\newlec

\textbf{Quantified Statements}

\begin{enumerate}
\item For every natural number $n$, $2n^{2} + 11n + 15$ is composite.
\item There is an integer $k$ such that $6 = 3k$.
\end{enumerate}
Symbolically, we write
\begin{enumerate}
\item $\forall n \in \mathbb{N}$, $2n^{2} + 11n + 15$ is composite.
\item $\exists k \in \mathbb{Z}$ such that $6 = 3k$.
\end{enumerate}
We call $\forall$ and $\exists$ quantifiers, $n$ and $k$ variables, $\mathbb{N}$ and $\mathbb{Z}$ domains and the rest are called an open sentence (usually involving the variable(s)).
\vskip1em
\note $\forall x \in S\, P(x)$ means for all $x$ in $S$, statment $P(x)$ holds. This is equivalent to $x \in S \Rightarrow P(x)$. 
\vskip1em
\pf (of number 1 above) Let $n$ be an arbitrary natural number. Then factoring gives $2n^{2} + 11n + 15 = (2n+5)(n + 3)$. Since $2n+5 > 1$ and $n+3 > 1$, we have $2n^{2} + 11n + 15$ is composite.
\vskip1em
\pf (of number 2 above) Since $3 \cdot 2 = 6$, we see that $k=2$ satisfies the given statement.
\vskip1em
\ex $S \subseteq T \equiv \forall x \in S\, x \in T$

\ic{This is the 7 minute mark}
\newpage
\hdc
\ex Prove that there is an $x \in \mathbb{R}$ such that $\frac{x^{2} + 3x - 3}{2x + 3} = 1$.
\vskip3em
\pf When $x=2$, note that $\frac{2^{2} + 3(2) -3}{2(2)+3} = \frac{7}{7} = 1$. \qed

\vskip1em
\note: The discovery of this proof is perhaps what is more interesting:
$$
\frac{x^{2} + 3x - 3}{2x + 3} = 1 \quad \Leftrightarrow \quad x^{2} + 3x - 3= 2x + 3 \quad \Leftrightarrow \quad x^{2} +x - 6 = 0 
$$
and the last equation factors as $(x-2)(x+3)=0$ and hence $x=2$.

\ic{This is the 17 minute mark}

\newpage
\note: Vacuously true statements $\forall x \in \emptyset, P(x)$. Since there is no element in the empty set, we define this statement to always be true as a matter of convention.

\vskip1em
\ex Let $a,b,c \in \mathbb{Z}$. If $\forall x \in \mathbb{Z}$, $a \mid (bx + c)$ then $a \mid (b+c)$.
\vskip1em
\pf Assume $\forall x \in \mathbb{Z}$, $a \mid (bx + c)$. Then, for example, when $x=1$, we see that $ a \mid (b(1) + c)$. Thus $a \mid (b+c)$.
\ic{Note: If you're running short on time, this next example can be omitted}
\ex $\exists m \in \mathbb{Z}$ such that $\frac{m-7}{2m+4} = 5$.
\vskip1em
\pf When $m=3$, note that $\frac{m-7}{2m+4} = \frac{-3-7}{2(-3)+4} = \frac{-10}{-2} = 5$
\ic{This should be the 26-30 minute mark}
\newpage
\hdc 

\ex Show that for each $x \in \mathbb{R}$, we have that $x^{2} + 4x + 7 > 0$.
\vskip3em
\ic{For the next two pages, you should give students say 5 minutes each (maybe more for the second handout) and then take them up as a class for 5 minutes each}
\pf Let $x \in \mathbb{R}$ be arbitrary. Then 
\begin{align*}
x^{2} + 4x + 7 &= x^{2} + 4x + 4 - 4 + 7\\
&= (x+2)^{2} + 3\\
&> 0
\end{align*}

\newpage
\hdc
Sometimes $\forall$ and $\exists$ are hidden! If you encounter a statement with quantifiers, take a moment to make sure you understand what the question is saying/asking.
\vskip0.1em
Examples:
\begin{enumerate}
\item $2n^{2} + 11n + 15$ is never prime when $n$ is a natural number.
\item If $n$ is a natural number, then $2n^{2} + 11n + 15$ is composite.
\item $\frac{m-7}{2m+4} = 5$ for some integer $m$.
\item $\frac{m-7}{2m+4} = 5$ has an integer solution.
\end{enumerate}

\vskip3em
\soln 

\begin{enumerate}
\item $\forall n \in \mathbb{N}, 2n^{2} + 11n + 15$ is not prime.
\item $\forall n \in \mathbb{N}, 2n^{2} + 11n + 15$ is composite.
\item $\exists m \in \mathbb{Z}, \frac{m-7}{2m+4} = 5$.
\item $\exists m \in \mathbb{Z}, \frac{m-7}{2m+4} = 5$.
\end{enumerate}

\ic{This should be about the 46 minute mark}

\newpage
\note: Domain is important!
\vskip1em
Let $P(x)$ be the statement $x^{2} = 2$ and let $S = \{\sqrt{2}, -\sqrt{2}\}$. Which of the following are true?
\begin{enumerate}
\item $\exists x \in \mathbb{Z}, P(x)$ 
\item $\forall x \in \mathbb{Z}, P(x)$ 
\item $\exists x \in \mathbb{R}, P(x)$ 
\item $\forall x \in \mathbb{R}, P(x)$ 
\item $\exists x \in S, P(x)$ 
\item $\forall x \in S, P(x)$ 
\end{enumerate}
\vskip3em
\soln
\begin{enumerate}
\item False
\item False
\item True
\item False
\item True
\item True
\end{enumerate}

\ic{This is the end of the lecture.}



