Download MATH CSE20 Test 2 Review Sheet Test Tuesday

MATH CSE20 Test 2 Review Sheet
Test Tuesday October 29 in lecture: CENTER 115, 3:30pm
Textbook sections: Unit Lo Sections 1 and 2
(1) All questions from Homeworks 3 and 4
(2) (Lo Review Question 4) The statement form (p ↔ r) → (q ↔ r) is equivalent to
From class (lecture 7) we proved that the biconditional or double implication
(p ↔ q) is equivalent to (p → q) ∧ (q → p). With this, we can convert the
statement to the following:
∼ (p ↔ r) ∨ (q ↔ r) by implication
∼ [(p → r) ∧ (r → p)] ∨ (q → r) ∧ (r → q) by double implication
∼ [(∼ p ∨ r) ∧ (∼ r ∨ p)] ∨ (∼ q ∨ r) ∧ (∼ r ∨ q) by implication
(e) ∼ [(∼ p ∨ r) ∧ (p ∨ ∼ r)] ∨ ([∼ q ∨ r) ∧ (q ∨ ∼ r)]
(3) (Epp 1.2.20) Write negations for each of the following statements. (Assume that all
variables represent fixed quantities or entities, as appropriate).
(a) If P is a square, then P is a rectangle.
Let Q be “P is a square” and R be “P is a rectangle.” Then we have:
Original: Q → R
=∼ Q ∨ R
Negation: ∼ (∼ Q ∨ R)
=Q∧∼R
Which translates to P is a square and not a rectangle.
Contrapositive:
∼R→∼Q
Which translates to If P is not a rectangle, then P is not a square.
Converse:
R→Q
Which translates to If P is a rectangle, then P is a square.
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Inverse:
∼Q→∼R
Which translates to If P is not a square, then P is not a rectangle.
We can apply the same logic to questions (b) and (c).
(b) If today is New Year’s Eve, then tomorrow is January.
Negation: Today is New Year’s Eve and tomorrow is not January.
Contrapositive: If tomorrow is not January, then today is not New Year’s Eve.
Converse: If tomorrow is January, then today is New Year’s Eve.
Inverse: If today is not New Year’s Eve, then tomorrow is not January.
(c) If the decimal expansion of r is terminating, then r is rational.
Negation: The decimal expansion of r is terminating and r is irrational.
Contrapositive: If r is not rational, then the decimal expansion of r is not terminating.
Converse: If r is rational, then the decimal expansion of r is terminating.
Inverse: If the decimal expansion of r is not terminating, then r is not rational.
(d) If n is prime, then n is odd or n is 2.
Let Q be “n is prime” and R be “n is odd or n is 2.” Then we have:
Original: Q → R
=∼ Q ∨ R
Negation: ∼ (∼ Q ∨ R)
=Q∧∼R
If we negate R, we have:
∼ (n is odd ∨ n is 2)
=∼ (n is odd) ∧ ∼ (n is 2)
Which translates to n is even and n is not 2.
Negation: n is prime and n is even and not 2.
Contrapositive: If n is not odd and not 2, then n is not prime.
Converse: If n is odd or n is 2, then n is prime.
Inverse: If n is not prime, then n is not odd and n is not 2.
(e) If x is nonnegative, then x is positive or x is zero.
Using the previous question’s logic, we have:
Negation: x is nonnegative and x is not positive nor zero.
Contrapositive: If x is not positive and x is not zero, then x is not nonnegative.
Converse: If x is positive or x is zero, then x is nonnegative.
Inverse: If x is not nonnegative, then x is not positive and x is not zero.
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(f) If Todd is Anna’s father, then Jim is her uncle and Sue is her aunt.
Let Q be “Todd is Anna’s father” and R be “Jim is her uncle and Sue is her aunt.”
Then, as before
∼ (Q → R) = Q ∧ ∼ R
If we negate R, we have:
∼ (Jim is her uncle ∧ Sue is her aunt)
∼ (Jim is her uncle) ∨ ∼ (Sue is her aunt)
Which translates to Jim is not her uncle or Sue is not her aunt.
Negation: Todd is Anna’s father and either Jim is not her uncle or Sue is not her
aunt.
Contrapositive: If Jim is not Anna’s uncle or Sue is not her aunt, then Todd is
not her father.
Converse: If Jim is her uncle and Sue is her aunt, then Todd is Anna’s father.
Inverse: If Todd is not Anna’s father, then Jim is not her uncle or Sue is not her
aunt.
(g) If n is divisible by 6, then n is divisible by 2 and n is divisible by 3.
Using the previous question’s logic, we have:
Negation: n is divisible by 6 and n is not divisible by 2 or n is not divisible by 3.
Contrapositive: If n is not divisible by 2 or n is not divisible by 3, then n is not
divisible by 6.
Converse: If n is divisible by 2 and n is divisible by 3, then n is divisible by 6.
Inverse: If n is not divisible by 6, then n is not divisible by 2 or n is not divisible
by 3.
(4) (Epp 1.2.21) Suppose that p and q are statements such that p → q is false. Find the
truth values of each of the following:
Based on the truth table for p → q, if the statement is false, p must be true and q
must be false. We can use these values of p and q to solve the questions.
(a) ∼ p → q
∼T →F
F → F is true.
(b) p ∨ q
T ∨ F is true.
(c) q → p
F → T is true.
(5) (Epp 2.3.9) Let D = E = {−2, −1, 0, 1, 2}. Explain why the following statements are
true.
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(a) ∀x ∈ D, ∃y ∈ E, x + y = 0.
To prove this is true, we can go through all five possible x’s and find a y ∈ E in
which x + y = 0.
Let x = −2 ∈ D. If y = 2 ∈ E, x + y = 0.
Let x = −1 ∈ D. If y = 1 ∈ E, x + y = 0.
Let x = 0 ∈ D. If y = 0 ∈ E, x + y = 0.
Let x = 1 ∈ D. If y = −1 ∈ E, x + y = 0.
Let x = 2 ∈ D. If y = 1 ∈ E, x + y = 0.
Thus, the statement is true because for all x’s in D, there is a y in E, such that
x + y = 0.
(b) ∃x ∈ D, ∀y ∈ E, x + y = y.
This statement is true because if we let x = 0 ∈ D, x + y = y, no matter the value
of y.
(6) (Spring 2013, Midterm 1) Consider the domain of real numbers. Define the predicate
P (x) to mean x is (strictly) positive, N (x) to mean x is (strictly) negative.
A. ∀x(P (x) ∨ N (x)).
C. ∀x(P (x) →∼ N (x)).
B. ∃x(∼ P (x)∧ ∼ N (x)).
D. ∃x(P (x) →∼ N (x)).
Before we begin answering the questions, let’s translate each statement to everyday language.
A. All numbers are strictly positive or strictly negative.
B. There is a number that is neither positive nor negative.
C. All positive numbers are not negative.
D. There is a positive number that is not negative.
(a) Which of the statements above is true? (List the letters.)
0 is a number that is neither positive nor negative. This disproves A and proves B
to be true. All positive numbers are in fact not negative so, C is true, leading to D
being true. B, C, and D are true.
(b) Which of the statements can be translated as
Every positive number is not negative.
From our translations above, C.
(c) Which pair of statements are negations of one another? (Give the two letters.)
From our translations, we can see that A and B are negations of each other.
(d) Give a counterexample which proves that
∀x(P (x) → N (x))
is false. Justify your answer briefly.
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We are looking for a counterexample that witnesses the existential statement
∃x(P (x)∧ ∼ N (x))
It’s enough to find one positive number that is not negative. But any
(strictly) positive number will work, for example, my favorite: 17.
(7) (Spring 2013 Midterm 1) Consider a proposition of the form
p → (r ∨ s).
Answer True or False.
(a) The proposition asserts that p is necessary for r ∨ s.
From the book (Lo Section 1, Example 5), “The statement ‘p is necessary
for q’ usually stands for the statement form ∼ p →∼ q” or “q → p.” As
we can see, r ∨ s is necessary for p but not vice versa, so the statement
is false.
(b) The proposition is equivalent to (∼ p ∨ r) ∨ s.
p → (r ∨ s), by the law of implication
∼ p ∨ (r ∨ s)
(∼ p ∨ r) ∨ s, by the Associative Rule
So the statement is true.
(c) In a direct proof, we may assume that r and s are both false.
False. We should only assume that p is true, and proving that at least
one of r and s is true. If we assume that r and s are both false, then
we are probably trying to prove the contrapositive (rather than using a
direct proof).
(8) (Spring 2013 Midterm 1) Fill in the blanks in the following contrapositive proof.
Any step that relies on something other than basic algebra must be justified/proved as
part of your proof.
√
√
• Thm. If 2 is not a rational number, then 1 + 2 is not a rational number either.
• Proof:
√
– Assume (by contrapositive) that
1
+
2 is a rational number
√
– WTS (for contrapositive) that 2 is a rational number.
Fill in the body of this proof . . .
By definition of rational numbers, there are integers a, b with b 6= 0 such that
√
a
1+ 2= .
b
√
We want to show that 2 is rational so we isolate it in the equation:
√
a
a−b
2= −1=
.
b
b
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The
√ integers are closed under subtraction so a − b ∈ Z. Thus, we’ve expressed
2 as a fraction of two integers, with nonzero denominator, hence it is a
rational number.
√
√
– Conclusion, therefore, if 2 is irrational, then so is 1 + 2QED.
(9) Let m, n be fixed integers. Prove that if m, n are both even then m + n is even. And,
prove that if m, n are both odd then m + n is even.
Proof: We give a direct proof of the implication. Assume that m and n are both even.
We WTS that m + n is even. By definition of even integers, there are integers C, D such
that
m = 2C
n = 2D.
Then, by factoring,
m + n = 2C + 2D = 2(C + D).
Since the integers are closed under addition, C + D ∈ Z and we have written m + n as
twice an integer. Thus, it is even and our proof is complete.
(10) Determine whether the property is true for all integers, true for no integers, or true for
some integers and false for other integers. Justify your answers.
(a) (a + b)2 = a2 + b2
True for some integers and false for others. For example, if a = 0 then the
property is true: (0 + b)2 = b2 = 0 + b2 = 02 + b2 . However, if a = 1, b = 1 then the
property is false: LHS = (1 + 1)2 = 22 = 4 and RHS = 12 + 12 = 1 + 1 = 2.
(b) The average of any two odd integers is odd.
True for some integers and false for others. For example the average of 3, 5 is
4, which is not an odd integer. But, the average of 1, 9 is 5, which is an odd integer.
(11) Find the truth set of each predicate.
(a) Predicate: d6 is an integer; Domain: Z
6
∈ Z} = {−6, −3, −2, −1, 1, 2, 3, 6}.
x
(b) Predicate: d6 is an integer; Domain: N
6
{x ∈ N | ∈ Z} = {1, 2, 3, 6}.
x
(c) Predicate: 1 ≤ x2 ≤ 4 ; Domain: R
{x ∈ Z |
x ∈ R | 1 ≤ x2 ≤ 4} = [−2, −1] ∪ [1, 2]
(d) Predicate: 1 ≤ x2 ≤ 4 ; Domain: Z
x ∈ Z | 1 ≤ x2 ≤ 4} = {−2, −1, 1, 2}