MAT102 Tutorial W2B

• 102_F2025_Tutorial1.pdf
• 1
  • $(P⟺Q)\wedge (\mathrm{\neg }P⟺Q)$
  • $\begin{array}{ccccc}P& Q& (P⟺Q)& (\mathrm{\neg }P⟺Q)& (P⟺Q)\wedge (\mathrm{\neg }P⟺Q)\\ T& T& T& F& F\\ T& F& F& T& F\\ F& T& F& T& F\\ F& F& T& F& F\end{array}$
  • This is a contradiction
• 2
  • $(P\vee Q)⟺(Q\vee R)⟺(R⟺P)$
  • $\begin{array}{ccccccc}P& Q& R& (P\vee Q)& (Q\vee R)& \begin{array}{c}(P\vee Q)⟺\\ (Q\vee R)\end{array}& \begin{array}{c}(P\vee Q)⟺\\ (Q\vee R)⟺\\ (R⟺P)\end{array}\\ T& T& T& T& T& T& T\\ T& T& F& T& T& T& F\\ T& F& T& T& T& T& T\\ T& F& F& T& F& F& F\\ F& T& T& T& T& T& F\\ F& T& F& T& T& T& T\\ F& F& T& T& T& T& F\\ F& F& F& F& F& T& F\end{array}$
  • Neither
• 3
  • Is a tautology
• Activity 2
  • $A=\{6k+2:k\in \mathbb{Z}\}$
  • $B=\{r\in \mathbb{R}:9\mid r\}$
  • Is $2\mathbb{Z}\times \mathbb{Q}\subseteq A\times B$
  • No because notice that in $\mathbb{Q}$ we have a lot of numbers below $9$, $B$ has numbers divisible by $9$.
  • But $\mathbb{Q}$ has a lot more numbers that aren't divisible by $9$.
  • So we extra numbers in $\mathbb{Q}\notin B$. So it's not a subset.
  • Notice that $\frac{3}{2}\in \mathbb{Q}$ but $\frac{3}{2}\notin B$ as $9\nmid \frac{3}{2}$
  • Can we show that $A\times B\subseteq 2\mathbb{Z}\times \mathbb{Q}$
  • $A$ contains evens only
  • So $A\subseteq 2\mathbb{Z}$
  • $B$ has numbers divisble by $9$, and that's in $\mathbb{Q}$
  • Gotta do a proof
    • Let $(a,b)\in A\times B$
    • So $a=6k+2,k\in \mathbb{Z}$
    • So $a=3k+1$, this shows that it is even
• Activity 3
  • 1
    • $(A\cap B)\cup (A\cap B^c)=A$
    • Let $x\in (A\cap B)\cup (A\cap B^c)$
    • So $x\in (A\cap B)$ or $x\in (A\cap B^c)$
    • So $x\in A$ and $x\in B$ and $x\in A$ but $x\in B^c$
  • 4
    • Let $(x,y)\in (A\setminus B)\times C$
    • So $x\in A,x\notin B$ and $y\in C$
    • So $(x,y)\in A\times C$
    • And $(x,y)\notin B\times C$
    • So $(x,y)\in (A\times C)\setminus (B\times C)$
    • Other way