CMSC 250 Fall 2001 Homework 9 solutions

  1. $A \cap B$ $\{2,4\}$
    $B \cup C$ $\{1,2,3,4,5,6,7,8,9,10\}$
    $A - C$ $\{2,4\}$
    $C - A$ $\{7,9\}$
    $D \times E$ $\{(1,2),(2,3),(2,2)\}$
    $D \times E \times F$ $\{(1,2,1),(1,2,4),(1,3,1),(1,3,4), $
      $(2,2,1),(2,2,4),(2,3,1),(2,3,4)\}$
    $D\times (E \times F)$ $\{(1,(2,1)),(1,(2,4)),(1,(3,1)),(1,(3,4)),$
      $(2,(2,1)),(2,(2,4)),(2,(3,1)),(2,(3,4))\}$
    $B^c $ $\{1,3,5,7,9\}$
    $(B^c)^c$ $\{2,4,6,8,10\}$

    1. $\sum^0$ = $\{ \varepsilon \}$

    2. $\sum^1$ = $\{q,x,z\}$

    3. $\sum^2 $ = $\{qq,qx,qz,xx,xq,xz,zz,zq,zx\}$

    4. $\sum^*$ = $\{ \varepsilon,q,x,z,qq,qx,qz,xx,xq,xz,zz,zq,zx,qqq,qqx,qxz..\}$

  3. $\{xzx,xwx,xyx,xxy,xwy,xzy,xyy,zzx,zwx,zyx,zxy,zwy,zzy,zyy\}$

    1. Same - Must fix picture.

      \includegraphics [height=3cm]{h9p4a.eps}

    2. Different

      \includegraphics [height=3.5cm]{h9a2.eps}

    3. Equal

      \includegraphics {h9a3.eps}

  5. Partition A into subsets $A_1(1 lines),A_2(2 lines),A_3(3 lines),A_4(4 lines)$ where,

    $A_1$=$\{I\}$ (This could also go in the 3 line set.)

    $A_2$=$\{L,T,X,V \}$

    $A_3$= $\{A,F,H,K,N,Y\}$

    $A_4$=$\{E,M,W\}$

  6. Partition the intergers from 2 to 20

    P= $\{ 2,3,5,7,11,13,17,19 \}$

    C= $\{ 4,6,8,9,10,12,14,15,16,18,20 \}$

  7. P($\{ 1,a,+,> \}$) = $\{\emptyset,\{a\},\{1\},\{+\},\{>\},\{1,a\},\{1,+\},\{1,>\},\{a,+\},\\ \{a,>\},\{+,>\},\{1,a,+\},\{1,a,>\},\{1,+,>\},\{a,+,>\},\{1,a,+,>\} \}\}$

    1. Proof:

      Assume that $A\subseteq B$. Hence, $\forall x\in U$ if $x\in A$, then $x\in B$.

      Let $y\in B^{c}$.

      Then, $y\not\in B$. (By the definition of complement.)

      Then, $y\not\in A$. (Because $A\subseteq B$.)

      Hence, $y\in A^{c}$. (By the definition of complement.)

    2. Proof:

      $A-B = A\cap B^{c}$. (By the alternate definition of $-$.)

      Hence, $(A-B)^{c} = (A\cap B^{c})^{c}$.

      Then, $(A\cap B^{c})^{c} = A^{c}\cup (B^{c})^{c}$. (By DeMorgan's laws.)

      And finally, $A^{c}\cup (B^{c})^{c} = A^{c}\cup B$ (By the double complement rule.)

  9. Proof:

    $\vert \sqrt{n({\bf {P}}(D))} - 12\vert < 2$

    which means that

    $-2 < \sqrt{n({\bf {P}}(D))} - 12 < 2$

    so that

    $10 < \sqrt{n({\bf {P}}(D))} < 14$.

    Then,

    $100 < n({\bf {P}(D)}) < 196$.

    Because power sets of sets are exact powers of 2, $n({\bf {P}(D)})$ is an exact power of 2, so it must be $128 = 2^{7}$, the only power of two between 100 and 196. Hence,

    $n(A\times B\times C) = 7$.

    The number of elements in $A\times B\times C$ is equal to the product of the number of elements in $A$, $B$, and $C$. Hence, $n(A)n(B)n(C) = 7$. Since $n(E)$ is an integer for all sets $E$, and there are at least two elements in $B$, $n(A) = n(C) = 1$, and $n(B) = 7$.

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Deep Saraf
2001-10-31