CMSC 250 Fall 2001 Homework 12 solutions
    1. $a R b$: $\{1R1, 1R2, 1R4, 2R1, 2R2, 2R4, 0R1\}$.

      $a \not{R} b$: $\{0\not{R}2, 0\not{R}4, 3\not{R}1, 3\not{R}2, 3\not{R}4\}$.

    2. $A \times B = \{(0,1),(0,2),(0,4),(1,1),(1,2),(1,4),(2,1),\\ (2,2),(2,4),(3,1),(3,2),(3,4)\}$

    3. R= $\{(1,1), (1,2), (1,4), (2,1), (2,2), (2,4), (0,1)\}$.

    1. $A\times A =\\ \{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),\\ (2,1),(2,2),(2,3),(2,... ...(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \}$

    2. $ R= \{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),\\ (2,2),(2,4),(2,6),(3,3),(3,6), (4,4),(5,5),(6,6) \}$

  3. $\{(2,6),(2,8),(3,6),(3,9),(4,6),\\ (4,8),(5,5),(6,6),(6,8),(6,9) \}$

    1. (Child,Parent), (Parent,Grandparent), (Grandparent,Uncle)

    2. (Child1,Parent1), (Parent1,Grandparent), (Grandparent,Parent2), (Parent2,Child2)
    3. (Grandparent,Parent), (Parent,child)

    1. The relationship is reflexive, symmetric and transitive.

    2. The relationship is not reflexive, not symmetric and not transitive. (However, symmetric and transitive also accepted due to ambiguity.)

    3. The relationship is reflexive and transitive, but not symmetric.

    4. The relationship is symmetric, but not reflexive or transitive.

    5. The relationship is not reflexive, not symmetric and is transitive.

  6. First, partition the set into the 5 subsets.

    1. $A_{0}=\{5,10,15,20\}$.

    2. $A_{1}=\{1,6,11,16\}$.

    3. $A_{2} = \{2,7,12,17\}$.

    4. $A_{3} = \{3,8,13,18\}.$

    5. $A_{4} = \{4,9,14,19\}.$

    Define $R:A\rightarrow A$ by $a R b$ if $a\equiv b\hbox{ mod }5.$

    1. $R$ is reflexive because $a\equiv a\hbox{ mod }5$ because $a-a = 0*5$, an integral multiple of $5$. for all $a\in A$.

    2. Given $a,b\in A$, if $a\equiv b\hbox{ mod }5$, then $a-b=5k$ for some $k\in{\bf {Z}}$, so $b-a=-5(-k)$, which means that $b-a$ is an integral multiple of $5$, so $b\equiv a\hbox{ mod }5$, which means that $bRa$. Hence, $R$ is symmetric.

    3. Let $a,b,c\in A$. If $a R b$ and $bRc$, then $a-b=5k$ and $b-c=5l$ for some $k,l\in{\bf {Z}}$. Then, $a-c=a-b+b-c = (a-b)+(b-c)=5k+5l=5(k+l)$, which means that $a-c=5m$ for some $m\in{\bf {Z}}$, which means that $a\equiv c\hbox{ mod }5$, which means that $aRc$. Hence, $R$ is transitive.

About this document ...

This document was generated using the LaTeX2HTML translator Version 99.1 release (March 30, 1999)

Copyright © 1993, 1994, 1995, 1996, Nikos Drakos, Computer Based Learning Unit, University of Leeds.
Copyright © 1997, 1998, 1999, Ross Moore, Mathematics Department, Macquarie University, Sydney.

The command line arguments were:
latex2html -show_section_numbers -split 0 -no_navigation -no_footnode h12ans.tex

The translation was initiated by Deep Saraf on 2001-11-29

Deep Saraf
2001-11-29