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  • MAT102 Final Exam Prep

MAT102 Worksheet 3 Review

Q1

Suppose $P,Q,R$ are predicates, you want to prove that $P⟹(Q\vee R)$

A

Construct the truth table for $P⟹(Q\vee R)$ and $(P\wedge \mathrm{\neg }Q)⟹R$.

B

• Prove that if $m,n\in \mathbb{Z}$ and $m^3+n^3$ is odd, then either $m$ is odd or $n$ is odd.
• If $m$ is odd, then we are done. So let's assume that $m$ is not odd.
• This must mean that $n$ is odd.
• If $m^3+n^3=2\ell +1$ for some $\ell \in \mathbb{Z}$.
• And $n=2k+1$ for some $k\in \mathbb{Z}$.
• If $m$ is not odd, then it's even, so $m=2a$ for some $a\in \mathbb{Z}$
• $(2a{)}^3+(2k+1{)}^3=2\ell +1$
• $(2a)(2a)(2a)+(2k+1)(2k+1)(2k+1)=2\ell +1$
• $(4a^2)(2a)+(2k+1)(2k+1)(2k+1)=2\ell +1$
• $8a^3+(4k^2+4k+1)(2k+1)=2\ell +1$
• $8a^3+(8k^3+4k^2+8k^2+6k+1)=2\ell +1$
• $8a^3+(8k^3+12k^2+6k+1)=2\ell +1$
• $8a^3+8k^3+12k^2+6k+1=2\ell +1$
• $2(4a^3+4k^3+6k^2+3k)+1=2\ell +1$
• Notice that we are able to express $m^3+n^3$ as an odd integer after expressing both $m,n$ as their specific parity representations.
• So therefore, we have shown that $n$ is odd if $m$ is not odd.

Q2

• A set $S\subseteq \mathbb{R}$ is open if for every $x\in S$ there exists $r>0$ such that the interval $(x-r,x+r)\subseteq S$.

A

• Show that $(0,1)$ is open.
• If $x\in (0,1)$
• For $x$ we can always create an interval $O=(x-r,x+r):O\subseteq S$
• Where $r=min(1-x,x)$
• Tests:
• $\{\begin{array}{ll}\text{if }x\in (0,\frac{1}{2}):& (x-x,x+x)\\ \text{if }x\in (\frac{1}{2},1):& (x-(1-x),x+(1-x))\\ & (x-1+x,x+1-x)\\ & (2x-1,1)\\ \\ \begin{array}{cc}x=& 0.3\\ r=& 0.3\end{array}⟹& (0.3-0.3,0.3+0.3)\\ \\ & (0,0.6)\\ \\ \begin{array}{cc}x& =0.7\\ r& =0.3\end{array}⟹& (0.7-0.7,0.7+0.3)\\ & ⟹(0,1)\end{array}$
• As you can see $(0,1)\subseteq (0,1)$.

B

• $\mathrm{\forall }x\in S,\mathrm{\exists }r>0,(x-r,x+r)\subseteq S$
• What's equivalent?
  • $\mathrm{\forall }x\in S,\mathrm{\exists }r>0,(x-r,x+r)\subset S$
    • This is not equivalent, because this is the $\subset$ sign, not $\subseteq$. So if we have an open interval that's equal to the original interval, then it would not count that. In our above example, $x=0.5,r=0.5$ would fail, because that results in $(x-r,x+r)$ of $(0,1)$ which is $=(0,1)$. But it is not a strict subset.
  • $\mathrm{\exists }r>0,\mathrm{\forall }x\in S,(x-r,x+r)\subseteq S$
    • This is not equivalent. Because there isn't an $r$ value where for all $x$ there can be an open interval.
  • $\mathrm{\forall }x\in S,\mathrm{\exists }r>0,[x-r,x+r]\subseteq S$
    • This isn't equivalent, because it's including the end point.
  • $\mathrm{\exists }x\in S,\mathrm{\exists }r>0,(x-r,x+r)\subseteq S$
    • This is true, but not equivalent.