Download MA 225-001 Test 1 Review Questions

MA 225-001 Test 1 Review Questions
S. Schecter
Expanded February 14, 2005
1. Use truth tables to show that P ⇒ Q and (∼ P ) ∨ Q are equivalent
propositional forms.
2. (1) Translate into a symbolic sentence with quantifiers, (2) write a clear
denial as a symbolic sentence, and (3) translate your denial into ordinary English. The universe for each statement is given in parentheses.
(a) There is a natural number that is less than or equal to every
natural number. (Natural numbers)
(b) Some people are good at both math and baseball.
(c) Everyone plays golf or tennis.
(d) Some people like Coke and some people like Pepsi.
3. Definitions:
(a) Let a and b be integers. We say a divides b if there is an integer k
such that ak = b.
(b) An integer a is even if there is an integer k such that a = 2k.
(c) An integer a is odd if there is an integer k such that a = 2k + 1.
(d) A real number x is rational if there exist integers a and b, with
b 6= 0, such that x = ab .
Using these definitions, prove the following. Do not use previously
proved results. For some statements, the universe is given in parentheses. A suggested method of proof is given for some statements.
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(a) If k is an integer, then k 2 + 3k is even. (Proof by cases.)
(b) If a divides b and b divides a + c, then a divides c. (Integers.)
(c) If ab does not divide ac, then b does not divide c. (Integers.
Contrapositive proof.)
(d) If x is rational and y is irrational, then 2x + y is irrational. (Real
numbers. Proof by contradiction.)
(e) There do not exist integers x and y such that x + y is even and
x − y is odd. (Proof by contradiction.)
(f) There exist integers m and n such that m + 2n = 1.
(g) For every natural number m the following is true: If there is a
natural number k such that m = 4k + 1, then there is a natural
number l such that m + 2 = 4l − 1. Suggestion: 3=4-1.)
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Some answers:
2. (a) There is a natural number that is less than or equal to every
natural number. (Natural numbers)
i. ∃x∀y(x ≤ y)
ii. ∼ ∃x∀y(x ≤ y)
∀x ∼ ∀y(x ≤ y)
∀x∃y ∼ (x ≤ y)
∀x∃y(x > y)
iii. Each natural number is bigger than some other natural number.
(b) Some people are good at both math and baseball.
i. ∃x(x is good at math ∧ x is good at baseball)
ii. ∼ ∃x(x is good at math ∧ x is good at baseball)
∀x ∼ (x is good at math ∧ x is good at baseball)
∀x(∼ x is good at math ∨ ∼ x is good at baseball)
iii. Everyone is bad at math or baseball.
(c) Everyone plays golf or tennis.
i. ∀x(x plays golf ∨ x plays tennis)
ii. ∼ ∀x(x plays golf ∨ x plays tennis)
∃x ∼ (x plays golf ∨ x plays tennis)
∃x(∼ x plays golf ∧ ∼ x plays tennis)
iii. Someone plays neither golf nor tennis.
(d) Some people like Coke and some people like Pepsi.
i. ∃x(x likes Coke) ∧ ∃x(x likes Pepsi)
ii. ∼ (∃x(x likes Coke) ∧ ∃x(x likes Pepsi))
∼ ∃x(x likes Coke)∨ ∼ ∃x(x likes Pepsi)
iii. No one likes Coke or no one likes Pepsi.
3. (a) If k is an integer, then k 2 + 3k is even. (Proof by cases.)
Proof. Let k be an integer.
Case 1. k is even. Then there is an integer m such that k = 2m.
Then k 2 + 3k = 4m2 + 6m = 2(2m2 + 3m). Since 2m2 + 3m is an
integer, k 2 + 3k is even.
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Case 2. k is odd. Then there is an integer n such that k = 2n + 1.
Then k 2 + 3k = (2n + 1)2 + 3(2n + 1) = 4n2 + 4n + 1 + 6n + 3 =
2(2n2 + 5n + 2). Since 2n2 + 5n + 2 is an integer, k 2 + 3k is even.
(b) If a divides b and b divides a + c, then a divides c. (Integers.)
Proof: Assume a divides b and b divides a + c. Then there exist
integers k and ℓ such that ak = b and bℓ = a + c. Therefore
akℓ = a + c, so akℓ − a = c, so a(kℓ − 1) = c. Since kℓ − 1 is an
integer, a divides c.
(c) If ab does not divide ac, then b does not divide c. (Integers.
Contrapositive proof.)
Proof: Suppose b does divide c. Then there exists an integer k
such that bk = c. Therefore abk = ac, so ab divides ac. Thus we
have proved the contrapositive of the given statement; therefore
the given statement is true.
(d) If x is rational and y is irrational, then 2x + y is irrational. (Real
numbers. Proof by contradiction.)
Proof: Assume x is rational and y is irrational. Suppose 2x + y
is rational. Since x is rational, there exist integers a and b, with
b 6= 0, such that x = ab . Since 2x+y is rational, there exist integers
c and d, with d 6= 0, such that 2x + y = dc . Therefore
y = (2x + y) − 2x =
a
bc − 2ad
c
−2 =
.
d
b
bd
Since bc − 2ad and bd are integers and bd 6= 0, y is rational. This
contradicts the assumption that y is irrational.
(f) m = 1, n = 0, for example.
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