Math 220
Exam 1 Sample Problems
Solution Guide
September 24, 2013
1.
(a) Statement
(b) Open sentence
(c) Open sentence
(d) Neither
(e) Statement
2.
(a) There exists an integer x such that x is even and x +1
2 is an integer.
(b) There exists a ∈ E and there exists b ∈ E such that ab is not a multiple of 4 or a + b is not a multiple of 4.
(c) There exists x ∈
R such that for all y ∈
R
, y
4
= x .
(d) ∃ x ∈
R
, ∀ y ∈
R
, y = 0 and xy ≥ 0.
(e) There exists n ∈
Z
+ such that n is a prime number and 2 n − 1 is not a prime number.
3.
(a) True
(b) False. A counterexample is a = 2 and b = 4. Then ab = 8, which is a multiple of 4, but a + b = 6, which is not a multiple of 4.
(c) False. Take x = − 1. For any y ∈ y
4
= − 1.
R
, we have y
4 ≥ 0, so there cannot be any y such that
(d) True. (For any x ∈
R
, take y = 0. Then the conclusion is satisfied.)
(e) False. Take n = 11. Then n is prime, but 2
11 − 1 = 2047 = 23 · 89 is not prime. (This is a nice problem, but it would be difficult to do on an exam, where you will not have a calculator.)
4.
(a) The first statement is true, and the second is false. To see that “ ∃ x ∈ [ − 1 , 1] so that
∀ y ∈
R
, sin( y ) = x ” is false, we consider its negation: “ ∀ x ∈ [ − 1 , 1] , ∃ y ∈
R so that sin( y ) = x .” Since sin( y ) takes on more than one value (in fact infinitely many in [ − 1 , 1]), for any given x ∈ [ − 1 , 1] we can find y ∈
R so that sin( y ) = x .
(b) This one is a little trickier, but the first statement is false, and the second is true. We want to show that the first statement is false. [Scratchwork: We solve the equation y
2
− 1 y +1
= x for y . So for a fixed x ∈
R
, y
2 − 1
= x ⇔ y
2
− 1 = x ( y + 1) ⇔ y
2
− xy − x = 0 .
y + 1
Using the quadratic formula, y = x ±
√ x 2 + 4 x
.
2
1

But this only has a solution if x
2
+4 x ≥ 0.] We would not need to include the scratchwork, but we could simply do the following. Let x = − 1. We wish to solve the equation y
2 − 1
= − 1 y + 1 for y . Simplifying this, we wish to solve y
2
+ y + 1 = 0, which has solutions y = by the quadratic formula. Therefore no such y ∈
R exists satisfying y
2 − 1 y +1
= − 1.
− 1 ±
√
− 3
2
5.
(a) ( ¬ P ∨ Q ) ∨ ( P ∧ ¬ Q )
(b) ( P ∨ ¬ Q ) ∧ ( Q ∧ ¬ P )
6.
(a)
P Q R P ∨ Q ¬ Q ∨ ¬ R ( P ∨ Q ) ∧ ( ¬ Q ∨ ¬ R )
T T T T F F
T T F
T F T
T F F
F T T
F T F
F F T
F F F
T
T
T
T
T
F
F
T
T
T
F
T
T
T
T
T
T
F
T
F
F
It is not necessary to include the fourth and fifth columns, but it helps in understanding the last column.
(b) Here it might be helpful, though not necessary, to use the fact that P ⇒ Q is logically equivalent to ¬ P ∨ Q .
P Q P ∧ ¬ Q P ⇒ Q ( P ∧ ¬ Q ) ∨ ( P ⇒ Q )
(= ¬ P ∨ Q )
T T F T T
T F
F T
F F
T
F
F
F
T
T
T
T
T
Since the last column is all True, we see that ( P ∧ ¬ Q ) ∨ ( ¬ P ∨ Q ) is a tautology .
(c) First of all, we have
P Q R P ∧ Q ¬ P ∨ R
T
T
T
T
T
F
T
F
T
T F F
F T T
F
F
F
T
F
F
F
T
F
T
T
F
F
F
F
F
F
T
F
T
F
T
T
T
T
R ⇒ ¬ Q
(= ¬ R ∨ ¬ Q )
F
T
T
T
T
T
F
T
( P ∧ Q ) ∧ ( ¬ P ∨ R ) ∧ ( R ⇒ ¬ Q )
F
F
F
F
F
F
F
F
2

It’s worth pointing out in passing (though not required by the directions of the problem) that the right-most statement form is a contradiction. In any case, we find
P Q R ( P ∧ Q ) ∧ ( ¬ P ∨ R ) ∧ ( R ⇒ ¬ Q ) ⇒ R
T T T
T T F
T F T
T
T
T
T F F
F T T
F T F
F F T
F F F
T
T
T
T
T
Thus the given statement form is a tautology .
7. Throughout these problems we use several times the following logical equivalences:
¬ ( P ∧ Q ) ⇔ ¬ P ∨ ¬ Q,
¬ ( P ∨ Q ) ⇔ ¬ P ∧ ¬ Q,
P ∧ ( Q ∨ R ) ⇔ ( P ∧ Q ) ∨ ( P ∧ R ) ,
P ∨ ( Q ∧ R ) ⇔ ( P ∨ Q ) ∧ ( P ∨ R ) ,
P ⇒ Q ⇔ ¬ P ∨ Q.
(a)
¬ (( P ∧ Q ) ∨ ( P ∧ ¬ R )) ⇔ ( ¬ ( P ∧ Q ) ∧ ¬ ( P ∧ ¬ R ))
⇔ ( ¬ P ∨ ¬ Q ) ∧ ( ¬ P ∨ R )
⇔ ¬ P ∨ ( ¬ Q ∧ R ) .
(b)
(c)
Q ∧ ( P ∨ ¬ Q ) ⇔ ( Q ∧ P ) ∨ ( Q ∧ ¬ Q )
⇔ Q ∧ P .
The last equality comes from the fact that Q ∧ ¬ Q is always false.
¬ ( ∃ x so that ( P ( x ) ∧ ¬ Q ( x )) ∨ ( P ( x ) ∧ R ( x )))
⇔ ∀ x, ¬ (( P ( x ) ∧ ¬ Q ( x )) ∨ ( P ( x ) ∧ R ( x )))
⇔ ∀ x, ¬ P ( x ) ∨ ( Q ( x ) ∧ ¬ R ( x )) .
(d)
The last step is a modification of part (a).
( P ⇒ Q ) ∧ ( R ∨ Q ) ⇔ ( ¬ P ∨ Q ) ∧ ( R ∨ Q )
⇔ ( ¬ P ∧ R ) ∨ Q .
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8.
(a) A = (3 , 5]
(b) B = ( −∞ , − 4) ∪ (4 , ∞ )
(c) A ∩ B = (4 , 5]
(d) A − B = (3 , 4]
(e) A ∩ ( B ∪ A ) = ( A ∩ B ) ∪ ( A ∩ A ) = ( A ∩ B ) ∪ ∅ = A ∩ B = (3 , 4]
9.
(a) (0 , 2]
(b) [1 , 2]
10. The elements are { 0 } , { 1 } , { 0 , 1 } , { 0 , 2 } , { 1 , 2 } , { 0 , 1 , 2 } .
11.
(a) By the inclusion-exclusion principle, | A ∪ B | = | A | + | B |−| A ∩ B | , so | A ∪ B | = 10+8 − 6 =
12. Likewise, | B ∪ C | = 8 + 14 − 3 = 19.
(b) Iterating the inclusion-exclusion rule,
| A ∪ B ∪ C | = | A ∪ B | + | C | − | ( A ∪ B ) ∩ C |
= | A | + | B | − | A ∩ B | + | C | − | ( A ∩ C ) ∪ ( B ∩ C ) |
= | A | + | B | + | C | − | A ∩ B | − ( | A ∩ C | + | B ∩ C | − | A ∩ B ∩ C | )
= | A | + | B | + | C | − | A ∩ B | − | A ∩ C | − | B ∩ C | + | A ∩ B ∩ C | .
From this we see that
| A ∪ B ∪ C | = 10 + 8 + 14 − 6 − 5 − 3 + | A ∩ B ∩ C |
= 18 + | A ∩ B ∩ C | .
Since 0 ≤ | A ∩ B ∩ C | ≤ 3, we see that the possible values for | A ∪ B ∪ C | are 18, 19, 20, and 21.
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