Download 3.3 Proofs Involving Quantifiers 1. In exercise 6 of Section 2.2 you

3.3 Proofs Involving Quantifiers
1. In exercise 6 of Section 2.2 you use logical equivalences to show that
∃x(P (x) → Q(x)) is equivalent to ∀xP (x) → ∃xQ(x). Now use the
methods of this section to prove that if ∃x(P (x) → Q(x)) is true,
then ∀xP (x) → ∃xQ(x) is true. (Note: The other direction of the
equivalence is quite a bit harder to prove. See exercise 12 of Section
3.5.)
(Solution)
Proof.
Suppose ∃x(P (x) → Q(x)).
Suppose ∀xP (x).
Since ∃x(P (x) → Q(x)) there exists some object x0 such that
P (x) → Q(x) is true. In other words, P (x0 ) → Q(x0 ). In order for
this statement to be true, one of the following conditions must be
met: P (x0 ) and Q(x0 ) are true, P (x0 ) is false and Q(x0 ) is true, or
P (x0 ) and Q(x0 ) are false.
But then, since ∀xP (x), P (x0 ) is true. Then it follows that Q(x0 ) is
true. Therefore, we have shown ∀xP (x) → ∃xQ(x).
2. Prove that if A and B\C are disjoint, then A ∩ B ⊆ C.
(Solution)
Proof. Suppose A and B\C are disjoint, and suppose x ∈ A ∩ B.
Then since A and B\C are disjoint, if x ∈ A, then x ∈
/ B\C.
By the definition of ∈,
/ x∈
/ B\C is equivalent to ¬x ∈ B\C.
By the definition of \, this is equivalent to ¬((x ∈ B) ∧ (x ∈
/ C)).
By DeMorgan’s law, this is equivalent to ¬(x ∈ B) ∨ ¬(x ∈
/ C).
By the definition of ¬, this is equivalent to x ∈
/ B ∨ x ∈ C.
By the conditional law, this is equivalent to x ∈ B → x ∈ C.
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Therefore, since x ∈ B → x ∈ C and x ∈ A ∩ B, x ∈ C.
Since x was an arbitrary element of A, we can conclude that
A ∩ B ⊆ C.
3. Suppose A ⊆ P(A). Prove that P(A) ⊆ P(P(A)).
(Solution)
Proof. Suppose A ⊆ P(A). Let x be an arbitrary element of A. Then
since A ⊆ P(A) and x ∈ A, x ∈ P(A). Then there exists y such that
∀y(y ∈ x → y ∈ A). Since y ∈ A and A ⊆ P(A), y ∈ P(A). Thus,
since x ∈ P(A) and y ∈ P(A), it follows that x ∈ P(P(A)).
Therefore, we have just shown P(A) ⊆ P(P(A)) as required.
4. The hypothesis of the theorem proven in exercise 3 is A ⊆ P(A).
(a) Can you think of a set A for which this hypothesis is true?
(Solution)
Definition 2.3.2. Suppose A is a set. The power set of A,
denoted P(A), is the set whose elements are all the subsets of
A. In otherwords,
P(A) = {x|x ⊆ A}.
Therefore, if A has an element, since A is a subset of itself, the
hypothesis is true.
If A is an empty set, since the empty set is a subset of every set,
the hypothesis is true.
Therefore, any set will do (such as R, Q, Z...).
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(b) Can you think of another?
(Solution)
There are countless examples of sets where the hypothesis is true.
5. Suppose x is a real number.
(a) Prove that if x 6= 1 then there is a real number y such that
y+1
= x.
y−2
(Solution)
Proof. Let x be an arbitrary real number, and suppose x 6= 1.
Let
2x + 1
y=
x−1
which is defined since x 6= 1. Then,
y+1
=
y−2
2x+1
x−1
2x+1
x−1
+1
−2
=
(2x+1)+(x−1)
x−1
(2x+1)−2(x−1)
x−1
=
3x
x−1
3
x−1
=
3x
= x.
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(b) Prove that if there is a real number y such that
x 6= 1.
y+1
y−2
= x, then
(Solution)
which is defined if y 6= 2. Solving this
Proof. Let x = y+1
y−2
equation for y, we have
y+1
x = y−2
x(y − 2) = y + 1
xy − 2x = y + 1
xy − y = 2x + 1
y(x − 1) = 2x + 1
y = 2x+1
.
x−1
Therefore, in order for y to be defined, x must not be 1.
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6. Prove that for every real number x, if x > 2 then there is a real
number y such that y + y1 = x.
(Solution)
Proof. Let x be an arbitrary number, and suppose x > 2. Let
√
x + x2 − 4
y=
2
which is defined since x > 2. Then,
√
x + x2 − 4
2
1
√
+
y+ =
y
2
x + x2 − 4
√
(x + x2 − 4)2 + 2 · 2
√
=
2(x + x2 − 4)
√
(x2 + 2x x2 − 4 + (x2 − 4)) + 4
√
=
2(x + x2 − 4)
√
2x2 + 2x x2 − 4
√
=
2(x + x2 − 4)
√
2x(x + x2 − 4)
√
=
2(x + x2 − 4)
= x.
7. Prove that if F is a family of sets and A ∈ F, then A ⊆ ∪F.
(Solution)
Proof. Suppose A ∈ F. Let x be an armitrary element of F. Since
A ∈ F and x ∈ A, x ∈ F. Therefore, clearly, x ∈ ∪F. But x was an
arbitrary element of F, so this shows that A ⊆ ∪F.
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8. Prove that if F is a family of sets and A ∈ F, then ∩F ⊆ A.
(Solution)
Proof. Suppose A ∈ F. Let x be an arbitrary element of ∩F. Then
by the definition 2.3.5., ∀A(A ∈ F → x ∈ A). Since A ∈ F and
∀A(A ∈ F → x ∈ A), it follows that x ∈ A. But x was an arbitrary
element of ∩F, so this shows that ∩F ⊆ A, as required.
9. Suppose F and G are families of sets. Prove that if F ⊆ G then
∪F ⊆ ∪G.
(Solution)
Proof. Suppose F ⊆ G. Let x be an arbitrary element of ∪F. By
definition 2.3.5., x ∈ ∪F means ∃A(A ∈ F ∧ x ∈ A). Then, since
A ∈ F and F ⊆ G, A ∈ G. Therefore, it follows that
∃A(A ∈ G ∧ x ∈ A). But x was an arbitrary element of ∪F, so this
shows that ∪F ⊆ ∪G, as required.
10. Suppose F and G are nonempty families of sets. Prove that if F ⊆ G
then ∩G ⊆ ∩F.
(Solution)
Proof. Suppose F ⊆ G. Let A be an arbitrary element of F. Then
since F ⊆ G and A ∈ F, A ∈ G.
Now, let x be an arbitrary element of ∩G, which is defined since G is
nonempty. Then, by the definition 2.3.5., ∀A(A ∈ G → x ∈ A). Thus,
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x ∈ A. Since A was an arbitrary element of F and x ∈ A, it follows
that x ∈ ∩F, which is defined since F is nonempty. But x was an
arbitrary element of ∩G, so this shows that ∩G ⊆ ∩F.
11. Suppose {Ai |i ∈ I} is an indexed family of sets. Prove that
∪i∈I P(Ai ) ⊆ P(∪i∈I Ai ). (Hint: First make sure you know what all
the notation means!)
(Solution)
Proof. Suppose {Ai |i ∈ I} is an indexed family of sets. Let x be an
arbitrary element of ∪i∈I P(Ai ). Let y be an arbitrary element of x.
Then, by the alternative notation of union of an indexed family of
sets, ∃i ∈ I(∀y(y ∈ x → y ∈ Ai )). In other words, there exists some
i0 ∈ I such that every element of x is an element of Ai0 . Therefore,
y ∈ ∪i∈I Ai . Since y was an arbirary element of x and y ∈ ∪i∈I Ai , we
can conclude that x ∈ P(∪i∈I Ai ). But x was an arbitrary element of
∪i∈I P(Ai ), so this shows that ∪i∈I P(Ai ) ⊆ P(∪i∈I Ai ), as required.
12. Prove the converse of the statement proven in Example 3.3.5. In
other words, prove that if F ⊆ P(B) then ∪F ⊆ B.
(Solution)
Proof. Suppose F ⊆ P(B). Let x be an arbitrary element of ∪F.
Then by definition 2.3.5., ∃A(A ∈ F ∧ x ∈ A). In other words, there
exists some set A such that A ∈ F and x ∈ A. Since A ∈ F and
F ⊆ P(B), A ∈ P(B). Then, by the definition of power set, all the
element of A is an element of B. Since x ∈ A, we can conclude x ∈ B.
But x was an arbitrary element of ∪F, so this shows that ∪F ⊆ B, as
required.
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13. Suppose F and G are nonempty families of sets, and every element of
F is a subset of every element of G. Prove that ∪F ⊆ ∩G.
(Solution)
Proof. Suppose F and G are nonempty families of sets, and every
element of F is a subset of every element of G. Let x be an arbitrary
element of ∪F. Then, by definition 2.3.5., ∃A(A ∈ F ∧ x ∈ A). In
other words, there exists some set A such that A ∈ F and x ∈ A.
Since A ∈ F and every element of F is a subset of every element of G,
A ∈ G. Therefore, since A ∈ G and x ∈ A, by the definition 2.3.5.,
x ∈ ∩G, which is defined since G is not an emptyset. But x was an
arbitrary element of ∪F, so this shows that ∪F ⊆ ∩G, as required.
14. In this problem all variables range over Z, the set of all integers.
(a) Prove that if a|b and a|c, then a|(b + c).
Proof. Let a, b, and c be arbitrary integers and suppose a|b and
a|c. Since a|b, we can choose some integer m such that ma = b.
Similarly, since a|c, we can choose an integer n such that na = c.
Therefore b + c = ma + na = (m + n)a, so since m + n is an
integer, a|(b + c).
(b) Prove that if ac|bc and c 6= 0, then a|b.
Proof. Let a, b, and c be arbitrary integers and suppose ac|bc and
c 6= 0. Since ac|bc, we can choose some integer m such that
mac = bc. Since c 6= 0, we can devide both sides by c. Therefore
ma = b, so since m is an integer, a|b.
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15. Consider the following theorem:
Theorem. For every real number x, x2 ≥ 0.
What’s wrong with the following proof of the theorem?
Proof. Suppose not. Then for every real number x, x2 < 0. In
particular, plugging in x = 3 we would get 9 < 0, which is clearly
false. This contradiction shows that for every number x, x2 ≥ 0.
(Solution)
We cannot let one example, x = 3 represent the whole set of real
numbers for the statement to a contradiction.
Proof.
There are three cases to be considered: x < 0, x = 0, and x > 0.
Case 1. x < 0.
Since x < 0, multiplying both sides by x, we would get x · x > 0 · x,
which is equivalent to x2 > 0.
Case 2. x = 0.
Multiplying both sides by x, we would get x · x = 0 · x, which is
equivalent to x2 = 0.
Case 3. x > 0.
Since x > 0, multiplying both sides by x, we would get x · x > 0 · x,
which is equivalent to x2 > 0.
Therefore, we have shown that for every real number x, x2 ≥ 0 (the
equality holds only when x = 0).
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16. Consider the following incorrect theorem:
Incorrect Theorem. If ∀x ∈ A(x 6= 0) and A ⊆ B then
∀x ∈ B(x 6= 0).
(a) What’s wrong with the following proof of the theorem?
Proof. Let x be an arbitrary element of A. Since ∀x ∈ A(x 6= 0),
we can conclude that x 6= 0. Also since A ⊆ B, x ∈ B. Since
x ∈ B, x 6= 0, and x was arbitrary, we can conclude that
∀x ∈ B(x 6= 0).
(Solution)
(p. 105) In particular, you must not assume that x is equal to
any other object already under discussion in the proof.
We cannot assume that x 6= 0 is also true for B without
appropriate argument.
(b) Find a counterexample to the theorem. In other words, find an
example of sets A and B for which the hypotheses of the
theorem are true but the conclusion is false.
(Solution)
Let A be a set of all positive integers (A = Z+ ), and B be a set
of all integers (B = Z). By the definition of Z+ , A does not have
0 as its element. Furthermore, all the elements in A belong to B,
so A ⊆ B is true. But then, by the definition of Z, 0 is an
element of B, so this theorem is not true.
17. Consider the following incorrect theorem:
Incorrect Theorem. ∃x ∈ R∀y ∈ R(xy 2 = y − x).
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What’s wrong with the following proofs of the theorem?
Proof. Let x = y/(y 2 + 1). Then
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y − x = y − y2y+1 = y2y+1 = y2y+1 · y 2 = xy 2 .
(Solution)
Depending on the value of y, x will change its value. Furthermore,
since the goal is to find x such that ∀y ∈ R(xy 2 = y − x), we cannot
start the proof by assigning some value to x.
18. Consider the following incorrect theorem:
Incorrect Theorem. Suppose F and G are families of sets. If ∪F
and ∪G are disjoint, then so are F and G.
(a) What’s wrong with the following proof of the theorem?
Proof. Suppose ∪F and ∪G are disjoint. Suppose F and G are
not disjoint. Then we can choose some set A such that A ∈ F
and A ∈ G. Since A ∈ F, by exercise 7, A ⊆ ∪F, so every
element of A is in ∪F. Similarly, since A ∈ G, every element of
A is in ∪G. But then every element of A is in both ∪F and ∪G,
and this is impossible since ∪F and ∪G are disjoint. Thus, we
have reached a contradiction, so F and G must be disjoint.
(Solution)
If A in the proof is not an empty set, then the contradiction
holds. But if A is an empty set, even though elements of other
sets in F and G do not duplicate each other, the contradiction
does not hold. In other words, there is an occasion such that
F ∩ G = A, but (∪F) ∩ (∪G) = ∅, even though none of the other
sets in F and G duplicates each other.
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(b) Find a counterexample to the theorem.
(Solution)
For example, suppose F = {A, E}, where A is an empty set, and
E is a set of all even numbers. Suppose G = {A, O}, where A is,
again, an emptyset, and O is a set of all odd numbers. Then
(∪F) ∩ (∪G) = ∅, but F ∩ G = A, which is a set with an empty
set as its element.
19. Prove that for every real number x there is a real number y such that
for every real number z, yz = (x + z)2 − (x2 + z 2 ).
Proof. We will consider two cases z = 0 and z 6= 0.
Case 1. z = 0.
yz = y · 0 = 0, and
(x + z)2 − (x2 + z 2 ) = (x + 0)2 − (x2 + 02 ) = x2 − x2 = 0. Therefore,
no matter what y is, the equality holds.
Case 2. z 6= 0.
Let x and z arbitrary real numbers, and y = 2x. Then,
yz = (2x)z = 2xz = x2 + 2xz + z 2 − (x2 + z 2 ) = (x + z)2 − (x2 + z 2 )
By Case 1, we know that the value of y does not matter if z = 0. So
we can let y = 2x for both cases.
(a) Comparing the various rules for dealing with quantifiers in
proofs, you should see a similarity between the rules for goals of
the form ∀xP (x) and givens of the form ∃xP (x). What is this
similarity? What about the rules for goals of the form ∃xP (x)
and givens of the form ∀xP (x)?
(Solution)
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To prove a goal of the form: ∀xP (x)
Let x stand for an arbitrary object, and prove P (x). (If the
letter x already stands for something in the proof, you will have
to use a different letter for the arbitrary object.)
To use a given of the form: ∃xP (x)
Introduce a new variable, say x0 , into the proof, to stand for a
particular object for which P (x0 ) is true.
The both of these rules above help us decide what fundamental
properties we assign in x.
To prove a goal of the form: ∃xP (x)
Find a value of x that makes P (x) true. Prove P (x) for this
value of x.
To use a given of the form: ∀xP (x)
You may plug in any value, say a for x, and conclude that P (a)
is true.
Both of these rules above are applied to add more properties to
x that will help us prove the theories.
(b) Can you think of a reason why these similarities might be
expected? (Hint: Think about how proof by contradiction works
when the goal starts with a quantifier.)
Proof by contradiction works by negating the goals and adding
them as givens and lead to a contradiction. If we negate our
goals involving quantifiers, then by the quantifier negation law,
existential quantifiers become universal, and vice versa, and,
goals change into givens. Naturally, these similarities are
expected.
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