MAT102 Tutorial W3

• 1

  • $A=\{4,8,9,13\}$
  • 1
    • $B=\{\dots ,-\frac{9}{2},-3,-\frac{3}{2},0,\frac{3}{2},3,\frac{9}{2},\dots \}$
    • $B=\{\frac{3x}{2}:x\in \mathbb{Z}\}$
  • 2
    • $C=\{\cdots \cup (-4,-2)\cup (-2,0)\cup (0,2)\cup \dots \}$
    • $C=\{x:x\in \mathbb{R}\wedge x\notin \{even\mathrm{\_}numbers\}\}$

• 2

  • 1
    • $A\times \mathrm{\varnothing }=\{(x,y):x\in A,y\in \mathrm{\varnothing }\}$
    • $A\times \mathrm{\varnothing }=\mathrm{\varnothing }$
  • 2
    • $\mathbb{Z}-\{2k+1\}$
      • Integers - Odd numbers = Even numbers
    • $\{2k:k\in \mathbb{Z}\}$
    • $\{x:x=2k,k\in \mathbb{Z}\}$
  • 3
    • $\{evens\}\times B\times \mathbb{Z}$
    • $\{(a,b,c):\{\begin{array}{l}a=2k\\ b=\frac{3}{2}l\\ c,k,l\in \mathbb{Z}\end{array}\}$

• sets

  • $A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$
    • $A\cup (B\cap C)$
    • If $x\in A$, then $x\in (A\cup B)$ and $x\in (A\cup C)$
    • $(A\cup B)\cap (A\cup C)$
    • We can see that $A\cup (B\cap C)\subseteq (A\cup B)\cap (A\cup C)$
    • Now in reverse
    • $(A\cup B)\cap (A\cup C)$
    • $x\in A\cup B$ and $x\in A\cup C$
    • $x\in A$ or it's in $x\in B\cap C$

    ---

    • We will proceed via double subset inclusion.
    • $\subseteq$
      • Let $x\in A\cup (B\cap C)$
      • $⟺x\in A$ or $x\in B\cap C$
      • $x\in A$ or $(x\in B\cap x\in C)$
        • $A\vee (B\wedge C)$
      • $(x\in A\cup x\in B)\cap (x\in A\cup x\in C)$
      • $(A\cup B)\cap (A\cup C)$
      • $Q.E.D$
      • The logic shows that these sets are subsets of each other. It shows the reverse with the $⟺$ symbol.
  • $(A\cap B{)}^c=A^c\cup B^c$
    • $(A\cap B{)}^c=A^c\cup B^c$
    • $x\notin A\cap x\notin B$, this is what the complement of a set is.
    • This means that $x\notin A\cup x\notin B$

    ---

    • Let $U$ be our universe of discourse.
    • $x\in (A\cup B{)}^c$
    • $⟹x\notin (A\cup B)$
    • $x\notin A$ or $x\notin B$
    • $x\in A^c$ or $x\in B^c$

• How to use if and only iff (iff)

• 5

  • $P=$ all bowls unbroken
  • $Q=$ there is an orange cat
  • $R=$ the soup is not spilled
  • $\mathrm{\neg }P⟹(Q\wedge R)$
  • If there is a broken bowl, then there is an orange cat and that soup is not spilled.
  • 8 row truth table
  • $\equiv$
    • $\mathrm{\neg }P⟹(Q\wedge R)$
    • $\mathrm{\neg }(\mathrm{\neg }P⟹(Q\wedge R))$
    • $\mathrm{\neg }(P\vee (Q\wedge R))$
    • $\mathrm{\neg }P\wedge (\mathrm{\neg }Q\vee \mathrm{\neg }R)$

    ---

    • $\mathrm{\neg }(Q\wedge R)⟹\mathrm{\neg }(\mathrm{\neg }P)$