Final Exam - Practice

Math 108 - Section 01 - Spring 2020

Tuesday, May 19, 2020

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• You have 135 minutes to complete this exam.

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• If you don’t know how to solve a problem, skip it, and come back to it later.

QUESTION VALUE SCORE

1 20

2 30

3 30

4 25

5 20

6 30

7 20

8 20

9 20

10 25

TOTAL 240

1

1. (20 points) ((a) = 10 points, (b) = 10 points)

(a) Suppose A = {b, c, d}, B = {a, b}. Find P(A)−P(B).

P(A) = {∅, {b} , {c} , {d} , {b, c} , {b, d} , {c, d} , {b, c, d}}

P(B) = {∅, {a} , {b} , {a, b}}

P(A)−P(B) = {{c} , {d} , {b, c} , {b, d} , {c, d} , {b, c, d}}

(b) Let (an)∞n=1 be an infinite sequence of real numbers. Negate the following statement:

For every ε > 0, there exists N ∈ N such that, for every m,n ∈ N, if m ≥ n ≥ N , then

|am − an| < ε.

There exists ε > 0 such that, for all N ∈ N, there exists m,n ∈ N with m ≥ n ≥ N , and

|am − an| ≥ ε.

2

2. (30 points) ((a) = 15 points, (b) = 15 points)

(a) Prove the following statement: If n ∈ Z, then n2 + 3n− 2 is even.

Proof: Let n ∈ Z. We consider two cases.

Case 1: n is even. Then n = 2k for some k ∈ Z. So n2 + 3n − 2 = 4k2 + 6k − 2 =

2(2k2 + 3k − 1) and 2k2 + 3k − 1 ∈ Z. Thus n2 + 3n− 2 is even.

Case 2: n is odd. Then n = 2k for some k ∈ Z. So n2+3n−2 = 4k2+4k+1+6k+3−2 =

2(2k2 + 5k + 1) and 2k2 + 5k + 1 ∈ Z. Thus n2 + 3n− 2 is even.

(b) Prove the following statement: Suppose a ∈ Z. If a3 is not divisible by 8, then a is odd.

Proof: We prove the contrapositive: If a is even, then 8 | a3. Assume a is even. Then

a = 2k for some k ∈ Z. So a3 = (2k)3 = 8k3 and k3 ∈ Z. Therefore 8 | a3.

3

3. (30 points) ((a) = 15 points, (b) = 15 points)

(a) Prove the following statement: For all sets A and B, A ∩ (B − A) = ∅.

Proof: We prove this by contradiction. Suppose A ∩ (B − A) is non-empty. So there

exists x ∈ A∩ (B−A). Then x ∈ A and x ∈ B−A. Thus x ∈ A and x ∈ B and x /∈ A.

But having x ∈ A and x /∈ A is a contradiction. Thus the assumption A ∩ (B − A) is

non-empty is false.

(b) Prove the following statement: Let a, b, c ∈ Z. Then abc is odd if and only if a, b, c are

all odd.

Proof: ⇐: Assume a, b, c are all odd. Then a = 2i+ 1 and b = 2j + 1 and c = 2k+ 1 for

some i, j, k ∈ Z. So abc = (2i+1)(2j+1)(2k+1) = 8ijk+4ij+4jk+4ik+2i+2j+2k+1

Thus abc is odd.

⇒: We want to prove: if abc is odd, then a, b, c are all odd. We prove the contrapositive:

If at least one of a, b, c is even, then abc is even. Without loss of generality assume, a is

even. Then a = 2k for some k ∈ Z. So abc = 2kbc = 2` where ` = kbc ∈ Z. Thus abc is

even.

4

4. (25 points) ((a) = 10 points, (b) = 15 points)

(a) Prove or disprove: There exists a b ∈ N such that bn > n! for all n ∈ N.

False. Disproof: Seeking a contradiction, suppose there exists a b ∈ N such that bn > n!

for all n ∈ N. Take n = bb. Then

n! = (bb)! = bb · (bb − 1) · · · 3 · 2 · 1 ≥ bb = bn

Contradiction.

(b) Prove: If A,B,C,D are sets, then (A×B)− (C ×D) = ((A−C)×B)∪ (A× (B−D))

True. Proof:

(A×B)− (C ×D) = {(x, y) : (x ∈ A ∧ y ∈ B)∧ ∼ (x ∈ C ∧ y ∈ D)}= {(x, y) : (x ∈ A ∧ y ∈ B) ∧ (x /∈ C ∨ y /∈ D)}= {(x, y) : (x ∈ A ∧ y ∈ B ∧ x /∈ C) ∨ (x ∈ A ∧ y ∈ B ∧ y /∈ D)}= {(x, y) : (x ∈ A− C ∧ y ∈ B) ∨ (x ∈ A ∧ y ∈ B −D)}= {(x, y) : (x, y) ∈ (A− C)×B ∨ (x, y) ∈ A× (B −D)}= ((A− C)×B) ∪ (A× (B −D)).

5

5. (20 points) Suppose (an)∞n=0 is an infinite sequence of real numbers that satisfy a0 ≤ 0,

a1 ≤ 1, and an ≤ an−1 + an−2. Prove that an ≤(53

)nfor every integer n ≥ 0.

Proof. We use strong induction. Let S(n) denote that statement that an ≤(53

)n.

Base Step: We verify that S(0) and S(1) are true.

We have a0 ≤ 0 ≤ 1(53

)0. So S(0) is true.

We have a1 ≤ 1 ≤ 53

=(53

)1. So S(1) is true.

Inductive Step: Let n be an integer such that n ≥ 2. Assume S(k) holds for every integer

k such that 0 ≤ k < n (this is the strong inductive hypothesis). We must show that S(n)

holds.

Since n ≥ 2, we have 0 ≤ n− 1 < n and 0 ≤ n− 2 < n. So the strong inductive hypothesis

implies that S(n− 1) and S(n− 2) hold. Thus an−1 ≤(53

)n−1and an−2 ≤

(53

)n−2. Since the

sequence satisfies an ≤ an−1 + an−2, it follows that

an ≤ an−1 + an−2 ≤(

5

3

)n−1

+

(5

3

)n−2

=

(5

3

)n−2(5

3+ 1

)=

(5

3

)n−2(8

3

)=

(5

3

)n−2(24

9

)<

(5

3

)n−2(25

9

)=

(5

3

)n−2(5

2

)2

=

(5

3

)n

Thus S(n) holds. This completes the proof by strong induction.

6

6. (30 points) ((a) = 15 points, (b) = 15 points)

(a) Let A = {0, 1, 2, 3, 4}. As a set, write the out the relation R that expresses > on A.

Then illustrate it with a diagram.

(x, y) ∈ R means x > y, so

R = {(1, 0), (2, 0), (3, 0), (4, 0), (2, 1), (3, 1), (4, 1), (3, 2), (4, 2), (4, 3)} .

(b) Give an example of a relation R on Z that is symmetric and reflexive but not transitive.

Justify your answer.

To make the relation R reflexive, it must include (x, x) for all x ∈ Z.

To be symmetric and not transitive, the

diagram for the relation R should include

a structure like the one to the right:

Thus we define R = {(x, x) : x ∈ Z} ∪ {(1, 2), (2, 1), (2, 3), (3, 2)}.

R is not transitive because (1, 2) ∈ R and (2, 3) ∈ R but (1, 3) /∈ R.

It is clear from inspection that R is reflexive: For all x ∈ Z, (x, x) ∈ R.

Likewise, it is clear R is symmetric: For all x, y ∈ Z, if (x, y) ∈ R, then (y, x) ∈ R.

Note: Other examples are possible.

7

7. (20 points) Let H = {2m : m ∈ Z}. A relation R is defined on the set Q+ of positive

rational numbers by xRy if and only if xy∈ H. Prove that R is an equivalence relation.

Describe the equivalence class of [3].

To be prove that R is an equivalence relation, we must prove that it is reflexive, transitive,

and symmetric.

Reflexive: Let x ∈ Q+. Then xx

= 1 = 20 ∈ H. So xRx.

Symmetric: Let x, y ∈ Q+. Assume xRy. So xy∈ H. Hence x

y= 2m for some m ∈ Z. Then

yx

= 2−m and −m ∈ Z. So yz∈ H. Thus yRx.

Transitive: Let x, y, z ∈ Q+. Assume xRy and yRz. So xy∈ H and y

z∈ H. Then x

y= 2m

and yz

= 2n for some m,n ∈ Z. Then

x

z=x

y· yz

= 2m · 2n = 2m+n

and m+ n ∈ Z. So xz∈ H. Thus xRz.

Now we consider the equivalence class of [3]. First, observe that xRy if and only if x = y ·2m

for some m ∈ Z. Thus the equivalence class of [y] is

[y] = {x : xRy} = {y · 2m : m ∈ Z} .

In particular, the equivalence class of [3] is

[3] = {3 · 2m : m ∈ Z} =

{. . . ,

3

8,3

4,3

2, 3, 6, 12, . . .

}.

8

8. (20 points) ((a) = 10 points, (b) = 10 points)

(a) Suppose f : Z2 → Z is defined as f(x, y) = 5x+ 10y. Write this function as a subset of

Z2 × Z and find its range.

f = {((x, y), 5x+ 10y) : (x, y) ∈ Z2}

The range is {5x+ 10y : (x, y) ∈ Z2} = {5n : n ∈ Z}.

(b) Prove or disprove: Given a function f : A→ B and subsets W,X ⊆ A, then f(W ∩X) =

f(W ) ∩ f(X).

False. Disprove: Define A = {1,−1}, B = {1}, f : A → B as f(−1) = f(1) = 1,

W = {−1}, X = {1}. Then f(W ∩X) = f(∅) = ∅ but f(W )∩ f(X) = {1}∩{1} = {1}.

9

9. (20 points) Show that the sets N × N and A = {(x, y) ∈ N× N : y ≤ x} have equal

cardinality by describing a bijection from one to the other. Describe your bijection with a

formula (not as a table); a piecewise formula is fine. Prove that the function you describe is

a bijection.

We claim that f(x, y) = (x+ y − 1, y) defines a bijection f : N× N→ A.

The idea behind this function f is given by the figure at the bottom of the page, which

illustrates that f is rotating each diagonal line by 45◦.

First we check that the range of f is actually contained in A. The range of f is

{f(x, y) = (x+ y − 1, y) : (x, y) ∈ N× N}. For every (x, y) ∈ N×N, we have (x+y−1, y) ∈N × N and y ≤ x + y − 1 because x ≥ 1. So (x + y − 1, y) ∈ A. Thus the range of f is a

subset of A.

Now we check that f is surjective. In other words, we check that the range of f equals A.

For all (x, y), (u, v) ∈ N× N, we see that:

f(x, y) = (u, v) ⇔ (x+ y − 1, y) = (u, v) ⇔ x+ y − 1 = u and y = v

⇔ x+ v − 1 = u and y = v ⇔ x = u− v + 1 and y = v

⇔ (x, y) = (u− v + 1, v)

So for any given (u, v) ∈ N, we can take (x, y) = (u − v + 1, v) to obtain f(x, y) = (u, v).

Thus f is surjective.

Finally, we show that f is injective. Let (x1, y1), (x2, y2) ∈ N × N. Assume f(x1, y1) =

f(x2, y2). Then (x1 + y1 − 1, y1) = (x2 + y2 − 1, y2). So x1 + y1 − 1 = x2 + y2 − 1 and

y1 = y2. Plugging the second equation into the first and rearranging gives x1 = x2. Thus

(x1, x2) = (y1, y2).

10

10. (25 points) ((a) = 10 points, (b) = 15 points)

(a) Prove or disprove: Every infinite set is a subset of a countably infinite set.

False. Disproof: R is an infinite set, but it is not contained in any countably infinite set.

Indeed, since R is actually uncountable, Theorem 14.9 implies that any set containing

R must be uncountable.

(b) Prove or disprove: The set A = {(a1, a2, a3, . . .) : ak ∈ Z} of infinite sequences of integers

is countably infinite.

False. In fact, A is uncountable.

Disproof: We mimic the proof that R is uncountable. It is clear that A is infinite. To

see that A is uncountable, we must show there is no bijection from N to A. In fact, we

will show there is no surjection from N to A. Seeking a contradiction, suppose there is a

surjection f : N→ A. For each n ∈ N, f(n) is some sequence (a1, a2, a3, . . .) ∈ A. We use

the notation f(n)k to stand for the k-th term of the sequence. That is, f(n)k = ak. Now

define a sequence (b1, b2, b3, . . .) in A by setting bk = 1 + f(k)k for each k ∈ {1, 2, 3, . . .}.Since f is a surjection, there exists some n ∈ N such that f(n) = (b1, b2, b3, . . .). Then

f(n)n = bn. But, by definition, bn = 1 + f(n)n. Thus f(n)n = 1 + f(n)n. Contradiction.

11