MATH 308 — Exam I Review

Exam I Review — Sets, Logic, Quantifiers, and Proof
MATH 308  Bridge to Advanced Mathematics  Chartrand Ch 1–4 (Lectures 1–7)

How to use this session. This is a consolidation, not new material. Everything here has appeared in Lectures 1–7. The exam asks you to do five things: name an object correctly, decide whether a statement is true, negate a statement, choose a proof technique, and carry it out. Each section below is organized that way — what you must know cold, the move that uses it, and the error that costs the most points.

What is covered.

Lectures	Chartrand	Topic
1–2	Ch 1	Sets: notation, subsets, operations, indexed families, partitions, products
3	Ch 2	Logic: connectives, implication, converse/inverse/contrapositive, De Morgan
4	Ch 2	Quantifiers: $\forall$∀, $\exists$∃, negation, mixed order
5–6	Ch 3	Proof: direct, contrapositive, cases, biconditional
7	Ch 4	Divisibility and congruence modulo $n$n

Sets (Lectures 1–2)

Know cold.

• Roster $\{1,2,3\}${1,2,3} versus set-builder $\left\{\, n \in \mathbb{N} \;:\; n \le 3 \,\right\}${n∈N:n≤3}.

• $A \subseteq B$A⊆B means $\ \forall x,\ (x \in A \Rightarrow x \in B)$ ∀x, (x∈A⇒x∈B). Every subset claim is a “for all” in disguise, which is why a proof of $A \subseteq B$A⊆B always opens “let $x \in A$x∈A be arbitrary.”

• $\varnothing \subseteq A$∅⊆A for every $A$A (vacuously). $\mathcal{P}(A) = \left\{\, X \;:\; X \subseteq A \,\right\}$P(A)={X:X⊆A}, and $\left\lvert \mathcal{P}(A) \right\rvert = 2^{n}$∣P(A)∣=2n when $\left\lvert A \right\rvert = n$∣A∣=n.

• $A \cup B$A∪B, $A \cap B$A∩B, $A - B$A−B, and the complement $A^{c}$Ac (never $\overline{A}$A — the overline is reserved for closure).

• Indexed families: $x \in \bigcup_{\alpha} A_\alpha$x∈⋃α​Aα​ means $x$x is in at least one; $x \in \bigcap_{\alpha} A_\alpha$x∈⋂α​Aα​ means $x$x is in every one.

• A partition of $S$S: nonempty cells, which cover $S$S, and are pairwise disjoint. All three conditions.

• $A \times B = \left\{\, (a,b) \;:\; a \in A,\ b \in B \,\right\}$A×B={(a,b):a∈A, b∈B}, and $\left\lvert A \times B \right\rvert = \left\lvert A \right\rvert\cdot\left\lvert B \right\rvert$∣A×B∣=∣A∣⋅∣B∣.

Trap. $\in$∈ versus $\subseteq$⊆. This is the single most common error in Chapter 1. $\{1\} \in \{1,2\}${1}∈{1,2} is false — the elements of $\{1,2\}${1,2} are the numbers $1$1 and $2$2, not the set $\{1\}${1}. But $\{1\} \subseteq \{1,2\}${1}⊆{1,2} is true. Read the symbol out loud: “is an element of” versus “is a subset of.” They are different relations and the exam will test whether you know which is which.

Trap. $\varnothing$∅ and $\{\varnothing\}${∅} are different. $\varnothing$∅ has no elements; $\{\varnothing\}${∅} has exactly one. So $\varnothing \in \{\varnothing\}$∅∈{∅} is true and $\varnothing \subseteq \{\varnothing\}$∅⊆{∅} is also true — for different reasons.

Trap. Checking a partition, students check “covers” and stop. Disjointness fails more often. If one element sits in two different cells, it is not a partition, no matter how well the cells cover.

Example 1 (Double containment — the method for proving two sets equal). To show $X = Y$X=Y you show $X \subseteq Y$X⊆Y and $Y \subseteq X$Y⊆X. Take $(A \cup B)^{c} = A^{c} \cap B^{c}$(A∪B)c=Ac∩Bc, promised in Lecture 1.

($\subseteq$⊆) Let $x \in (A\cup B)^{c}$x∈(A∪B)c. Then $x \notin A \cup B$x∈/A∪B, so $x$x is in neither $A$A nor $B$B. Hence $x \in A^{c}$x∈Ac and $x \in B^{c}$x∈Bc, i.e. $x \in A^{c} \cap B^{c}$x∈Ac∩Bc.

($\supseteq$⊇) Let $x \in A^{c} \cap B^{c}$x∈Ac∩Bc. Then $x \notin A$x∈/A and $x \notin B$x∈/B, so $x$x is in no part of $A \cup B$A∪B, i.e. $x \in (A \cup B)^{c}$x∈(A∪B)c.

Both containments hold, so the sets are equal. Notice the engine: “not (in $A$A or in $B$B)” became “not in $A$A and not in $B$B” — De Morgan for logic (§2) driving De Morgan for sets. This is why Chapter 1 waits for Chapter 2.

For a nested family $A_1 \subseteq A_2 \subseteq \cdots$A1​⊆A2​⊆⋯ the pattern is always the same: the union is everything eventually reached, and the intersection is the first member.

Logic (Lecture 3)

Know cold. The implication is the whole chapter:

$\begin{array}{cc|c|c|c} P & Q & P \Rightarrow Q & (\sim\! P) \vee Q & \sim\!(P \Rightarrow Q)\\\hline T & T & T & T & F\\ T & F & \mathbf{F} & \mathbf{F} & \mathbf{T}\\ F & T & T & T & F\\ F & F & T & T & F \end{array}$PTTFF​QTFTF​P⇒QTFTT​(∼P)∨QTFTT​∼(P⇒Q)FTFF​​

$P \Rightarrow Q$P⇒Q is false in exactly one row: $P$P true, $Q$Q false. Everything else follows from that one line.

• $P \Rightarrow Q \ \equiv\ (\sim\! P) \vee Q$P⇒Q ≡ (∼P)∨Q and $\sim\!(P \Rightarrow Q) \ \equiv\ P \wedge (\sim\! Q)$∼(P⇒Q) ≡ P∧(∼Q).

• Contrapositive $(\sim\! Q) \Rightarrow (\sim\! P)$(∼Q)⇒(∼P) is equivalent to $P \Rightarrow Q$P⇒Q.

• Converse $Q \Rightarrow P$Q⇒P and inverse $(\sim\! P) \Rightarrow (\sim\! Q)$(∼P)⇒(∼Q) are not — though they are equivalent to each other.

• De Morgan: $\sim\!(P \wedge Q) \equiv (\sim\! P) \vee (\sim\! Q)$∼(P∧Q)≡(∼P)∨(∼Q) and $\sim\!(P \vee Q) \equiv (\sim\! P) \wedge (\sim\! Q)$∼(P∨Q)≡(∼P)∧(∼Q).

Trap. The negation of an implication is not an implication. $\sim\!(P \Rightarrow Q)$∼(P⇒Q) is $P \wedge (\sim\! Q)$P∧(∼Q) — a conjunction. “It is false that if it rains the game is cancelled” means it rained and the game went ahead, not “if it rains the game happens.”

Trap. Negating a compound statement without flipping the connective. The negation of “$3 \mid n$3∣n and $n$n is odd” is “$3 \nmid n$3∤n or $n$n is even.” The “and” must become “or.” A negation that keeps the connective is simply wrong.

Trap. Confusing converse with contrapositive. “If $n$n is a multiple of $6$6 then $n$n is even” is true; its converse (“every even number is a multiple of $6$6”) is false, witness $n = 2$n=2. Its contrapositive (“if $n$n is odd then $n$n is not a multiple of $6$6”) is true, automatically, because it says the same thing.

Trap. “Or” in mathematics is inclusive. “$a$a is even or $b$b is even” is satisfied when both are.

Quantifiers (Lecture 4)

Know cold.

Statement	To prove it	To disprove it
$\forall x \in S,\ P(x)$∀x∈S, P(x)	argue for an arbitrary $x$x	exhibit one counterexample
$\exists x \in S,\ P(x)$∃x∈S, P(x)	exhibit one witness	argue no $x$x can work

The negation rules — each quantifier flips, and the predicate is negated:

$\sim\!\bigl(\forall x,\ P(x)\bigr) \ \equiv\ \exists x,\ \sim\! P(x), \qquad \sim\!\bigl(\exists x,\ P(x)\bigr) \ \equiv\ \forall x,\ \sim\! P(x).$∼(∀x, P(x)) ≡ ∃x, ∼P(x),∼(∃x, P(x)) ≡ ∀x, ∼P(x).

Trap. Order of mixed quantifiers. In $\forall x\, \exists y$∀x∃y, the $y$y is chosen after $x$x and may depend on it. In $\exists y\, \forall x$∃y∀x, one $y$y must work for every $x$x. Over $\mathbb{R}$R: $\forall x\, \exists y\,(x+y=0)$∀x∃y(x+y=0) is true (take $y=-x$y=−x), while $\exists y\, \forall x\,(x+y=0)$∃y∀x(x+y=0) is false. Same symbols, opposite answers.

Trap. Forgetting the domain. “$x^{2} \ge x$x2≥x” is true for all $x \in \mathbb{Z}$x∈Z but false for all $x \in \mathbb{R}$x∈R — take $x = \tfrac 12$x=21​. Always read the domain before deciding.

Trap. Negating “both.” The negation of “both $a$a and $b$b are odd” is “at least one of them is even” — not “neither is odd,” and not “both are even.” This is De Morgan with quantifier flavour, and it is exactly what you need to set up a contrapositive in §4.

Proof technique (Lectures 5–6)

Know cold. $n$n is even if $n = 2k$n=2k for some $k \in \mathbb{Z}$k∈Z; odd if $n = 2k+1$n=2k+1 for some $k \in \mathbb{Z}$k∈Z. Every proof in this chapter starts by writing that down.

How each proof opens and closes. Choosing the technique is half the exam.

Technique	Open with	Use when
Direct	“Assume $P$P.” $\ \to$ → reason forward $\to$→ “therefore $Q$Q.”	$P$P gives you something concrete to compute with.
Contrapositive	“We prove the contrapositive: if $\sim\! Q$∼Q then $\sim\! P$∼P.”	$\sim\! Q$∼Q is easy to write down but $P$P is not — typically when $P$P is about $n^2$n2 and $Q$Q is about $n$n.
Cases	“Every integer is even or odd, so there are two cases.”	the hypothesis splits naturally, or you need parity of an unknown.
Biconditional	“($\Rightarrow$⇒) …($\Leftarrow$⇐) …”	the statement says “if and only if.” Two proofs, always.

Trap. Proving the converse by accident. If asked to show $P \Rightarrow Q$P⇒Q and you open with “assume $Q$Q,” you have proved something else. Check which letter you assumed.

Trap. Reusing a variable. If $a = 2k$a=2k and $b$b is also even, then $b = 2j$b=2j for a new letter $j$j — not $b = 2k$b=2k. Writing $b=2k$b=2k secretly assumes $a = b$a=b, and it invalidates the proof even when the final line looks right.

Trap. Non-exhaustive cases. State why your cases cover everything (“every integer is even or odd”). A proof by cases that omits a case proves nothing.

Trap. Half a biconditional. “If and only if” needs both directions, each with its own opening and closing. Proving one direction twice, in different words, is the most common way to lose half the credit.

Example 2 (Choosing the technique). Consider: if $n^{2}$n2 is even then $n$n is even. Direct is hopeless — from $n^2 = 2k$n2=2k you cannot extract $n$n. The contrapositive is one line: if $n$n is odd, $n = 2k+1$n=2k+1, then $n^{2} = 4k^{2}+4k+1 = 2(2k^{2}+2k)+1$n2=4k2+4k+1=2(2k2+2k)+1 is odd. The shape of the hypothesis tells you which technique to reach for.

Divisibility and congruence (Lecture 7)

Know cold.

• $a \mid b$a∣b means there exists $r \in \mathbb{Z}$r∈Z with $b = ar$b=ar. It is a statement, true or false — not a number.

• Transitivity: $a \mid b$a∣b and $b \mid c$b∣c $\Rightarrow$⇒ $a \mid c$a∣c.

• Linear combinations: $a \mid b$a∣b and $a \mid c$a∣c $\Rightarrow$⇒ $a \mid (bx + cy)$a∣(bx+cy) for all $x,y \in \mathbb{Z}$x,y∈Z.

• $a \equiv b \pmod{n}$a≡b(modn) means $n \mid (a-b)$n∣(a−b).

• Congruence is preserved by $+$+ and $\times$×: if $a \equiv b$a≡b and $c \equiv d \pmod n$c≡d(modn), then $a+c \equiv b+d$a+c≡b+d and $ac \equiv bd \pmod n$ac≡bd(modn). Hence congruent bases may be raised to the same power.

Trap. $a \mid b$a∣b versus $a / b$a/b. The first is a statement (“$a$a divides $b$b”); the second is a number. Writing “$a \mid b = 3$a∣b=3” is not a sentence. And the definition runs $b = ar$b=ar — students reverse it to $a = br$a=br under pressure.

Trap. Every divisibility proof opens by unpacking the definition. If you do not write “$b = ar$b=ar for some $r \in \mathbb{Z}$r∈Z” in the first line, there is nothing to compute with. The whole proof is: unpack, substitute, factor out $a$a, observe the cofactor is an integer.

Trap. Reducing to a remainder. $-1 \equiv 4 \pmod 5$−1≡4(mod5). An answer of $-1$−1 to “what is the remainder” is not finished — a remainder lives in $\{0, 1, \dots, n-1\}${0,1,…,n−1}.

Trap. $a \equiv b \pmod n$a≡b(modn) and $a \equiv b \pmod m$a≡b(modm) do not give $a \equiv b \pmod{nm}$a≡b(modnm). Take $a = 4$a=4, $b = 0$b=0, $n = 2$n=2, $m = 4$m=4: then $4 \equiv 0 \pmod 2$4≡0(mod2) and $4 \equiv 0 \pmod 4$4≡0(mod4), but $4 \not\equiv 0 \pmod 8$4≡0(mod8). (It does follow when $n$n and $m$m share no common factor — a result you will meet as the Chinese Remainder Theorem. The counterexample needs $n$n and $m$m to overlap, as $2$2 and $4$4 do.)

Example 3 (The $-1$−1 trick). Find the remainder of $3^{101}$3101 on division by $4$4. Since $3 \equiv -1 \pmod 4$3≡−1(mod4) and congruence survives multiplication, $3^{101} \equiv (-1)^{101} = -1 \equiv 3 \pmod 4$3101≡(−1)101=−1≡3(mod4). The remainder is $3$3. Look for a base congruent to $\pm 1$±1 before doing anything harder.

Practice problems

Most of these problems are drawn from the homework problems of Lectures 1–7, so nothing here is new. Work them without notes first.

A. Say it precisely

1. Write down, from memory and precisely, the definition of: (a) $A \subseteq B$A⊆B;  (b) a partition of $S$S;  (c) $a \mid b$a∣b;  (d) $a \equiv b \pmod{n}$a≡b(modn). Then say which of the four are statements (true or false) and which are objects.

2. Explain the difference between $\varnothing$∅, $\{\varnothing\}${∅}, and $\{\{\varnothing\}\}${{∅}}. How many elements does each have?

B. Find the error

Each argument below is wrong. Say exactly where, and what it actually proves (if anything).

1. “Let $a$a and $b$b be even. Then $a = 2k$a=2k and $b = 2k$b=2k, so $a + b = 4k = 2(2k)$a+b=4k=2(2k), which is even.”

2. “The negation of ‘$n$n is prime and $n > 2$n>2’ is ‘$n$n is not prime and $n \le 2$n≤2.’ ”

3. “To prove: if $n^{2}$n2 is odd then $n$n is odd. Proof: assume $n$n is odd. Then $n = 2k+1$n=2k+1, so $n^{2} = 2(2k^{2}+2k)+1$n2=2(2k2+2k)+1 is odd. $\blacksquare$■”

4. “To prove: $n^{2} - n$n2−n is even for all $n \in \mathbb{Z}$n∈Z. Proof: if $n$n is even, $n = 2k$n=2k, so $n^{2} - n = 2(2k^{2}-k)$n2−n=2(2k2−k) is even. $\blacksquare$■”

C. Sets past the basics

1. List every element of $\mathcal{P}(\{a,b\})$P({a,b}), and of $\mathcal{P}(\mathcal{P}(\varnothing))$P(P(∅)). If $\left\lvert A \right\rvert = 5$∣A∣=5, how many elements has $\mathcal{P}(A)$P(A)? How many has $\mathcal{P}(\mathcal{P}(A))$P(P(A))?

2. Let $A = \{0,1\}$A={0,1} and $B = \{x,y,z\}$B={x,y,z}. List $A \times B$A×B. Is $A \times B = B \times A$A×B=B×A? Is $\left\lvert A \times B \right\rvert = \left\lvert B \times A \right\rvert$∣A×B∣=∣B×A∣?

3. Prove by double containment that $A - (B \cap C) = (A - B) \cup (A - C)$A−(B∩C)=(A−B)∪(A−C).

D. Logic

1. Is “for every $x \in \varnothing$x∈∅, $x > 5$x>5” true or false? Explain in one sentence. (This is vacuous truth — the same mechanism that makes $\varnothing \subseteq A$∅⊆A hold for every $A$A.)

2. Build the full truth table for $P \Rightarrow (Q \vee R)$P⇒(Q∨R). In how many of the eight rows is it true?

3. Of the four statements $P \Rightarrow Q$P⇒Q,  $(\sim\! P) \vee Q$(∼P)∨Q,  $Q \Rightarrow P$Q⇒P,  $(\sim\! Q) \Rightarrow (\sim\! P)$(∼Q)⇒(∼P), group them into classes of logically equivalent statements. Justify with one truth table.

E. Quantifiers

1. Translate into symbols, stating the domain, then decide the truth of each: “every integer has an additive inverse” and “there is an integer that is an additive inverse of every integer.” Which is $\forall\exists$∀∃ and which is $\exists\forall$∃∀?

2. Recall $\lim_{x \to a} f(x) = L$limx→a​f(x)=L means: $\forall \varepsilon > 0,\ \exists \delta > 0,\ \forall x,\ \bigl(0 < \left\lvert x - a \right\rvert < \delta \Rightarrow \left\lvert f(x) - L \right\rvert < \varepsilon\bigr)$∀ε>0, ∃δ>0, ∀x, (0<∣x−a∣<δ⇒∣f(x)−L∣<ε). Write $\lim_{x\to a} f(x) \neq L$limx→a​f(x)=L as a fully quantified sentence with no $\sim$∼ in front of any quantifier.

F. Prove or disprove

For each: decide first, then either prove it or give a counterexample. Deciding is part of the exercise.

1. The sum of two odd integers is odd.

2. For $n \in \mathbb{Z}$n∈Z: if $n^{3}$n3 is even, then $n$n is even.

3. For every $n \in \mathbb{Z}$n∈Z, $n^{3} - n$n3−n is even.  (Hint: $n^{3}-n = (n-1)n(n+1)$n3−n=(n−1)n(n+1).)

4. For $n \in \mathbb{Z}$n∈Z: $n^{2}$n2 is odd if and only if $n$n is odd.

G. Divisibility and congruence

1. Prove transitivity: if $a \mid b$a∣b and $b \mid c$b∣c, then $a \mid c$a∣c.

2. Let $a,b,c \in \mathbb{Z}$a,b,c∈Z with $a \neq 0$a=0. Prove that if $a \mid (b+c)$a∣(b+c) and $a \mid b$a∣b, then $a \mid c$a∣c.

3. Test each, showing the arithmetic: (a) $23 \equiv 5 \pmod 6$23≡5(mod6)?  (b) $-4 \equiv 11 \pmod 5$−4≡11(mod5)?  (c) $100 \equiv 3 \pmod 7$100≡3(mod7)?

4. Prove that for every $n \in \mathbb{Z}$n∈Z, exactly one of $n \equiv 0$n≡0, $n \equiv 1$n≡1, $n \equiv 2 \pmod 3$n≡2(mod3) holds. Which idea from Lecture 2 is this an instance of?

Answers (sketch). 1. (a), (c), (d) are statements — each is true or false. (b) names an object: a partition is a collection of subsets, not an assertion. The point of the question is that “$a \mid b$a∣b” is a statement while “$a/b$a/b” would be a number. 2. $0$0, $1$1, and $1$1 element respectively. 3. Reusing $k$k for both — it secretly assumes $a = b$a=b. It proves only that $a+a$a+a is even. 4. De Morgan: “$n$n is not prime or $n \le 2$n≤2.” 5. They proved the converse. Check which letter was assumed. 6. Only one case; the odd case is missing, so nothing is proved. 7. $4$4; $\mathcal{P}(\varnothing)=\{\varnothing\}$P(∅)={∅} so $\mathcal{P}(\mathcal{P}(\varnothing))=\{\varnothing,\{\varnothing\}\}$P(P(∅))={∅,{∅}}; $32$32; $2^{32}$232. 8. Six pairs; $A\times B \neq B \times A$A×B=B×A, but the cardinalities agree. 9. Both containments; De Morgan again drives it. 10. True, vacuously — there is no $x$x to violate it. 11. True in $7$7 of $8$8 rows (false only when $P$P is true and $Q,R$Q,R both false). 12. $\{P\Rightarrow Q,\ (\sim\!P)\vee Q,\ (\sim\!Q)\Rightarrow(\sim\!P)\}${P⇒Q, (∼P)∨Q, (∼Q)⇒(∼P)} is one class; $Q \Rightarrow P$Q⇒P stands alone. 13. $\forall n \exists m\,(n+m=0)$∀n∃m(n+m=0) true; $\exists m \forall n\,(n+m=0)$∃m∀n(n+m=0) false. 14. $\exists \varepsilon > 0,\ \forall \delta > 0,\ \exists x,\ \bigl(0 < \left\lvert x-a \right\rvert < \delta \ \wedge\ \left\lvert f(x)-L \right\rvert \ge \varepsilon\bigr)$∃ε>0, ∀δ>0, ∃x, (0<∣x−a∣<δ ∧ ∣f(x)−L∣≥ε). 15. Disprove — it is even: $(2k+1)+(2j+1) = 2(k+j+1)$(2k+1)+(2j+1)=2(k+j+1). 16. True; contrapositive ($n$n odd $\Rightarrow n^3$⇒n3 odd) is the clean route. 17. True; among any two consecutive integers one is even, so $(n-1)n$(n−1)n is even. 18. True; both directions, the ($\Rightarrow$⇒) by contrapositive. 19. $b=ar$b=ar, $c=bs$c=bs $\Rightarrow c = a(rs)$⇒c=a(rs). 20. $b+c = ar$b+c=ar, $b = as$b=as $\Rightarrow c = a(r-s)$⇒c=a(r−s). 21. (a) yes, $23-5=18=6\cdot 3$23−5=18=6⋅3; (b) yes, $-4-11=-15=5\cdot(-3)$−4−11=−15=5⋅(−3); (c) no — $100-3 = 97$100−3=97 and $7 \nmid 97$7∤97. (Note $100 \equiv 2 \pmod 7$100≡2(mod7), since $98 = 7\cdot 14$98=7⋅14.) 22. Division algorithm gives a unique remainder in $\{0,1,2\}${0,1,2}; this is the partition of $\mathbb{Z}$Z into three cells from Lecture 2.

Suggested textbook exercises (Chartrand Ch 1–4): the end-of-chapter review sets for Chapters 2 and 3 are the best use of your time; add §1.5 (partitions), §2.10 (quantifiers), and §4.2 (congruence) if you want more drill on the pieces you missed above.