como se desee.
Las pruebas para\(-_{n} \text { and } \times_{n}\) son similares.
Se podría tratar de definir una operación de exponenciación mediante:\[\bar{a} \wedge_{n} \bar{b}=\overline{a^{b}} \quad \text { for } \bar{a}, \bar{b} \in \mathbb{Z}_{n} .\]
Desafortunadamente, esto no funciona, porque n no está bien definido:
Encontrar\(a_{1}, a_{2}, b_{1}, b_{2} \in \mathbb{Z}\), tal que\(\left[a_{1}\right]_{3}=\left[a_{2}\right]_{3}\) y\(\left[b_{1}\right]_{3}=\left[b_{2}\right]_{3}\), pero\(\left[a_{1}^{b_{1}}\right]_{3} \neq\left[a_{2}^{b_{2}}\right]_{3}\).
Asumir\(m, n \in \mathbb{N}^{+}\).
- Mostrar que si\(n > 2\), entonces el valor absoluto no proporciona una función bien definida de\(\mathbb{Z}_{n}\) a\(\mathbb{Z}_{n}\). Es decir, mostrar que existen\(a, b \in \mathbb{Z}\), tal que\([a]_{n}=[b]_{n}, \text { but }[|a|]_{n} \neq[|b|]_{n}\).
- Mostrar que si\(m \mid n\), entonces hay una función bien definida\[f: \mathbb{Z}_{n} \rightarrow \mathbb{Z}_{m}, \text { given by } f\left([a]_{n}\right)=[a]_{m} .\]
- Mostrar que si tratamos de definir una función\(g: \mathbb{Z}_{3} \rightarrow \mathbb{Z}_{2}\) por\ (g\ left ([a] _ {3}\ right) = [a] _ {2}), entonces el resultado no está bien definido.