Math 301 Exam 1 Study GuideSolutions

The exam will cover sections 1.1 - 1.2, 2.1 - 2.4, and 3.1 - 3.5. The problems listed in this studyguide are just some of the types of problems that might appear on the exam. To fully preparefor the exam, in addition to making sure you can do these problems, you should review yourhomework and class notes. Please note the following: At least one exam question will betaken directly from your homework.

Practice Problems

1. Which of the following are statements?

(a) 3 + 5 = 1 Yes, this is a statement.

(b) 3x+5y = 2 No, this is not a statement, because it is not clear what x and y represent.

(c) Does x + 1 = 0 have a solution? No, this is not a statement because it is a question.

(d) There exists an integer x such that 2x = 1. Yes, this is a statement.

Note: As it’s a bit objective (and the book isn’t clear on this) whether or not an “if...then” sentence with variables (but no quantifiers) is a statement, you can assume that all“if... then” sentences such as “if x = 1, then 2x = 1” or “if x is an integer, then 2x iseven” are in fact statements (think of there being an imaginary “for all” quantifier at thebeginning of these sentences).

2. Which of the following statements are true and which are false? Explain.

(a) There exists a real number x such that x2 + 4 = 0. This is a false statement becausethere are no real solutions to x2 = −4.

(b) For all integers x and y, x2 + y2 = z2 for some integer z.This is a false statement. For example, let x = 1 and y = 1. Then x2 + y2 = 2 butthere is no integer z such that z2 = 2.

(c) If x < 1, then (x− 1)2 > 0.This is a true statement. Suppose x < 1. Then x− 1 < 0. Therefore, (x− 1)2 > 0.

(d) If x2 ≥ 0, then x ≥ 0.This is a false statement. Let x = −2. Then x2 = 4 ≥ 0. However, x 6≥ 0.

3. Suppose that Susie says, “If it is cold this weekend, then I will not go camping.”

(a) Is Susie lying or not lying in the event that it is not cold this weekend and she does notgo camping?

No, Susie is not lying, because she didn’t say what she would do if it was not cold thisweekend.

(b) Is Susie lying or not lying in the event that it is cold this weekend and she does gocamping?

Yes, Susie is lying.

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4. Write a truth table for the statement P → Q. When is P → Q a false statement?

Below is a truth table for P → Q. Note that P → Q is only false when P is true and Q isfalse, and it is true in all other cases, even when P is false.

P Q P → Q

T T T

T F F

F T T

F F T

5. This question is about closure properties of number systems.

(a) Show that the natural numbers are closed under multiplication but not division.

Let a, b ∈ N. Note a and b are integers and thus ab is an integer since the integers areclosed under multiplication. Also, note since a > 0 and b > 0, ab > 0. Hence, ab is anatural number. Thus, the natural numbers are closed under multiplication.

Now note 1 and 2 are natural numbers but 12

is not a natural number. Thus, thenatural numbers are not closed under division.

(b) Show that the rational numbers are closed under addition and multiplication but notdivision.

Let x and y be two rational numbers. Then x = p1q1

and y = p2q2

where p1, q1, p2, q2 ∈ Zand q1 6= 0 and q2 6= 0. Note that

x + y =p1q1

+p2q2

=p1q2 + p2q1

q1q2.

Since the integers are closed under addition and multiplication, note that p1q2+p2q1 ∈ Zand q1q2 ∈ Z. Also, since q1 and q2 are nonzero, q1q2 6= 0. Hence, x + y is a rationalnumber.

Also notice that

xy =p1q1· p2q2

=p1p2q1q2

.

Note p1p2 ∈ Z and as we said before q1q2 is a nonzero integer. Thus, xy is a rationalnumber.

Thus, we have showed that the rational numbers are closed under addition andmultiplication. Now note that 1 ∈ Q and 0 ∈ Q. However, note that 1

06∈ Q. Thus, the

rational numbers are not closed under division.

(c) Show that the nonzero rational numbers are closed under division.

This was a homework problem.

(d) Show that the irrational numbers are not closed under addition nor under multiplica-tion.

Note√

2 and −√

2 are irrational numbers. However, note√

2 + (−√

2) = 0 is notirrational. Also, note that (

√2)(√

2) = 2 which is not irrational. Thus, the irrationalnumbers are not closed under addition nor under multiplication.

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Note: For showing closure properties of number systems, the one axiom you can use is thatthe integers are closed under addition, multiplication, and subtraction. Anything else youhave to prove.

6. Write down the definition of what it means for an integer n to be odd. Write down thedefinition of what it means for an integer n to be even.

An integer n is odd provided that n = 2k + 1 where k is an integer. An integer n is evenprovided that n = 2k where k is an integer.

7. Prove or disprove the following.

(a) If x is odd, then x3 + 1 is even.

Theorem. If x is odd, then x3 + 1 is even.

Proof. Suppose x is odd. Then x = 2k + 1 for some integer k. Thus,

x3 + 1 = (2k + 1)(2k + 1)(2k + 1) + 1

= (4k2 + 4k + 1)(2k + 1) + 1

= 8k3 + 4k2 + 8k2 + 4k + 2k + 1 + 1

= 8k3 + 12k2 + 6k + 2

= 2(4k3 + 6k2 + 3k + 1).

Note 4k3 + 6k2 + 3k + 1 is an integer since the integers are closed under addition andmultiplication. Thus, x3 + 1 is even.

(b) If x is even, then 3x2 + 4 is even.

Theorem. If x is even, then 3x2 + 4 is even.

Proof. Suppose x is even. Then x = 2k for some integer k. Thus,

3x2 + 4 = 3(2k)2 + 4

= 3(4k2) + 4

= 12k2 + 4

= 2(6k2 + 2).

Note that 6k2 + 2 is an integer. Thus, 3x2 + 4 is even.

(c) If x is even and y is odd, then xy is even.

Theorem. If x is even and y is odd, then xy is even.

Proof. Suppose x is even and y is odd. Then there exist integers n and m such thatx = 2n and y = 2m + 1. Thus,

xy = (2n)(2m + 1) = 4nm + 2n = 2(2nm + n).

Note that 2nm + n is an integer since the integers are closed under addition and mul-tiplication. Thus, xy is even.

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(d) If x is odd and y is odd, then xy is odd.

Theorem. If x is odd and y is odd, then xy is odd.

Proof. Suppose x is odd and y is odd. Then x = 2n+1 and y = 2m+1 where n,m ∈ Z.Thus,

xy = (2n + 1)(2m + 1)

= 2nm + 2n + 2m + 1

= 2(nm + n + m) + 1.

Note that nm + n + m is an integer. Thus, xy is odd.

(e) If a is an odd integer and b and c are integers, then a(b + c) is odd.

A counterexample can be found by taking a = 1, b = 2, and c = 0. Note that a is oddand b and c are integers. However, a(b + c) = 1(2 + 0) = 2 which is even, not odd.Thus, the statement is false.

(f) If a is an even integer and b and c are integers, then a(b + c) is even.

Theorem. If a is an even integer and b and c are integers, then a(b + c) is even.

Proof. Suppose a is an even integer and b and c are integers. Since a is even, a = 2kfor some integer k. Thus,

a(b + c) = 2k(b + c) = 2(bk + ck).

Note that bk + ck ∈ Z. Thus, a(b + c) is even.

8. What do the symbols ∨, ∧, and ∼ (or ¬) stand for?

The symbol ∨ stands for “or,” the symbol ∧ stands for “and,” and the symbol ∼ (or ¬)stands for “not.”

9. Write truth tables for ∼ P , P ∧Q, and P ∨Q.

P ∼ P

T F

F T

P Q P ∧Q

T T T

T F F

F T F

F F F

P Q P ∨Q

T T T

T F T

F T T

F F F

10. Construct a truth table for the following statements. Assume P,Q, and R are statements.

(a) (P ∨Q)→ P

P Q P ∨Q (P ∨Q)→ PT T T TT F T TF T T FF F F T

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(b) (P ∧Q)→ (Q ∨ ∼ P )

P Q P ∧Q Q ∨ ∼ P (P ∧Q)→ (Q ∨ ∼ P )T T T T TT F F F TF T F T TF F F T T

(c) P → (∼ Q ∨R)

P Q R ∼ Q ∨R P → (∼ Q ∨R)T T T T TT T F F FT F T T TT F F T TF T T T TF T F F TF F T T TF F F T T

(d) ∼ (P ∧Q)→ ∼ R

P Q R P ∧Q ∼ (P ∧Q) ∼ (P ∧Q)→ ∼ RT T T T F TT T F T F TT F T F T FT F F F T TF T T F T FF T F F T TF F T F T FF F F F T T

11. Are any of the statements in the previous problems tautologies or contradictions? Explain.

Yes, (P ∧ Q) → (Q ∨ ∼ P ) is a tautology since it is always true regardless of the truthvalues of P and Q.

12. Construct a truth table for P ∨ (Q ∧ P ). Is this statement a tautology, a contradiction, orneither?

P Q Q ∧ P P ∨ (Q ∧ P )T T T TT F F TF T F FF F F F

It is neither.

13. Explain what it means for two statements to be logically equivalent. Give an example oftwo logically equivalent statements.

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Two statements P and Q are logically equivalent, written P ≡ Q, provided that they havethe same truth values for all possible combinations of truth values for all variables in thestatements.

An example of two logically equivalent statements are the statements P → Q and ∼ P ∨Q.

14. Explain how to construct a truth table for P ↔ Q and then construct one.

P ↔ Q means that both P → Q and Q→ P , so we can construct a truth table for P ↔ Qby constructing one for (P → Q) ∧ (Q→ P ).

P Q P → Q Q→ P (P → Q) ∧ (Q→ P )T T T T TT F F T FF T T F FF F T T T

Thus, the truth table for P ↔ Q looks like the following:

P Q P ↔ QT T TT F FF T FF F T

15. Construct a truth table for (∼ P ∨Q) ↔ (P ∧ ∼ Q). Is this statement a contradiction, atautology, or neither?

P Q ∼ P ∨Q P ∧ ∼ Q (∼ P ∨Q)↔ (P ∧ ∼ Q)T T T F FT F F T FF T T F FF F T F F

The statement is a contradiction.

16. Prove De Morgan’s Laws:

(a) ∼ (P ∨Q) ≡ ∼ P ∧ ∼ Q, and

(b) ∼ (P ∧Q) ≡ ∼ P ∨ ∼ Q

The truth tables below show that ∼ (P ∨Q) ≡ ∼ P ∧ ∼ Q and ∼ (P ∧Q) ≡ ∼ P ∨ ∼ Q.

P Q P ∨Q ∼ (P ∨Q) ∼ P ∧ ∼ QT T T F FT F T F FF T T F FF F F T T

P Q P ∧Q ∼ (P ∧Q) ∼ P ∨ ∼ QT T T F FT F F T TF T F T TF F F T T

17. Show that P → Q ≡ ∼ P ∨ Q. Then show that ∼ (P → Q) ≡ P ∧ ∼ Q.

The truth table below shows that P → Q ≡ ∼ P ∨ Q:

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P Q P → Q ∼ P ∨ QT T T TT F F FF T T TF F T T

Now note that, with the use of DeMorgan’s Laws,

∼ (P → Q) ≡ ∼ (∼ P ∨Q) ≡ P ∧ ∼ Q.

18. Is P → Q logically equivalent to its converse Q → P? Justify your answer with a truthtable.

No, P → Q is not logically equivalent to its converse Q → P , as shown in the truth tablebelow.

P Q P → Q Q→ PT T T TT F F TF T T FF F T T

19. Write down the contrapositive of P → Q. Is P → Q logically equivalent to its contrapositive?Explain your answer using a truth table.

The contrapositive of P → Q is ∼ Q → ∼ P which is logically equivalent to its P → Q, asshown below.

P Q ∼ P ∼ Q P → Q ∼ Q→ ∼ PT T F F T TT F F T F FF T T F T TF F T T T T

20. Write down the converse and the contrapositive of the following statements.

(a) For all integers x, if x is even, then 6x2 − 1 is odd.

The converse is “For all integers x, if 6x2−1 is odd, then x is even.” The contrapositiveis “For all integers x, if 6x2 − 1 is even, then x is odd.”

(b) If x ≥ 3, then x2 ≥ 9.

The converse is “If x2 ≥ 9, then x ≥ 3.” The contrapositive is “If x2 < 9, then x < 3.”

(c) If David does not come to class, he will do poorly on the exams.

The converse is “If David does poorly on the exams, then he does not come to class.”The contrapositive is, “If David does well on the exams, then David comes to class.”

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21. Write an example of a statement in English (one that does not involve math) where theconverse is clearly not true.

An example is the sentence, “If you live in Pullman, WA, then you live in the USA.” Theconverse is, “If you live in the USA, then you live in Pullman, WA” which is clearly not true.

22. Use (a) the roster method and (b) set-builder notation to specify the following sets:

(a) The set of all integers greater than -4 and less than 5.

Using the roster method, we can write the set as {−3,−2,−1, 0, 1, 2, 3, 4}. Using set-builder notation, we can write the set as {n ∈ Z | − 4 < n < 5}.

(b) The set of all natural numbers greater than -4 and less than 5.

Using the roster method, we can write the set as {1, 2, 3, 4}. Using set-builder notation,we can write the set as {n ∈ N | 0 < n < 5}.

(c) The set of all multiples of 3.

Using the roster method, we can write the set as {. . . ,−6,−3, 0, 3, 6, . . .}. Using set-builder notation, we can write the set as {x ∈ R | x = 3n where n ∈ Z}, or we canwrite {3n | n ∈ Z}.

(d) The set of all odd integers.

Using the roster method, we can write the set as {. . . ,−5,−3,−1, 1, 3, 5, . . .}. Usingset-builder notation, we can write the set as {x ∈ R | x = 2n+ 1 where n ∈ Z}, or wecan write {2n + 1 | n ∈ Z}.

23. Write the following sets using set-builder notation:

(a) {1, 4, 7, 10, 13, . . .} {1 + 3n | n = 0, 1, 2, . . .} or {−2 + 3n | n ∈ N}(b) {1,−1, 1,−1, 1,−1, . . .} {(−1)n+1 | n ∈ N}(c) {1, 2, 4, 8, 16, 32, . . .} {2n | n = 0, 1, 2, . . .} or {2n−1 | n ∈ N}

24. Let A = {1, {1}, a, b}.

(a) How many elements does A have? Explain what it means to write that x ∈ A.

A has 4 elements. To write that x ∈ A means that x is an element of A. In this case,the 4 elements of A are 1, {1}, a, and b.

(b) If B is some other set, explain what it means for B ⊆ A.

If B is a set, then B ⊆ A means that every element of B is also an element of A.

(c) Is 1 ∈ A? Is 1 ⊆ A? Yes, 1 ∈ A but 1 6⊆ A.

(d) Is {1} ∈ A? Is {1} ⊆ A? Yes, {1} ∈ A. Also note that {1} ⊆ A as well since {1}contains the element 1 which is also an element of A.

(e) Is ∅ ⊆ A? Is ∅ ∈ A? Yes, ∅ ⊆ A since the empty set is always a subset of any set.No, ∅ 6∈ A.

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(f) Is {1, a} ∈ A? Is {1, a} ⊆ A? No, {1, a} 6∈ A, but {1, a} ⊆ A since {1, a} containsthe elements 1 and a, both of which are also elements of A.

25. Let A = {1, 4, 7, 10, 14, . . .}. Which of the following sets are subsets of A? Which are equalto A, if any?

(a) {31, 34, 37, 40, . . .} This set is a subset of A since every element of this set is also anelement of A.

(b) {1, 7, 14, 21, 28, . . . , } This set is not a subset of A since for example 21 6∈ A as21 6= 1 + 3k for some integer k.

(c) {1, 4, 4, 7, 10, 10} This set is a subset of A.

(d) {1 + 3k | k ∈ Z} This set is not a subset of A. Note that it also contains negativeintegers, whereas A only contains positive integers.

(e) {1, 1, 4, 4, 7, 7, 10, 10, 14, 14, . . .} This set is equal to A since it contains exactly thesame elements as A (it doesn’t matter that elements have been repeated).

26. What is an open sentence or predicate?

An open sentence or predicate is a sentence P (x1, x2, . . . , xn) that is not true or false butbecomes a statement (so is either true or false) once values for the variables are plugged intoP .

27. Let P (x) be the sentence “(x + 1)2 = x2 + 1.” Is P (x) an open sentence? If so, what is thetruth set of P (x)?

Yes, P (x) is an open sentence. To find the truth set for P (x), let us solve the equation forx. Note

(x + 1)2 = x2 + 1 ⇒ x2 + 2x + 1 = x2 + 1 ⇒ 2x = 0 ⇒ x = 0.

Thus, the truth set for P (x) is {0}.

28. Write the following quantified statements in symbolic form. Which of the statements aretrue?

(a) There exists a real number x such that x2 − 2x = −1.

In symbolic form: (∃x ∈ R)(x2 − 2x = −1). This statement is is true since x = 1 is areal number that satisfies the equation.

(b) The square of every even integer is even.

In symbolic form: (∀n ∈ Z)(If n is even, then n2 is even.) This statement is true sinceif n is even, then n = 2k for some integer k and so n2 = (2k)2 = 4k2 = 2(2k2) is alsoeven.

(c) The equation 2x2 − x = 0 has an integer solution.

In symbolic form: (∃x ∈ Z)(2x2 − x = 0). This statement is true since x = 0 is aninteger solution of the equation.

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(d) For every integer n, there exists an integer m such that n + m = 0.

In symbolic form: (∀n ∈ Z)(∃m ∈ Z)(n + m = 0). This statement is true. To showthis, let n ∈ Z. Let m = −n. Then n + m = 0.

(e) There exists an integer n such that for all integers m, n + m = 0.

In symbolic form: (∃n ∈ Z)(∀m ∈ Z)(n + m = 0). This statement is false. Note forany n ∈ Z that I choose, it is not possible for n + m = 0 for all integers m. Anotherway to show this is false is to show its negation is true. Its negation is (∀n ∈ Z)(∃m ∈Z)(n+m 6= 0). Take any n ∈ Z. Let m = −n+1. Note that n+m = n−n+1 = 1 6= 0.

(f) There are integers a and b such that 2a + 3b = 0.

In symbolic form: (∃a ∈ Z)(∃b ∈ Z)(2a + 3b = 0). This statement is true since a = −3and b = 2 are integer solutions to the equation.

29. Write the following statements in English. Then determine which are true and which arefalse.

(a) (∀x ∈ Z)(4x2 − 2x is even).

In English: For all integers x, 4x2−2x is even. This statement is true. Note 4x2−2x =2(2x2 − x). Since the integers are closed under multiplication and subtraction, 2x2 − xis an integer. Thus, 4x2 − 2x is even.

(b) (∃n ∈ N)(3n = 1).

In English: There exists a natural number n such that 3n = 1. This statement is falsesince the only real number that satisfies 3n = 1 is n = 0 which is not a natural number.

(c) (∀x ∈ R)(∃y ∈ R)(xy = 1).

In English: For every real number x, there exists a real number y such that xy = 1.This statement is false since if x = 0, there is no real number y such that xy = 1.

(d) (∃x ∈ R)(∀y ∈ R)(xy = 0).

In English: There exists a real number x such that for all real numbers y, xy = 0. Thisstatement is true since x = 0 is such a real number.

30. Explain what it means for an nonzero integer a to divide an integer b.

A nonzero integer a divides an integer b, written a | b, provided b = ak for some integer k.

31. Does 4 | 60? Does 4 | 30? Explain.

Yes, 4 | 60 since 60 = 4(15). However, 4 - 30 since 30 6= 4k for any integer k.

32. Prove or disprove the following.

(a) Let a, b, c, and d be integers with a and c nonzero. If a | (bc) and c | d, then a | d.

This statement is false. A counterexample can be found by letting a = 2, b = 0, c = 1,and d = 3. Note bc = 0 and so a | (bc). Also note c | d since 3 = 1(3). However, notea - d since 3 6= 2k for any integer k.

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(b) For each integer a, if 6 divides a2 − 1, then 6 divides a− 1.

Consider a = −1. Then a2−1 = 0 and so 6 | (a2−1). However, note a−1 = −1−1 = −2and so 6 - (a− 1). This counterexample shows the conjecture is false.

(c) Let a, b, c ∈ Z where a 6= 0 and b 6= 0. If a | b and b | c, then a | c.

Theorem. Let a, b, c ∈ Z where a 6= 0 and b 6= 0. If a | b and b | c, then a | c.

Proof. Let a, b, c ∈ Z where a 6= 0 and b 6= 0 and suppose a | b and b | c. Then thereexist integers k1 and k2 such that b = ak1 and c = bk2. Thus, notice that

c = bk2 = (ak1)(k2) = a(k1k2).

Notice that k1k2 is an integer since the integers are closed under multiplication. Thus,a | c.

(d) Let n ∈ Z. If 2 | (a + 1), then 4 | (a2 − 1).

Theorem. Let a ∈ Z. If 2 | (a + 1), then 4 | (a2 − 1).

Proof. Let a ∈ Z and suppose 2 | (a + 1). Then a + 1 = 2k for some integer k. Thus,a = 2k − 1. It follows that

a2 − 1 = (2k − 1)2 − 1

= 4k2 − 4k + 1− 1

= 4k2 − 4k

= 4(k2 − k).

Note k2−k is an integer since the integers are closed under multiplication and subtrac-tion. Thus, 4 | (a2 − 1).

(e) For all integers a and b, if 8 | (ab), then 8 | a or 8 | b.

This statement is false. A counterexample can be found by letting a = 4 and b = 2.Then ab = 8 and so 8 | (ab). However, note that 8 - 4 and 8 - 2.

(f) Let a and b be nonzero integers. If a | b and b | a, then a = b.

Consider a = 1 and b = −1. Then a | b and b | a, but a 6= b. This counterexampleshows the conjecture is false.

33. Complete the following definition: Let a, b ∈ Z and let n ∈ N. Then a ≡ b (mod n) providedthat ... n | (a− b).

34. Specify the set of all integers that are congruent to 3 (mod 6) using (a) the roster methodand (b) set-builder notation (but without using notion of congruence).

(a) {. . . ,−9,−3, 3, 9, 15, . . .}(b) {3 + 6k | k ∈ Z}

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35. Suppose it is currently 3:00 (am/pm doesn’t matter). Explain how to use the notion ofcongruence to determine where the hour hand will be (on a normal 12 hour clock) 35 hoursfrom now.

We need to find an integer a where 1 ≤ a ≤ 12 such that a ≡ (3 + 25) (mod 12), i.e. a ≡ 28(mod 12). Note that the set of all integers congruent to 28 (mod 12) are those in the set

{. . . ,−8, 4, 16, 28, . . .}

Thus 4 ≡ 28 (mod 12). Hence, it is 4:00.

36. Prove or disprove the following. In each of the following, a, b, and c are integers and n is anatural number.

(a) Suppose a ≡ 7 (mod 5) and b ≡ 3 (mod 5). Then a + b ≡ 0 (mod 5).

Theorem. If a ≡ 7 (mod 5) and b ≡ 3 (mod 5), then a + b ≡ 0 (mod 5).

Proof. Suppose a ≡ 7 (mod 5) and b ≡ 3 (mod 5). Then 5 | (a − 7) and 5 | (b − 3).Thus, there exist integers k1 and k2 such that a − 7 = 5k1 and b − 3 = 5k2. Thus,a = 7 + 5k1 and b = 3 + 5k2. It follows that

a + b = 7 + 5k1 + 3 + 5k2

= 10 + 5k1 + 5k2

= 5(2 + k1 + k2).

Note that 2 + k1 + k2 is an integer. Thus, 5 | (a+ b− 0). Hence, a+ b ≡ 0 (mod 5).

(b) If a ≡ 2 (mod 5) and b ≡ 2 (mod 5), then ab ≡ 9 (mod 5).

Theorem. If a ≡ 2 (mod 5) and b ≡ 2 (mod 5), then ab ≡ 9 (mod 5).

Proof. Suppose a ≡ 2 (mod 5) and b ≡ 2 (mod 5). Then 5 | (a − 2) and 5 | (b − 2).Hence, there exist integers k1 and k2 such that a − 2 = 5k1 and b − 2 = 5k2. Thus,a = 5k1 + 2 and b = 5k2 + 2. It follows that

ab− 9 = (5k1 + 2)(5k2 + 2)− 9

= 25k1k2 + 10k1 + 10k2 + 4− 9

= 25k1k2 + 10k1 + 10k2 − 5

= 5(5k1k2 + 2k1 + 2k2 − 1).

Note that 5k1k2 + 2k1 + 2k2 − 1 is an integer. Thus, 5 | (ab− 9). Hence, ab ≡ 9 (mod5).

(c) If a2 ≡ 9 (mod 12), then a ≡ 3 (mod 12).

This statement is false. Consider a = −3. Note that a2 − 9 = 0 and so 12 | (a2 − 9).Thus, a2 ≡ 9 (mod 12). However, note that a− 3 = −6 and so 12 - (a− 3). So, a 6≡ 3(mod 12).

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(d) If a ≡ b (mod n), then b ≡ a (mod n).

Theorem. If a ≡ b (mod n), then b ≡ a (mod n).

Proof. Suppose a ≡ b (mod n). Then n | (a− b). Thus, a− b = nk for some integer k.Thus, b−a = −(a− b) = −nk = n(−k). Note that −k is an integer. Hence, n | (b−a).It follows that b ≡ a (mod n).

37. Let x ∈ Z. Use the contrapositive to prove that if x3 − 1 is even, then x is odd.

Theorem. Let x ∈ Z. If x3 − 1 is even, then x is odd.

Proof. We will prove the contrapositive. Suppose x is an even integer. Then x = 2k forsome integer k. Thus,

x3 − 1 = (2k)3 − 1

= 8k3 − 1

= 8k3 − 2 + 1

= 2(4k3 − 1) + 1.

Note that 4k3 − 1 is an integer. Thus, x3 − 1 is odd.

38. Prove or disprove the following biconditional statements.

(a) Let x ∈ R. Then x2 = 4x if and only if x = 4 or x = 0.

Theorem. Let x ∈ R. Then x2 = 4x if and only if x = 4 or x = 0.

Proof. (⇒) Suppose x2 = 4x. Then x(x− 4) = 0. Thus, x = 0 or x = 4.

(⇐) Suppose x = 4 or x = 0. In the case where x = 4, note that x2 = (4)2 = 16and 4x = 4(4) = 16. Hence, x2 = 4x. In the case where x = 0, note that x2 = 02 = 0and 4x = 4(0) = 0. Hence, x2 = 4x.

(b) Let x ∈ R. Then x2 > 4 if and only if x > 2.

This statement is false. The backward direction is certainly true. However, the forwarddirection is not, as can be seen by the counterexample x = −3. Note when x = −3,x2 = 9 > 4, but x 6> 2.

(c) For each integer a, a ≡ 3 (mod 2) if and only if 3a ≡ 3 (mod 6).

Theorem. For each integer a, a ≡ 3 (mod 2) if and only if 3a ≡ 3 (mod 6).

13

Proof. Let a ∈ Z. First we will prove that if a ≡ 3 (mod 2), then 3a ≡ 3 (mod 6).Suppose a ≡ 3 (mod 2). Then 2 | (a− 3). Thus, a− 3 = 2k for some integer k. Hence,a = 2k + 3. It follows that

3a− 3 = 3(2k + 3)− 3

= 6k + 9− 3

= 6k + 6

= 6(k + 1).

Note that k + 1 ∈ Z. Thus, 6 | (3a− 3). Hence, 3a ≡ 3 (mod 6).

Now we will prove that if 3a ≡ 3 (mod 6), then a ≡ 3 (mod 2). Suppose 3a ≡ 3(mod 6). Then 6 | (3a− 3). Thus, 3a− 3 = 6k for some integer k. Hence, 3a = 6k + 3and so a = 2k + 1. It follows that

a− 3 = 2k + 1− 3

= 2k − 2

= 2(k − 1)

Hence, 2 | (a− 3), and so a ≡ 3 (mod 2).

(d) For all integers a and b where a > 0, a | (b + 2) if and only if b ≡ −2 (mod a).

Theorem. For all integers a and b where a > 0, a | (b+ 2) if and only if b ≡ −2 (moda).

Proof. Suppose a, b ∈ Z where a > 0. Notice that

a | (b + 2) ⇔ a | (b− (−2))

⇔ b ≡ −2 (mod a)

which proves the theorem.

(e) For all integers x and y, xy is even if and only if x is even or y is even.

Theorem. For all integers x and y, xy is even if and only if x is even or y is even.

Proof. We will prove the contrapositive. Suppose x is odd and y is odd. Then x = 2n+1and y = 2m + 1 for integers n and m. Thus,

xy = (2n + 1)(2m + 1)

= 4nm + 2n + 2m + 1

= 2(2nm + n + m) + 1.

Notice 2nm + n + m is an integer. Thus, xy is odd.

14

(f) For all integers n, n2 + 3n + 1 is odd if and only if n is even.

It is true that if n is even, then n2 + 3n + 1 is odd. However, it is not true thatif n3 + 3n + 1 is odd, then n is even. To show this is not true, we can prove thatthe contrapositive is not true. The contrapositive is the statement, “If n is odd, thenn3 + 3n + 1 is even.” Consider n = 1. Note n is odd but n3 + 3n + 1 = 5 is not even.Thus, the statement is false.

39. Prove that there exists an integer x such that 4x2 − 17x− 15 = 0.

Theorem. There exists an integer x such that 4x2 − 17x− 15 = 0.

Proof. Let x = 5. Note that x is an integer and

4x2 − 17x− 15 = (4x + 3)(x− 5) = (4(5) + 3)(5− 5) = 0

which proves the theorem.

40. Prove that for any integer a, there exists an integer b such that b | a.

Theorem. For any integer a, there exists an integer b such that b | a.

Proof. Let a be any integer and let b = a. Then clearly b | a since a = a(1).

41. Prove that for every nonzero x ∈ Q, there exists a nonzero y ∈ Q such that xy = 2.

Theorem. For every nonzero x ∈ Q, there exists a nonzero y ∈ Q such that xy = 2.

Proof. Let x be any nonzero rational number. Then x = pq

where p, q ∈ Z and p, q 6= 0.

Choose y = 2qp

. Note 2q ∈ Z and p ∈ Z. Also, note that both 2q 6= 0 and p 6= 0. Thus, y isa nonzero rational number. Also, note that

xy =p

q· 2q

p=

2pq

pq= 2

which proves the result.

42. Prove that there is a real number solution to x5 − 2x3 + 1 = 0.

Theorem. There is a real number solution to x5 − 2x3 + 1 = 0.

Proof. Let f(x) = x5 − 2x3 + 1. Note that f is continuous since it is a polynomial. Also,note that f(−2) = (−2)5 − 2(−2)3 + 1 = −32 + 16 + 1 = −15 < 0 and that f(0) = 1 > 0.Thus, by the Intermediate Value Theorem, there is a real number c ∈ (−2, 0) such thatf(c) = 0.

43. Use a proof by contradiction to prove the following.

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(a) For every integer n, n3 is odd or n is even.

Theorem. For every integer n, n3 is odd or n is even.

Proof. We will use a proof by contradiction. Suppose there exists an integer n such thatn3 is even and n is odd. Since n is odd, there exists an integer k such that n = 2k + 1.Thus,

n3 = (2k + 1)3

= (4k2 + 4k + 1)(2k + 1)

= 8k3 + 4k2 + 8k2 + 4k + 2k + 1

= 8k3 + 12k2 + 6k + 1

= 2(4k3 + 6k2 + 3k) + 1.

Note that 4k3+6k2+3k is an integer. Thus, n3 is odd, a contradiction since we assumedn3 was even.

(b) For each integer a, if a ≡ 2 (mod 3), then a 6≡ 3 (mod 6).

Theorem. For each integer a, if a ≡ 2 (mod 3), then a 6≡ 3 (mod 6).

Proof. We will use a proof by contradiction. Suppose a ≡ 2 (mod 3) and a ≡ 3 (mod6). Since a ≡ 2 (mod 3), 3 | (a− 2). Thus, there is an integer k1 such that a− 2 = 3k1.Thus a = 3k1 + 2. Now, since a ≡ 3 (mod 6), 6 | (a− 3). Thus, there is an integer k2such that a− 3 = 6k2. Thus, a = 6k2 + 3.

Since a = 3k1 + 2 and a = 6k2 + 3, we obtain that 3k1 + 2 = 6k2 + 3. Thus,3k1 − 6k2 = 1. It follows that 3(k1 − 2k2) = 1 and so k1 − 2k2 = 1

3. But this is a

contradiction since k1 − 2k2 is an integer by the closure properties of integers.

(c) For all nonzero integers x and y, if x | y and y | 3, then x 6= 6.

Theorem. For all nonzero integers x and y, if x | y and y | 3, then x 6= 6.

Proof. We will use a proof by contradiction. Suppose x = 6 and x | y and y | 3 where yis a nonzero integer. Since x = 6 and x | y, y = 6n for some integer n. Now, since y | 3,3 = ym for some integer m. Thus, 3 = ym = (6n)(m). Hence nm = 1

2, a contradiction

since nm is an integer.

(d) Let x be a positive real number. If x2 is irrational, then x2/3 is irrational.

Theorem. Let x be a positive real number. If x2 is irrational, then x2/3 is irrational.

Proof. We will use a proof by contradiction. Suppose x2 is irrational and x2/3 is rationalwhere x is a positive real number. Since x2/3 is rational, there exist integers p and qwhere q 6= 0 such that

x2/3 =p

q.

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Raising both sides to the third power, we obtain

x2 =p3

q3.

Notice that p3 and q3 are integers since integers are closed under multiplication. Also,notice that since q 6= 0, q3 6= 0. Hence, x2 is rational, a contradiction since we assumedx2 is irrational.

44. Prove or disprove the following.

(a) If x is irrational and y is any real number, then x + y is irrational.

This statement is false. A counterexample is x =√

2 and y = −√

2. Notice that x isirrational and y is a real number. However, x + y =

√2−√

2 = 0 which is rational.

(b) The number√

2 +√

10 is irrational. (Use only the fact that√

5 is an irrational numberin this proof and note a valid proof for this is NOT something like: Note since the sumof two irrational numbers are irrational,

√2 +√

10 is irrational. This does not workbecause irrational numbers are not closed under addition. Also we are assuming weonly know that

√5 is irrational and nothing about

√2 and

√10.)

Theorem. The number√

2 +√

10 is irrational.

Proof. We will use a proof by contradiction. Suppose√

2+√

10 is rational. Then thereexist integers p and q where q 6= 0 such that

√2 +√

10 = p/q. Squaring both sides, weobtain

(√

2 +√

10)2 =

(p

q

)2

which gives

2 + 2√

20 + 10 =p2

q2.

We rewrite this as

√20 =

p2

2q2− 6.

Now notice that√

20 =√

4 · 5 = 2√

5. Thus, we obtain

√5 =

p2

4q2− 3 =

p2 − 12q2

4q2.

Notice that p2 − 12q2 and 4q2 are integers and that, since q 6= 0, 4q2 6= 0. Thus,√

5 isrational, a contradiction.

(c) For all real numbers x and y, if x 6= 0 and y 6= 0, then x2 + y2 6= (x + y)2.

Theorem. For all real numbers x and y, if x 6= 0 and y 6= 0, then x2 + y2 6= (x + y)2.

17

Proof. We will use a proof by contradiction. Suppose x 6= 0 and y 6= 0 but x2 + y2 =(x + y)2. We can rewrite the equation x2 + y2 = (x + y)2 as x2 + y2 = x2 + 2xy + y2.Thus, 0 = 2xy. Hence 0 = xy. Thus, either x = 0 or y = 0, a contradiction since weassumed x 6= 0 and y 6= 0.

45. Prove the following using cases.

(a) If n is an integer, then n3 + n is even.

Theorem. If n is an integer, then n3 + n is even.

Proof. Suppose n is an integer. Then n is even or n is odd.

In the case where n is even, n = 2k for some integer k. Thus, n3+n = (2k)3+(2k) =8k3 + 2k = 2(4k3 + k). Note 4k3 + k is an integer and so n3 + n is even.

In the case where n is odd, n = 2k + 1 for some integer k. Thus,

n3 + n = (2k + 1)3 + (2k + 1)

= 8k3 + 12k2 + 8k + 2

= 2(4k3 + 6k2 + 4k + 1).

Note that 4k3 + 6k2 + 4k + 1 is an integer and so n3 + n is even.

In both cases, n3 + n is even, and so we have proved the result.

(b) If a - (bc), then a - b and a - c.

Theorem. If a - (bc), then a - b and a - c.

Proof. We will prove the contrapositive. Suppose a | b or a | c. We want to show thata | (bc).

In the case where a | b, note that b = ak for some integer k. Thus, bc = (ak)c =a(kc). Note that kc is an integer. Hence, a | (bc).

In the case where a | c, note that c = ak for some integer k. Thus, bc = b(ak) =a(bk). Note that bk is an integer. Thus, a | (bc).

We have shown that a | (bc) in either case, and thus the proof is completed.

(c) For all real numbers x, if x > 1 or x < −1, then |x| > 1.

Theorem. For all real numbers x, if x > 1 or x < −1, then |x| > 1.

Proof. Suppose that x > 1 or x < −1. We want to show that |x| > 1.

Case 1: Suppose x > 1. Then by definition of absolute value, |x| = x > 1.

Case 2: Suppose x < −1. Then by definition of absolute value, |x| = −x > 1.

Thus, in either case, |x| > 1. Hence the result is proved.

(d) Let a, x ∈ R. If a > 0, then |ax| = a|x|.

Theorem. Let a, x ∈ R. If a > 0, then |ax| = a|x|.

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Proof. Suppose a, x ∈ R and suppose a > 0. We will show that |ax| = a|x| by consid-ering two cases: (1) x ≥ 0 and (2) x < 0.

Case 1: Suppose x ≥ 0. Then, since a > 0, ax ≥ 0. Thus, by definition of absolutevalue, |ax| = ax. Also, |x| = x. Thus, |ax| = a|x|.Case 2: Suppose x < 0. Then since a > 0, ax < 0. Thus, by definition of absolutevalue, |ax| = −(ax). Also, note that |x| = −x. Thus, |ax| = a|x|.Since we have shown that in both cases |ax| = a|x|, we have proven the result.

46. What does the Division Algorithm say?

The Division Algorithm says that for any two integers a and b with b > 0, there exist uniqueintegers q and r where 0 ≤ r ≤ b− 1 such that a = bq + r.

47. For each of the following, find the quotient and remainder and then explain how the twointegers are related by writing an equation of the form a = bq + r where 0 ≤ r ≤ b− 1 as inthe Division Algorithm.

(a) when 13 is divided by 7.

As can be seen by using long division, the quotient is 1 and the remainder is 6. Thus,13 = 7(1) + 6.

(b) when 26 is divided by 3.

The quotient is 8 and the remainder is 2. Thus, 26 = 3(8) + 2.

(c) when −26 is divided by 3.

We have to be careful because we are working with a negative integer and so we cannotdo long division like normal to find the quotient and remainder. We need to find integersq and r such that −26 = 3q+r where 0 ≤ r ≤ 2. Note the only two integers that satisfythese conditions are q = −9 and r = 1. Thus, the quotient is −9 and the remainder is1. We can write −26 = 3(−9) + 1.

(d) when 1 is divided by 2.

The remainder is 1 and the quotient is 0. Note 1 = 2(0) + 1.

(e) when 0 is divided by 2.

The remainder is 0 and the quotient is 0. Note that 0 = 2(0) + 0.

48. Let a ∈ Z and suppose a has a remainder of 3 when divided by 5.

(a) Prove that a ≡ 3 (mod 5).

Theorem. Let a ∈ Z and suppose a has a remainder of 3 when divided by 5. Thena ≡ 3 (mod 5).

Proof. Suppose a ∈ Z and that a has a remainder of 3 when divided by 5. By theDivision Algorithm, there exists a unique integer q such that a = 5q+3. Thus, a−3 = 5qand so 5 | (a− 3). Hence, a ≡ 3 (mod 5).

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(b) Can you say that a2 ≡ r (mod 5) for some integer r where 0 ≤ r ≤ 4? If so, what is rand can you prove your answer is correct?

From part (a), a ≡ 3 (mod 5) and so a = 3 + 5k for some integer k. Thus,

a2 = (3 + 5k)2 = 9 + 30k + 25k2

Notice that a2 − 4 = 5 + 30k + 25k2 = 5(1 + 6k + 5k2) and so we see that a2 ≡ 4 (mod5). Let us write this result as a theorem and write up a formal proof.

Theorem. Let a ∈ Z and suppose a has a remainder of 3 when divided by 5. Thena2 ≡ 4 (mod 5).

Proof. Suppose a ∈ Z and that a has a remainder of 3 when divided by 5. From part(a), a ≡ 3 (mod 5) and thus a = 3 + 5k for some integer k. Thus,

a2 − 4 = (3 + 5k)2 − 4

= 9 + 30k + 25k2 − 4

= 5 + 30k + 25k2

= 5(1 + 6k + 5k2).

Thus, 5 | (a2 − 4). It follows that a2 ≡ 4 (mod 5).

49. Prove that if a is an odd integer that is not a multiple of 3, then a2 ≡ 1 (mod 6).

Suppose a is an odd integer that is not a multiple of 3. By the Division Algorithm (considerdividing a by 6), there exist unique integers q and r where 0 ≤ r ≤ 5 and a = 6q + r. Notethat we cannot have r = 0, r = 2, or r = 4, since a is odd. Also, since a is not a multiple of3, we cannot have r = 3. Thus, either r = 1 or r = 5. Thus, there are two possible cases:(1) a = 6q + 1, and (2) a = 6q + 5.

Case 1: Suppose a = 6q + 1. Then

a2 − 1 = (6q + 1)2 − 1

= 36q2 + 12q + 1− 1

= 6(6q2 + 2q)

and so 6 | (a2 − 1). Thus, a2 ≡ 1 (mod 6).

Case 2: Suppose a = 6q + 5. Then

a2 − 1 = (6q + 5)2 − 1

= 36q2 + 60q + 25− 1

= 36q2 + 60q + 24

= 6(6q2 + 10q + 4)

and so 6 | (a2 − 1). Thus, a2 ≡ 1 (mod 6).

Since a2 ≡ 1 (mod 6) in each of the two possible cases, we have proven the result.

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50. Prove that if a is an even integer that is not a multiple of 4, then a2 ≡ 4 (mod 8).

Theorem. If a is an even integer that is not a multiple of 4, then a2 ≡ 4 (mod 8).

Proof. Suppose a is an even integer that is not a multiple of 4. Using the Division Algorithm,note that there exist unique integers q and r where 0 ≤ r ≤ 7 and a = 8q + r. Since a iseven, note that r ∈ {0, 2, 4, 6}. Also, since a is not a multiple of 4, note that r 6= 0 andr 6= 4. Thus, either r = 2 or r = 6.

Case 1: Suppose r = 2. Then a = 8q + 2. Thus,

a2 − 4 = (8q + 2)2 − 4

= 64q2 + 16q + 4− 4

= 8(8q2 + 2q).

Note that 8q2 + 2q is an integer. Thus, 8 | (a2 − 4) and hence a2 ≡ 4 (mod 8).

Case 2: Suppose r = 6. In this case, a = 8q + 6. Thus,

a2 − 4 = (8q + 6)2 − 4

= 64q2 + 96q + 36− 4

= 64q2 + 96q + 32

= 8(8q2 + 12q + 4).

Note 8q2 + 12q + 4 is an integer. Thus, 8 | (a2 − 4). Hence, a2 ≡ 4 (mod 8).

Since in both cases a2 ≡ 4 (mod 8), we have proven the result.

51. Prove that if n ∈ Z and 3 - n, then n2 + n ≡ 2 (mod 3) or n2 + n ≡ 0 (mod 3).

Theorem. If n ∈ Z and 3 - n, then n2 + n ≡ 2 (mod 3) or n2 + n ≡ 0 (mod 3).

Proof. Suppose n ∈ Z and 3 - n. Using the Division Algorithm, note that there exist uniqueintegers q and r where n = 3q + r and 0 ≤ r ≤ 2. Since 3 - n, note that r 6= 0. Thus, eitherr = 1 or r = 2.

Case 1: r = 1. In this case, note that n = 3q + 1. Thus,

n2 + n = (3q + 1)2 + (3q + 1)

= 9q2 + 6q + 1 + 3q + 1

= 9q2 + 9q + 2.

Hence, n2 + n− 2 = 3(3q2 + 3q). Note 3q2 + 3q is an integer. Thus, 3 | (n2 + n− 2) and son2 + n ≡ 2 (mod 3).

Case 2: r = 2. In this case, note that n = 3q + 2. Thus,

n2 + n = (3q + 2)2 + (3q + 2)

= 9q2 + 12q + 4 + 3q + 2

= 9q2 + 15q + 6

= 3(3q2 + 5q + 2).

21

Note that 3q2 + 5q + 2 is an integer. Thus, 3 | (n2 + n) and hence n2 + n ≡ 0 (mod 3).

Thus, we have shown that either n2 + n ≡ 2 (mod 3) or n2 + n ≡ 0 (mod 3).

22