Math I

Sets and Propositions: Practice Problems and Methods

Use Venn diagrams and number lines for sets, and think of conditions as sets: deciding necessary and sufficient conditions then becomes a question of inclusion.

Basic, Standard, Advanced: Grade 10 · Term 1

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Set operations and counting

$A \cap B$ (intersection) means "in $A$ and in $B$ ", $A \cup B$ (union) means "in $A$ or in $B$ ", and $\overline{A}$ (complement) is everything in the universal set not in $A$ .

De Morgan's laws and counting

\overline{A \cup B} = \overline{A} \cap \overline{B},\qquad \overline{A \cap B} = \overline{A} \cup \overline{B}

n(A \cup B) = n(A) + n(B) - n(A \cap B)

Example: among the integers 1 to 100, $50 + 33 - 16 = 67$ are divisible by 2 or 3.

Necessary and sufficient conditions

If " $p \Rightarrow q$ " is true, then $p$ is a sufficient condition for $q$ , and $q$ is a necessary condition for $p$ . If both directions hold, the condition is necessary and sufficient.

Think in sets

If $P$ and $Q$ are the sets satisfying $p$ and $q$ , then " $p \Rightarrow q$ is true" $\iff P \subset Q$ . For $p$ : $x > 3$ and $q$ : $x > 1$ , $P \subset Q$ , so $p$ is sufficient for $q$ .

To show an implication is false, one counterexample is enough: $x^2 = 4 \Rightarrow x = 2$ fails for $x = -2$ .

Negation, converse, inverse and contrapositive

The negation of " $p$ and $q$ " is "not $p$ or not $q$ ", and the negation of "for every $x$ , $p$ " is "for some $x$ , not $p$ ".

For " $p \Rightarrow q$ ", the converse is " $q \Rightarrow p$ ", the inverse is "not $p \Rightarrow$ not $q$ " and the contrapositive is "not $q \Rightarrow$ not $p$ ". A statement and its contrapositive are always both true or both false, so a hard statement can be proved through its contrapositive. Proving that $\sqrt{2}$ is irrational by assuming the opposite and reaching a contradiction is proof by contradiction.

Q1BasicGr 10 · Term 1Sets and propositions

Let $U = \{1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8\}$ be the universal set, with subsets $A = \{1,\ 2,\ 7,\ 8\}$ and $B = \{3,\ 7,\ 8\}$ . List all the elements of each set.

(1) $A \cap B$
(2) $A \cup B$
(3) $\overline{A} \cup \overline{B}$

Answer

(1) $\{7,\ 8\}$
(2) $\{1,\ 2,\ 3,\ 7,\ 8\}$
(3) $\{1,\ 2,\ 3,\ 4,\ 5,\ 6\}$

Solution

(1)
$A \cap B = \{7,\ 8\}$
(2)
$A \cup B = \{1,\ 2,\ 3,\ 7,\ 8\}$
(3) $\overline{A} \cup \overline{B} = \overline{A \cap B}$
$\overline{A} \cup \overline{B} = \{1,\ 2,\ 3,\ 4,\ 5,\ 6\}$

Q2StandardGr 10 · Term 1Sets and propositions

How many integers from 1 to 250 are divisible by neither 3 nor 8?

Answer

146

Solution

There are 83 multiples of 3, 31 multiples of 8, and 10 common multiples (multiples of 24).
104 numbers are divisible by 3 or 8, so
$250 - 104 = 146$

Q3AdvancedGr 10 · Term 1Sets and propositions

Prove by contradiction that $\sqrt{2}$ is irrational.

Answer

Suppose $\sqrt{2}$ is rational. Then $\sqrt{2} = \frac{m}{n}$ for coprime natural numbers $m$ and $n$ . Squaring gives $m^2 = 2n^2$ , so $m^2$ is even and hence $m$ is even. Writing $m = 2k$ gives $4k^2 = 2n^2$ , so $n^2 = 2k^2$ and $n$ is even as well. This contradicts the assumption that $m$ and $n$ are coprime. Therefore $\sqrt{2}$ is irrational.

Solution

Suppose $\sqrt{2}$ is rational. Then $\sqrt{2} = \frac{m}{n}$ for coprime natural numbers $m$ and $n$ . Squaring gives $m^2 = 2n^2$ , so $m^2$ is even and hence $m$ is even. Writing $m = 2k$ gives $4k^2 = 2n^2$ , so $n^2 = 2k^2$ and $n$ is even as well. This contradicts the assumption that $m$ and $n$ are coprime. Therefore $\sqrt{2}$ is irrational.

Q1BasicGr 10 · Term 1Sets and propositions

Let $U = \{1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8\}$ be the universal set, with subsets $A = \{1,\ 2,\ 4,\ 5,\ 8\}$ and $B = \{2,\ 3,\ 5,\ 7\}$ . List all the elements of each set.

(1) $A \cap B$
(2) $A \cup B$
(3) $\overline{A} \cap B$

Q2StandardGr 10 · Term 1Sets and propositions

Let $x$ and $y$ be real numbers. Choose the correct negation of " $-4 \leqq x < 0$ ".

A $x \leqq -4$ or $x > 0$
B $x \leqq -4$ or $x \geqq 0$
C $x < -4$ and $x \geqq 0$
D $x < -4$ or $x \geqq 0$

Q3AdvancedGr 10 · Term 1Sets and propositions

Let $a$ be an integer, and let $A = \{2,\ a - 3,\ a^2 - 2a + 1\}$ and $B = \{8,\ a - 1,\ a^2 + 2a - 1\}$ . Suppose $A \cap B = \{0,\ 2\}$ .

(1)Find the value of $a$ .
(2)List all the elements of $A \cup B$ .

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