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\usepackage[utf8]{inputenc}
\title{Math 220.201 Class 4 Questions}
\begin{document}
\maketitle
\begin{enumerate}
\item Show that $\sim (P \iff Q) \equiv (P \wedge \sim Q) \vee (Q \wedge \sim P)$.
\vskip 0.1in
{\bf Solution:} You can do this by a truth table in $P$ and $Q$. Or, you can use the rules of logic to prove it, as follows (the middle step is De Morgan's law):
$$
\begin{aligned}
\sim (P \iff Q) &\equiv \sim ((P\implies Q) \wedge (Q\implies P))\\
&\equiv \sim (P\implies Q) \vee \sim (Q\implies P)\\
&\equiv (P\wedge \sim Q) \vee (Q \wedge \sim P)
\end{aligned}
$$
\vskip 0.1in
\item Let $P(n)$ be the open sentence
$$P(n):\frac{5n-3}{6}\text{ is an integer}$$
over the domain $n \in \mathbb{Z}$.
\begin{enumerate}
\item Is the following true or false? $\exists n\in\mathbb{Z}, P(n)$. Explain.
\vskip 0.1in
{\bf Solution:} True, because $P(3)$ is true. (i.e. $\frac{5\cdot 3-3}{6}$ is an integer.)
\vskip 0.1in
\item Is the following true or false? $\forall n\in\mathbb{Z}, P(n)$. Explain.
\vskip 0.1in
{\bf Solution:} False, because $P(1)$ is false. (i.e. $\frac{5\cdot 1-3}{6}$ is not an integer.)
\vskip 0.1in
\end{enumerate}
\item Rewrite the following using quantifiers:
\begin{enumerate}
\item There is a real number such that $1/y=y+1$.
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{\bf Solution:} $\exists y\in\mathbb{R}, (1/y=y+1)$.
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\item For every natural number $z$, there is a natural number $w$ such that $z^2<w$.
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{\bf Solution:} $\forall z\in\mathbb{N}, (\exists w\in\mathbb{N}, (z^2<w))$.
\vskip 0.1in
\end{enumerate}
\item Let $A, B$ be subsets of a universal set $U$. Rewrite the following using quantifiers involving the elements. For example, $\forall x\in U, \ldots$
\begin{enumerate}
\item $A \subset B$
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{\bf Solution:} $(\forall x\in A, (x\in B)) \wedge (\exists x\in B, (x\notin A))$.
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\item $A \cap B=\emptyset$
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{\bf Solution:} $\forall x\in A, (x\notin B)$. Another way is $\forall x\in U, ((x\notin A) \vee (x\notin B))$.
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\item $A\not\subset B$
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{\bf Solution:} Just negate (a).
$$
\sim \left((\forall x\in A, (x\in B)) \wedge (\exists x\in B, (x\notin A))\right)$$
$$\equiv (\not\forall x\in A, (x\in B)) \vee (\not\exists x\in B, (x\notin A))
$$
$$\equiv (\exists x\in A, (x\notin B)) \vee (\forall x\in B, (x\in A))$$
\vskip 0.1in
\end{enumerate}
\item Is the following sentence true or false?
$$\forall \epsilon \in \mathbb{R}_{>0}, \exists \delta \in \mathbb{R}_{>0}, |(3+\delta)^2-3^2|<\epsilon$$
Write its negation.
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{\bf Solution:} The sentence is true. Here is an argument why. It is enough to show that for any $\epsilon \in \mathbb{R}_{>0}$, there exists $\delta\in \mathbb{R}_{>0}$ such that $6\delta+\delta^2<\epsilon$. If we pick $\delta$ such that $(7\delta<\epsilon) \wedge (7\delta^2<\epsilon)$, then $6\delta+\delta^2<\epsilon$, because $6\delta+\delta^2$ is between $7\delta$ and $7\delta^2$. $\delta$ can be chosen to be any real number less than $\text{min}(\epsilon/7, \sqrt{\epsilon/7})$. Letting $\delta=\frac{1}{2}\text{min}(\epsilon/7, \sqrt{\epsilon/7})$ suffices.
\vskip 0.1in
The negation is
$$\exists \epsilon \in \mathbb{R}_{>0}, \forall \delta \in \mathbb{R}_{>0}, |(3+\delta)^2-3^2|\ge\epsilon$$
\vskip 0.1in
\item For what values of $x$ is the following open sentence (in $x$) true?
$$\forall \epsilon \in \mathbb{R}_{>0}, \exists \delta \in \mathbb{R}_{>0}, |(x+\delta)^2-x^2|<\epsilon$$
How about this one?
$$\exists \delta \in \mathbb{R}_{>0},\forall \epsilon \in \mathbb{R}_{>0}, |(x+\delta)^2-x^2|<\epsilon$$
\vskip 0.1in
{\bf Solution:} The first sentence is true for all $x$. For any $x$, one can use an argument identical to that in the previous question to find that $\delta=|\frac{1}{2}\text{min}\left(\frac{\epsilon}{2x+1}, \sqrt{\frac{\epsilon}{2x+1}}\right)|$ satisfies the required condition.
\vskip 0.1in
The second sentence is false because its negation
$$\forall \delta \in \mathbb{R}_{>0},\exists \epsilon \in \mathbb{R}_{>0}, |(x+\delta)^2-x^2|\ge\epsilon$$
is true. For any $\delta$, taking $\epsilon=\frac{1}{2}|\delta^2+2x\delta|$ satisfies the required condition.
\end{enumerate}
\end{document}