Download Midterm 2 Review Answers

Midterm
2
Review
Answers
A. Are the following propositions true of false? (N includes 0)
1. ∃!x ∈ Z[(x − 2)2 = 9]. FALSE
2. ∃!x ∈ N[(x − 2)2 = 9]. TRUE
3. [∀x∃yP (x, y)] → [∃x∀yP (x, y)]. FALSE
4. [A ⊆ B] → [A − B = ∅]. TRUE
5. ∀a, b, c ∈ Z[(a | b) ∧ (b | c) → (a | c)]. TRUE
6. Z − N is infinite. TRUE
7. Z and R are disjoint. FALSE
B. Provide counterexamples disproving the following propositions. (Explain
why they are counterexamples)
1. ∃xP (x) ∧ ∃xQ(x) ≡ ∃x(P (x) ∧ Q(x)).
Define P (x) ≡ “x is even”, Q(x) ≡ “x is odd”. Then ∃xP (x) is true for
x = 2 and ∃xQ(x) is true for x = 3. However, if the universe is restricted to
be Z, then P (x) ≡ ¬Q(x), meaning that P (x)∧Q(x) ≡ ¬Q(x)∧Q(x) ≡ F ,
so the right-hand side is always false.
2. ∀x∃!y[x = y 2 ].
Let x = 4. Since 4 = 22 = (−2)2 , there are actually two values of y, both
2 and −2, which solve the equation. Therefore the uniqueness claim is
false, and the overall proposition is false.
3. A − (B ∪ C) = (A − B) ∪ (A − C).
Set A = {1, 2, 3, 4}, B = {1, 2}, C = {3, 4}. Then A − (B ∪ C) =
{1, 2, 3, 4} − ({1, 2} ∪ {3, 4}) = {1, 2, 3, 4} − {1, 2, 3, 4} = ∅, which is not
equal to (A − B) ∪ (A − C) = ({1, 2, 3, 4} − {1, 2}) ∪ ({1, 2, 3, 4} − {3, 4}) =
{3, 4} ∪ {1, 2} = {1, 2, 3, 4}.
4. Suppose that x and y are real numbers and x 6= 3. If x2 y = 9y then y = 0.
Set x = −3. Then x2 y = (−3)2 y = 9y. Therefore y can be any real
number, such as 1. It does not need to be 0.
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C. Enumerate every element of the following sets. Assume A = {1, a, B}, B =
{a, b}, C = {1, 2}. (N includes 0)
1. P (A) = {∅, {1}, {a}, {B}, {1, a}, {1, B}, {a, B}, {1, a, B}}.
2. A × B = {(1, a), (1, b), (a, a), (a, b), (B, a), (B, b)}.
3. A × C × A =
{(1, 1, 1), (1, 1, a), (1, 1, B), (1, 2, 1), (1, 2, a), (1, 2, B),
(a, 1, 1), (a, 1, a), (a, 1, B), (a, 2, 1), (a, 2, a), (a, 2, B),
(B, 1, 1), (B, 1, a), (B, 1, B), (B, 2, 1), (B, 2, a), (B, 2, B)}.
4. P ({x ∈ N|x < 2}) = {∅, {0}, {1}, {0, 1}}.
5. B×P (B) = {(a, ∅), (a, {a}), (a, {b}), (a, {a, b}), (b, ∅), (b, {a}), (b, {b}), (b, {a, b})}.
6. P (A) − P (B) = {{1}, {B}, {1, a}, {1, B}, {a, B}, {1, a, B}}
7. P (P (B)) =
{∅, {∅}, {{a}}, {{b}}, {{a, b}},
{∅, {a}}, {∅, {b}}, {∅, {a, b}}, {{a}, {b}}, {{a}, {a, b}}, {{b}, {a, b}},
{∅, {a}, {b}}, {∅, {a}, {a, b}}, {∅, {b}, {a, b}}, {{a}, {b}, {a, b}},
{∅, {a}, {b}, {a, b}}}.
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D. Prove the following theorems.
1. For every integer n, n3 is even iff n is even.
Proof:
Case 1: Start with the proof of (n is even) → (n3 is even).
Assume the antecedant, and prove the consequent:
1.
≡
2.
≡
3.
≡
4.
n is even
{Def. of “even”}
∃k ∈ Z[n = 2k]
{Cube both sides of equation}
∃k ∈ Z[n3 = (2k)3 ]
{Simplify right-hand side, and separate a 2}
∃k ∈ Z[n3 = 2(4k 3 )]
5.
6.
7.
8.
Define i = 4k 3 .
i is an integer because Z is closed under multiplication.
Therefore, ∃i ∈ Z[n3 = 2i] {substitution of i}.
By the def. of “even”, n3 is even.
End Case 1: (n is even) → (n3 is even) is proven.
Case 2: Prove (n3 is even) → (n is even).
Prove the contrapositive:
9. (n is not even) → (n3 is not even)
≡ {n ∈ Z (given) and n3 ∈ Z (via closure), so “not even” means “odd”}
10. (n is odd) → (n3 is odd)
Assume the antecedant, and prove the consequent:
11. n is odd
≡ {Def. of “odd”}
12. ∃k ∈ Z[n = 2k + 1]
≡ {Cube both sides of equation}
13. ∃k ∈ Z[n3 = (2k + 1)3 ]
≡ {Expand right-hand side}
14. ∃k ∈ Z[n3 = 8k 3 + 12k 2 + 6k + 1]
≡ {Factor out a 2 on the right-hand side}
15. ∃k ∈ Z[n3 = 2(4k 3 + 6k 2 + 3k) + 1]
16.
17.
18.
19.
Define i = 4k 3 + 6k 2 + 3k.
i is an integer because Z is closed under multiplication and addition.
Therefore, ∃i ∈ Z[n3 = 2i + 1] {substitution of i}.
By the def. of “odd”, n3 is odd.
20. Contrapositive proven, therefore (n3 is even) → (n is even) as well.
End Case 2: (n3 is even) → (n is even) is proven.
21. Combining the results from cases 1 and 2 proves:
For every integer n, n3 is even iff n is even.
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2. A ⊆ B iff P (A) ⊆ P (B).
Proof:
Start with Lemma: A ⊆ B ∧ B ⊆ C → A ⊆ C.
1. Let x ∈ A
⇒ {From the given A ⊆ B and the def. of ⊆}
2. x ∈ B
⇒ {From the given B ⊆ C and the def. of ⊆}
3. x ∈ C
4. Therefore, x ∈ A → x ∈ C
≡ {Def. of ⊆}
5. A ⊆ C
Lemma proven: A ⊆ B ∧ B ⊆ C → A ⊆ C.
Move on to proof of: A ⊆ B iff P (A) ⊆ P (B).
Case 1: Prove A ⊆ B → P (A) ⊆ P (B).
Assume A ⊆ B and prove P (A) ⊆ P (B).
6. Let x ∈ P (A)
⇒ {Def. powerset}
7. x ⊆ A
⇒ {Assumption that A ⊆ B, and Lemma}
8. x ⊆ B
⇒ {Def. powerset}
9. x ∈ P (B)
10. Therefore, x ∈ P (A) → x ∈ P (B)
≡ {Def. of ⊆}
11. P (A) ⊆ P (B)
End Case 1: A ⊆ B → P (A) ⊆ P (B) proven.
Case 2: Prove P (A) ⊆ P (B) → A ⊆ B.
Assume P (A) ⊆ P (B) and prove A ⊆ B.
Note that every set is an element of its own powerset:
12. A ∈ P (A)
⇒ {From the given P (A) ⊆ P (B) and the def. of ⊆}
13. A ∈ P (B)
⇒ {Def. powerset}
14. A ⊆ B
End Case 2: P (A) ⊆ P (B) → A ⊆ B proven.
15. Combining the results from cases 1 and 2 proves:
A ⊆ B iff P (A) ⊆ P (B).
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3. The length of the hypotenuse of a right triangle is less than the sum of
the lengths of the two legs.
Proof:
1. Let a, b, and c be different lengths of sides of a right triangle, in which
c corresponds to the hypotenuse.
2. To prove that c < a + b, assume the opposite, i.e. c ≥ a + b, and seek a
contradiction.
3.
4.
5.
6.
7.
8.
Because a, b, and c are triangle side lengths, they are positive numbers.
Therefore, the assumed inequality can be squared to get c2 ≥ (a + b)2 .
Expand the right-hand side to get c2 ≥ a2 + 2ab + b2 .
Because a and b are positive, a2 + 2ab + b2 > a2 + b2 .
Combining the inequalities in lines 5 and 6 yields c2 > a2 + b2 .
However, according to the Pythagorean Theorem: c2 = a2 + b2 .
9. Line 8 contradicts line 7, so the assumption in line 2 must have been
false.
10. ∴ c < a + b.
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E. Look at the proofs of the following theorems. Each one has a flaw that makes
the proof invalid. Identify the step that is flawed and explain why it is flawed.
(Proof steps numbered for convenience)
1. Incorrect Theorem: 0 = 1.
Proof:
1. Let x = 0, and modify the equation as follows:
2.
≡
3.
≡
4.
≡
5.
≡
6.
x=0
{Multiply both sides by (x − 1), 0 dominates in multiplication}
x(x − 1) = 0
{Divide both sides by x, 0 divided by any number is still 0}
x−1=0
{Add 1 to both sides}
x=1
{Substitute the known value of x}
0=1
Line 4 is an illegal operation because it is known that x = 0 (line 2), yet
to divide by 0 as in line 4 is not allowed. Everything derived from this
point is invalid.
2. Incorrect Theorem: A ∪ (B ∩ C) ∪ C C = A ∪ B.
Proof:
Start from the left-hand side of the equation and use set identities to work
towards the right-hand side.
1.
=
2.
=
3.
=
4.
=
5.
=
6.
A ∪ (B ∩ C) ∪ C C
{∪ distributes over ∩}
(A ∪ B) ∩ (A ∪ C) ∪ C C
{∪ associativity}
(A ∪ B) ∩ (A ∪ (C ∪ C C ))
{∪ complement}
(A ∪ B) ∩ (A ∪ U )
{∪ domination}
(A ∪ B) ∩ U
{∩ identity}
A∪B
The error can be thought of as occurring either in line 2 or line 3. The
mistake is due to improperly grouping sub-expressions, i.e. being lazy
about the use of parentheses. In line 2 the sets (A ∪ B) and (A ∪ C)
should be grouped together by yet another pair of parentheses due to
the ∩ between them. This ∩ operation separates these two sets from the
adjacent set C C , around which an improper use of ∪ associativity occurs
in line 3. The error in line 3 is a direct result of the laziness with the
parentheses in line 2.
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3. Incorrect Theorem: Suppose x and y are real numbers and x + y = 10.
Then x 6= 3 and y 6= 8.
Proof:
Assume the conclusion of the theorem is false and seek a contradiction.
1. Then x = 3 and y = 8.
2. Add both values to get x + y = 3 + 8 = 11.
3. However, x + y = 10 was given, and 10 6= 11.
4. Because the given equation was contradicted, the theorem’s conclusion
must be true.
∴ x 6= 3 and y 6= 8.
The flaw is in line 1: The conclusion of the theorem is x 6= 3 ∧ y 6= 8,
therefore its negation (via De Morgan) is x = 3 ∨ y = 8, but on line 1 this
is improperly taken to be x = 3 ∧ y = 8.
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F. Decide whether or not the following propositions are true, and then either
prove or disprove them.
1. Suppose A ⊆ B ⊆ C. Then C − B ⊆ C − A. (TRUE)
Proof:
Consider the goal of the proof:
1. C − B ⊆ C − A
≡ {Def. of ⊆}
2. x ∈ C − B → x ∈ C − A
3. So assume x ∈ C − B and prove x ∈ C − A.
4. Assume x ∈
/ C − A and seek a contradiction.
5.
≡
7.
≡
8.
x∈
/ C −A
{Rewrite the abbreviation for ∈,
/ def. of set difference}
¬(x ∈ C ∧ x ∈
/ A)
{De Morgan, ∈
/ abbreviation}
x∈
/ C ∨x∈A
There are now two cases. Each case must lead to a contradiction.
Case 1:
9. Assume x ∈
/ C.
Take the assumption from line 3:
10. x ∈ C − B
≡ {Def. set difference}
11. x ∈ C ∧ x ∈
/B
⇒ {simplification}
12. x ∈ C
13. Line 12 contradicts the assumption on line 9, so the assumption must
have been wrong. Therefore a contradiction has been found in case 1.
Case 2:
14. Assume x ∈ A.
⇒ {From the given that A ⊆ B, and the def. of ⊆}
15. x ∈ B
Now take the assumption from line 3:
16. x ∈ C − B
≡ {Def. set difference}
17. x ∈ C ∧ x ∈
/B
⇒ {simplification}
18. x ∈
/B
19. Line 18 contradicts the conclusion on line 15, so the conclusion must
have been wrong. Therefore a contradiction has been found in case 2.
20. Since both cases lead to a contradiction, the assumption on line 4
must have been wrong. Therefore, x ∈ C − A.
21. Therefore C − B ⊆ C − A.
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2. P (A − B) − (P (A) − P (B)) = ∅. (FALSE)
Counterexample:
Let A = {1}, B = {2}. Then P (A) = {∅, {1}} and P (B) = {∅, {2}}. So
P (A) − P (B) = {{1}}. A − B = {1}, so P (A − B) = {∅, {1}}. Finally,
P (A − B) − (P (A) − P (B)) = {∅, {1}} − {{1}} = {∅}.
{∅} =
6 ∅, thus disproving the claim that P (A − B) − (P (A) − P (B)) = ∅.
3. Suppose a, b and c are integers. If a | b and a | c then a | (b + c). (TRUE)
Proof:
1.
2.
3.
4.
5.
Assume a | b and a | c and prove a | (b + c).
a | b ≡ ∃k ∈ Z[b = ka] {Def. of “divides”}
a | c ≡ ∃i ∈ Z[c = ia] {Def. of “divides”}
Add the equations in lines 2 and 3 to get b + c = ka + ia.
Factor a out from the right-hand side of the equation: b + c = (k + i)a.
6.
7.
8.
9.
Define q = k + i.
q ∈ Z because Z is closed under addition.
Therefore, ∃q ∈ Z[b + c = qa]. {Substitution of q}
So by the def. of “divides”, a | (b + c).
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