MATH 308 — Exam II Review

Exam II Review — Proof Techniques, Induction, and Relations
MATH 308  Bridge to Advanced Mathematics  Chartrand Ch 4–9 (Lectures 8–14)

How to use this note. This is a consolidation, not new material. Exam I asked whether you could carry out a proof once the technique was named. This exam asks something harder: whether you can choose the technique, and whether you can tell the difference between proving a statement, disproving it, and proving something uniquely exists. Each section below gives what you must know cold, the move that uses it, and the error that costs the most points.

What is covered.

Lectures	Chartrand	Topic
8	Ch 4	Proofs about real numbers and about sets
9	Ch 5	Existence proofs; proof by contradiction
10	Ch 5	Disproof by counterexample; uniqueness
11	Ch 6	Mathematical induction
12	Ch 6–7	Strong induction; choosing among techniques
13–14	Ch 8–9	Relations; equivalence relations and equivalence classes

Proofs about reals and sets (Lecture 8)

Know cold. To prove $A \subseteq B$A⊆B: take an arbitrary $x \in A$x∈A and show $x \in B$x∈B. To prove $A = B$A=B: do that twice, in both directions (double containment).

The move (reals). Almost every inequality proof in this chapter starts from something you already know is non-negative — usually a square. If you must show $P \ge Q$P≥Q, try to show $P - Q$P−Q is a square, or a sum of squares, or a product of things whose signs you control.

Trap. Dividing an inequality by a quantity without knowing its sign. Dividing by $x$x reverses the inequality when $x<0$x<0, and is illegal when $x=0$x=0. Say which case you are in.

Trap. One containment is not equality. If the problem says “prove $A=B$A=B” and you prove only $A \subseteq B$A⊆B, you have done half the problem and earn half the points at best.

Example 1 (Start from a square). To show $a^{2}+b^{2} \ge 2ab$a2+b2≥2ab for all real $a,b$a,b: note $(a-b)^{2}\ge0$(a−b)2≥0, expand to $a^{2}-2ab+b^{2}\ge0$a2−2ab+b2≥0, and add $2ab$2ab to both sides. Equality exactly when $a=b$a=b.

Existence and contradiction (Lecture 9)

Know cold. An existence claim $\exists x, P(x)$∃x,P(x) is proved by exhibiting one $x$x and verifying $P(x)$P(x). Nothing more is required — and nothing less suffices.

The move (contradiction). Write down the negation of the statement, assume it, and derive something false. The hardest step is the first one: negating correctly. Recall $\sim(\forall x, P)$∼(∀x,P) is $\exists x, \sim\! P$∃x,∼P, and $\sim(P \Rightarrow Q)$∼(P⇒Q) is $P \wedge \sim\! Q$P∧∼Q.

Trap. Assuming the wrong thing. To prove “$P \Rightarrow Q$P⇒Q” by contradiction you assume $P$P and $\sim\! Q$∼Q — not $\sim\! P$∼P, and not just $\sim\! Q$∼Q.

Trap. Proof by contradiction is not a licence to stop early. You must arrive at a genuine contradiction and say what it contradicts.

Trap. “I could not find one” is not a disproof of existence, and “it works for the numbers I tried” is not a proof of a universal statement.

Example 2 (Exhibit and verify). There exist irrational $a$a and rational $b$b with $a+b$a+b irrational. Take $a=\sqrt2$a=2​ and $b=1$b=1. Then $b \in \mathbb{Q}$b∈Q, $a \notin \mathbb{Q}$a∈/Q, and $a+b = \sqrt2+1$a+b=2​+1 is irrational — for if it were rational, subtracting $1$1 would make $\sqrt2$2​ rational. Done: one witness, fully verified.

Counterexamples and uniqueness (Lecture 10)

Know cold. To disprove “for all $x$x, $P(x) \Rightarrow Q(x)$P(x)⇒Q(x)” you need exactly one $x$x for which $P(x)$P(x) is true and $Q(x)$Q(x) is false. Both halves must be checked.

The move (uniqueness). “There is at most one $x$x with $P(x)$P(x)” is proved by assuming $a$a and $b$b both satisfy $P$P and deriving $a=b$a=b. “Exactly one” means you must also prove existence — read which one is being asked.

Trap. A counterexample that fails the hypothesis proves nothing. If the claim is “$x^{2}>4 \Rightarrow x>2$x2>4⇒x>2” then $x=1$x=1 is useless, because $1^{2}>4$12>4 is false. You need $x=-3$x=−3.

Trap. Showing two particular solutions are equal is not a uniqueness proof. Start from two arbitrary solutions $a$a and $b$b.

Example 3 (Uniqueness, the standard shape). At most one real $x$x satisfies $5x^{3}+x = 7$5x3+x=7. Suppose $5a^{3}+a = 5b^{3}+b$5a3+a=5b3+b. Then $5\left(a^{3}-b^{3}\right)+(a-b)=0$5(a3−b3)+(a−b)=0, so $(a-b)\left[5\left(a^{2}+ab+b^{2}\right)+1\right]=0$(a−b)[5(a2+ab+b2)+1]=0. The bracket is at least $1$1 (since $a^{2}+ab+b^{2} = \left(a+\tfrac{b}{2}\right)^{2}+\tfrac{3b^{2}}{4} \ge 0$a2+ab+b2=(a+2b​)2+43b2​≥0), so $a-b=0$a−b=0.

Induction and strong induction (Lectures 11–12)

Know cold. Every induction proof has exactly three visible parts, and you lose points for omitting any of them:

1. Base case. Verify the statement at the smallest value — which is not always $1$1.

2. Inductive hypothesis. Write it out: “assume the statement holds for $k$k.”

3. Inductive step. Prove it for $k+1$k+1, using the hypothesis, and say where you used it.

The move (which induction?). If the statement at $n$n depends only on $n-1$n−1, ordinary induction suffices. If it reaches further back — a recurrence using $a_{n-1}$an−1​ and $a_{n-2}$an−2​, or a factoring argument that splits $n$n into two smaller pieces — you need strong induction, assuming the statement for all $k<n$k<n. A recurrence that looks back two steps needs two base cases.

Trap. Starting at $n=1$n=1 when the claim only becomes true later. $2^{n}>n^{2}$2n>n2 is false at $n=2,3,4$n=2,3,4; its base case is $n=5$n=5.

Trap. Never using the hypothesis. If your inductive step would read the same with the hypothesis deleted, you have not done an induction — and that is usually a sign you proved something else.

Trap. One base case for a two-step recurrence. With only $a_{1}$a1​ verified, the step cannot even start at $n=3$n=3, because it needs $a_{2}$a2​ as well.

Example 4 (Where the hypothesis enters). To prove $\displaystyle\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$i=1∑n​i=2n(n+1)​: base $n=1$n=1 gives $1 = \frac{1\cdot2}{2}$1=21⋅2​. Assuming it for $k$k,

$\underbrace{\sum_{i=1}^{k} i}_{=\,k(k+1)/2 \ \text{by hypothesis}} + (k+1) = \frac{k(k+1)+2(k+1)}{2} = \frac{(k+1)(k+2)}{2},$=k(k+1)/2 by hypothesisi=1∑k​i​​+(k+1)=2k(k+1)+2(k+1)​=2(k+1)(k+2)​,

which is the statement for $k+1$k+1. Mark where the hypothesis was used, as done here.

Relations and equivalence classes (Lectures 13–14)

Know cold. A relation $R$R on $S$S is a subset of $S \times S$S×S. It is an equivalence relation when it is:

• reflexive: $a\,R\,a$aRa for every $a \in S$a∈S;

• symmetric: $a\,R\,b \Rightarrow b\,R\,a$aRb⇒bRa;

• transitive: $a\,R\,b$aRb and $b\,R\,c$bRc $\Rightarrow$⇒ $a\,R\,c$aRc.

Then $[a] = \left\{\, x \in S \;:\; x\,R\,a \,\right\}$[a]={x∈S:xRa}, and the distinct classes partition $S$S: every element lies in exactly one.

The move. To verify a divisibility-style relation, produce the integer witness explicitly every time. If $a\,R\,b$aRb means $n \mid (\text{something})$n∣(something), then write that something as $n$n times an integer you name.

Trap. Reflexivity is a promise about every element. On a finite set, one missing pair $(c,c)$(c,c) kills it.

Trap. Different representatives can name the same class: $[0]=[3]=[-6]$[0]=[3]=[−6] modulo $3$3. Listing $[0],[1],[2],[3],\dots$[0],[1],[2],[3],… as though they were all different is the most common error on this material — count the distinct classes.

Trap. Checking a few numerical cases is not a proof of transitivity. Argue with arbitrary $a,b,c$a,b,c.

Example 5 (Produce the witness). On $\mathbb{Z}$Z, let $a\,R\,b$aRb mean $4 \mid (a-b)$4∣(a−b). Symmetry: if $a-b = 4k$a−b=4k then $b-a = 4(-k)$b−a=4(−k), and $-k$−k is an integer — that named integer is the proof. Transitivity: $a-b=4k$a−b=4k and $b-c=4\ell$b−c=4ℓ give $a-c = 4(k+\ell)$a−c=4(k+ℓ). The classes are $[0],[1],[2],[3]$[0],[1],[2],[3]: four of them, not infinitely many.

Practice problems

Deliberately not shaped like the exam — the point is to stretch the same skills over different ground. Work them without notes first.

A. Say it precisely

1. From memory, state: what must be shown to prove $A=B$A=B for sets; what must be assumed to prove “$P \Rightarrow Q$P⇒Q” by contradiction; what a counterexample to “$\forall x,\ P(x)\Rightarrow Q(x)$∀x, P(x)⇒Q(x)” must satisfy; the three parts of an induction proof; the three properties of an equivalence relation.

2. Explain the difference between “at most one,” “at least one,” and “exactly one,” and say what must be proved in each case.

B. Find the error

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

1. “To prove $x^{2} \ge x$x2≥x for all real $x$x: divide both sides by $x$x to get $x \ge 1$x≥1, which is what we wanted.”

2. “$R$R on $\{1,2,3\}${1,2,3} given by $R=\{(1,1),(2,2),(1,2),(2,1)\}$R={(1,1),(2,2),(1,2),(2,1)} is an equivalence relation, since it is symmetric and transitive.”

3. “Claim: $n^{2}+n+41$n2+n+41 is prime for every $n \in \mathbb{N}$n∈N. Proof: it is prime for $n=1,2,3,4,5$n=1,2,3,4,5.”

4. “To prove $\sum_{i=1}^{n} (2i) = n(n+1)$∑i=1n​(2i)=n(n+1): assume it for $k$k. Then $\sum_{i=1}^{k+1}(2i) = (k+1)(k+2)$∑i=1k+1​(2i)=(k+1)(k+2), which is the statement for $k+1$k+1. $\blacksquare$■”

5. “The classes of congruence mod $5$5 are $[0],[1],[2],[3],[4],[5],[6],\dots$[0],[1],[2],[3],[4],[5],[6],…, one for each non-negative integer.”

C. Reals and sets

1. Prove that for all real $a,b$a,b:  $a^{2}+b^{2}+1 \ \ge\ ab+a+b$a2+b2+1 ≥ ab+a+b, and say exactly when equality holds.  (Double both sides first, then look for squares.)

2. Let $A = \left\{\, x \in \mathbb{R} \;:\; x^{2}<4 \,\right\}$A={x∈R:x2<4} and $B = \left\{\, x\in\mathbb{R} \;:\; -2<x<2 \,\right\}$B={x∈R:−2<x<2}. Prove $A=B$A=B by double containment.

3. Prove that for all real $a,b$a,b: $\left\lvert a+b \right\rvert \le \left\lvert a \right\rvert+\left\lvert b \right\rvert$∣a+b∣≤∣a∣+∣b∣.  (Consider cases on signs, or square both sides.)

D. Existence, contradiction, counterexample

1. Prove there exist integers $m,n$m,n with $12m+18n = 6$12m+18n=6.

2. Prove by contradiction that $\sqrt{3}$3​ is irrational.

3. Prove by contradiction that there is no largest even integer.

4. Prove or disprove: for all $n \in \mathbb{Z}$n∈Z, if $n^{2}$n2 is divisible by $4$4 then $n$n is divisible by $4$4.

5. Prove or disprove: for all real $x$x, if $x^{3}>x$x3>x then $x>1$x>1.

E. Uniqueness

1. Prove there is at most one real $x$x with $x^{5}+x = 4$x5+x=4.

2. Prove that a real number $x$x with $x + c = x$x+c=x for every real $c$c does not exist, but that an additive identity is unique once it does exist. (Two different claims — say which is which.)

F. Induction

1. Prove by induction: $\displaystyle\sum_{i=1}^{n} i^{2} = \frac{n(n+1)(2n+1)}{6}$i=1∑n​i2=6n(n+1)(2n+1)​.

2. Prove by induction: $3 \mid \left(n^{3}-n\right)$3∣(n3−n) for every $n \in \mathbb{N}$n∈N.

3. Prove by induction: $n! > 2^{n}$n!>2n for every integer $n \ge 4$n≥4.  (Where does the base case go, and why not at $n=1$n=1?)

4. Let $b_{1}=2$b1​=2, $b_{2}=4$b2​=4, and $b_{n}=2b_{n-1}-b_{n-2}$bn​=2bn−1​−bn−2​ for $n\ge3$n≥3. Prove $b_{n}=2n$bn​=2n.  (How many base cases, and why?)

G. Relations and classes

1. On $\mathbb{Z}$Z, define $a\,R\,b$aRb to mean $5 \mid (a-b)$5∣(a−b). Prove $R$R is an equivalence relation and list its distinct classes.

2. On $\mathbb{R}$R, define $x\,R\,y$xRy to mean $x^{2}=y^{2}$x2=y2. Prove $R$R is an equivalence relation and describe $[3]$[3] and $[0]$[0]. How many elements does a typical class have?

3. For each, decide which of reflexive, symmetric, transitive hold — and justify each failure with a specific counterexample: (a) $a\,R\,b$aRb iff $a<b$a<b on $\mathbb{Z}$Z;  (b) $a\,R\,b$aRb iff $a\cdot b>0$a⋅b>0 on $\mathbb{Z}\setminus\{0\}$Z∖{0};  (c) $a\,R\,b$aRb iff $a+b$a+b is even, on $\mathbb{Z}$Z.

4. On $S=\{1,2,3,4,5,6\}$S={1,2,3,4,5,6}, the partition $\bigl\{\{1,2\},\{3,4,5\},\{6\}\bigr\}${{1,2},{3,4,5},{6}} induces an equivalence relation. Write $[3]$[3] and $[6]$[6], decide whether $1 \sim 5$1∼5, and say how many ordered pairs the relation contains.