Set roster notation
List the elements of the set $\{x\in \mathbb{Z}: 1 < |x| < 5\}$.
Solution. $\{-4, -3, -2, 2, 3, 4\}$
List the elements of the set $\{x\in \mathbb{Q}: x^5+x^3−2=0\}$.
Solution. By the rational roots test, a rational root $\frac{p}{q}$ of $f(x)=x^5+x^3-2$ must have its numerator $p$ divide the constant term $-2$ and its denominator $q$ divide $1$. So the possibilities for $p$ are $\pm 1$ or $\pm 2$; the possibilities for $q$ are $\pm 1$. Overall, $\frac{p}{q}$ could be $\pm 2$ or $\pm 1$.
We check $f(1)=1^5+1^3-2=0$ so $1$ is a root. Also, $f(-1)=-4$ and $|f(\pm2)|$ is to large to equal $0$. So $\{x\in \mathbb{Q}: x^5+x^3−2=0\}=\{1\}$.
Operations on finite sets
List the elements of the set $(A\cup B)\cup (B\cap C)$ where $A=\{1, 2, 3, a, b, c, d\}$, $B=\{a, d, e, 1, 3, 5\}$, and $C=\{1, 2, 4, b, b, a\}$.
Solution. $$ \begin{align*} A\cup B &= \{1, 2, 3, 5, a, b, c, d, e\}\\ B\cap C &= \{a, 1\}\\ \Longrightarrow\quad(A\cup B)\cup (B\cap C) &= \{1, 2, 3, 5, a, b, c, d, e\} \end{align*} $$
List the elements of the set $(A\cup B)- (B\cap C)$ where $A=\{1, 2, 3, a, b, c, d\}$, $B=\{a, d, e, 1, 3, 5\}$, and $C=\{1, 2, 4, b, b, a\}$.
Solution. $$ \begin{align*} A\cup B &= \{1, 2, 3, 5, a, b, c, d, e\}\\ B\cap C &= \{a, 1\}\\ \Longrightarrow\quad(A\cup B)- (B\cap C) &= \{2, 3, 5, b, c, d, e\} \end{align*} $$
Let $U=\{a, b, c, d, e, f, g, h\}$ be the universe. List the elements of the set $$(\sim\{a, c, e, g\})-\{a,b,c,d\}$$
Solution. Recall that $\sim A$ is the complement of $A$. $$ \begin{align*} \sim\{a, c, e, g\}&=\{b,d,f,h\}\\ \Longrightarrow\quad(\sim\{a, c, e, g\})-\{a,b,c,d\} &= \{f, h\} \end{align*} $$
Let $U=\{1, 2, 3,\dots,10\}$ be the universe. List the elements of the set $$\sim P-\sim O$$ where $P=\{2, 3, 5, 7\}$ and $O=\{1, 3, 5, 7, 9\}$
Solution. $$ \begin{align*} \sim P - \sim O&=\,\sim\{2, 3, 5, 7\}-\sim \{1, 3, 5, 7, 9\}\\ &= \{1,4,6,8,9,10\} - \{2,4,6,8,10\}\\ &= \{1, 9\} \end{align*} $$
Intervals
Let $A=(0,3)$ and $B=[1,2]$.
Graph the set $A-(\sim B)$ on the number line.
Solution 1.
Since $B=[1,2]$, its complement $\sim B=\;\sim[1,2]$ looks like:
Solution 2. In general, $S-T=S\cap (\sim T)$ for any two sets $S$ and $T$. So here we have $A-(\sim T)=A\cap (\sim\sim B)=A\cap B=(0,3)\cap [1,2]=[1,2]$.
Cardinality
Find the cardinality of the set $\{7n:n\in \mathbb{Z}\}\cap [0, 91]$.
Answer:Solution.
Elements of this set are bound by $0\le 7n\le 91$. Dividing by $7$ gives $$ 0\le n\le 13 $$ So the cardinality of this set is $$ \text{max integer}-\text{min integer}+1 = 13-0+1=\boxed{14} $$
Find the cardinality of the set $\{n\in \mathbb{Z}:n\equiv 2\pmod{13}\}\cap [0, 100]$.
Answer:Solution.
Elements of this set leave remainder $2$ when divided by $13$ so have the form $n=13q+2$. The bound additionally reads as $0\le 13q+2\le 100$. Subtracting by $2$ gives $$ -2\le 13q\le 98 $$ Then dividing by $13$ gives $$ -\frac{2}{13}\le q\le \frac{98}{13}=7+\frac{7}{13} $$ So $q$ can be the integers $0$, $1$, $2$, ..., $7$. The size of this set is therefore $8$.
Find the cardinality of the set $\{1_2, 11_2, 111_2, 1111_2\}\cup \{1_3, 11_3, 111_3, 1111_3\}$. As a reminder $111_2$ is a number written in base 2, and $1111_3$ is a number written in base 3.
Answer:Solution.
We can write elements in both sets in base 10:
$$ \begin{align*} \{1_2, 10_2, 100_2, 1000_2\} & = \{1, 2, 4, 8\}\\ \{1_3, 11_3, 111_3, 1111_3\} & = \{1, 4, 13, 40\} \end{align*} $$So their union is $\{1, 2, 4, 8, 13, 40\}$ which has $6$ elements.
Find the cardinality of the set $A\cap B\cap C$ where $A = \{7n:n\in \mathbb{Z}\}$, $B=[-50, 700]$, and $C = \{5n:n\in \mathbb{Z}\}$.
Answer:Solution.
To find the elements simultaneously in all three sets, we find elements common to both $A$ and $C$ first. Since $A$ is the set of multiples of $7$ and $C$ is the set of multiples of $5$, their intersection $A\cap C$ is the set of the multiples of $7\cdot 5 = 35$. The intersection $A\cap B\cap C=(A\cap C)\cap [-50, 700]$ now consists of the only negative multiple $-35$ as well as $700/35+1=21$ nonnegative multiples of $35$, so $|A\cap B\cap C|=21+1=22$.