Download Math 109. Instructor: Chow Homework #2 Hints Problem 1: Prove

Math 109. Instructor: Chow
Homework #2 Hints
Problem 1: Prove, using logical argument from the definitions, that
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) .
That is, prove that x ∈ A ∪ (B ∩ C) if and only if x ∈ (A ∪ B) ∩ (A ∪ C) .
Hint: You may use the fact that: ‘P or (Q and R)’ is logically equivalent
to ‘(P or Q) and (P or R)’.
Solution:
x ∈ A ∪ (B ∩ C) ⇔ x ∈ A or (x ∈ B and x ∈ C) ⇔ (x ∈ A or x ∈ B)
and (x ∈ A or x ∈ C) ⇔ x ∈ (A ∪ B) ∩ (A ∪ C) . Here we used the fact that:
‘P or (Q and R)’ is logically equivalent to ‘(P or Q) and (P or R)’.
Problem 2:
(i) Prove: ∀x ∈ Z+ , ∃y ∈ Z+ , y > 3x + 2.
(ii) Prove: ∀x ∈ Z, ∃y ∈ Z, y > x2 + 9.
(iii) DISProve: ∃x ∈ R, ∀y ∈ R, y < x2 .
(iv) Disprove: ∀n ∈ Z+ , ∃m ∈ Z+ , n ≤ m2 ≤ n + 39.
Solution:
(i) Let x be an arbitrary positive integer. Then let y = 3x + 3. Clearly
y > 3x + 2, and y is also a positive integer. This proves the existence of y.
(ii) Let x be an arbitrary integer. Then let y = x2 + 10. Clearly
y > x2 + 9, and y is also an integer. This proves existence of y.
(iii) Want to show, ∀x ∈ R, ∃y ∈ R, y ≥ x2 .
Let x be an arbitrary real number, let y = x2 + 1. Clearly y ∈ R and y ≥ x2 .
This proves the existence of y, and hence disproves the original statement.
(iv) Want to show, ∃n ∈ Z+ , ∀m ∈ Z+ , either m2 < n or m2 > n + 39.
Let n = 1601 ∈ Z+ , ∀m ∈ Z+ , either m ≤ 40 or m ≥ 41, i.e.either m2 ≤
1600 < 1601 = n or m2 ≥ 1681 > 1640 = n + 39. This proves the claim,
hence disproves the original statement.
Problem 3:
(i) Prove: ∃x ∈ R, ∀y ∈ R, 2xy = x3 + 2x2 .
(ii) Disprove: ∃x ∈ R, ∀y ∈ R, 2xy > 1.
(iii) Prove: ∀x ∈ R, ∃y ∈ R, x2 + 1 y = x3 + 2x2 .
(iv) Prove: ∀x ∈ R − Q, ∃y ∈ R − Q, xy = 1.
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Solution:
(i) Let x = 0 ∈ R, then ∀y ∈ R, 2xy = 0 = x3 + 2x2 , this proves the
existence of x.
(ii) Want to show, ∀x ∈ R, ∃y ∈ R, 2xy ≤ 1.
Let x be an arbitrary real number, Let y = 0, then 2xy = 0 ≤ 1. This
proves the existence of y and hence disproves the original statement.
(iii) Let x be an arbitrary real number, then x2 +1 ≥ 1 > 0, in particular
3
2
x3 +2x2
2
+ 1 6= 0. Let y = xx+2x
(x2 + 1) =
2 +1 , then y ∈ R and y(x + 1) =
x2 +1
(x3 + 3x2 ). This proves the existence of y.
x2
(iv) Let x be an arbitrary irrational number, then x 6= 0 ∈ Q. Let
y = x1 , then y ∈ R and xy = 1. Need only to show y ∈ R − Q.
Suppose this is not true, i.e.y ∈ Q. Then y = m
n for some nonzero integers
n
m,n (notice here y 6= 0 since y = x1 ), it follows that x = m
∈ Q, contradicting
that x ∈ R − Q, this proves y ∈ R − Q and hence the statement.
Problem 4: Let n ∈ Z.
Let P (n) be the statement: There exists q ∈ Z such that n = 5q + 3.
Let Q(n) be the statement: There exists p ∈ Z such that n2 = 5p + 4.
Prove that P (n) implies Q(n).
Hint: This is a direct argument, not a proof by induction.
Solution: P (n) is true ⇒ ∃q ∈ Z, such that n = 5q + 3 ⇒ n2 =
25q 2 + 30q + 9 = 5(5q 2 + 6q + 1) + 4 ⇒ ∃p = 5q 2 + 6q + 1, such that
n2 = 5p + 4 ⇒ Q(n) is true.
Problem 5: Let f : [a, b] → R be a differentiable function. The mean
(a)
value theorem says that there exists c ∈ (a, b) such that f 0 (c) = f (b)−f
.
b−a
Use the mean value theorem to prove that the function f : R → R defined by
f (x) = 13 x3 + 3x2 + 10x is strictly increasing. (Recall that strictly increasing
means that for any a < b, f (a) < f (b).)
Solution: Let a, b be two arbitrary real numbers such that a < b,
(a)
mean value theorem implies that, for some c ∈ (a, b) f (b)−f
= f 0 (c) =
b−a
(a)
c2 +6c+10 = (c+3)2 +1 ≥ 1. Since b−a > 0, f (b)−f (a) = f (b)−f
(b−a) ≥
b−a
1(b − a) > 0. Thus f (b) > f (a) and f is strictly increasing.
Problem 6: We say that limx→∞ f (x) = ∞ if for any M ∈ R there
exists N ∈ R such that if x ≥ N , then f (x) ≥ M .
The intermediate value theorem says that if f : [a, b] → R is a continuous
function and if y is between f (a) and f (b), then there exists x ∈ (a, b) such
that f (x) = y.
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(a) Define, analogously to the above, what it means for limx→−∞ f (x) =
−∞.
(b) Let f : R → R be a continuous function with limx→−∞ f (x) = −∞
and limx→∞ f (x) = ∞. Prove: If y ∈ R, then there exists x ∈ R such that
f (x) = y.
Solution:
(a) For any M ∈ R there exists N ∈ R such that if x ≤ N , then
f (x) ≤ M .
(b) Let y be an arbitrary real number, let N1 , N2 be such that f (x) ≤
y − 1 whenever x ≤ N1 , and f (x) ≥ y + 1 whenever x ≥ N2 , by definition
of limx→±∞ f (x) = ±∞. Claim that N1 < N2 . Suppose this is not true,
then ∃x0 , such that N2 ≤ x0 ≤ N1 , hence f (x0 ) ≤ y − 1 and f (x0 ) ≥
y + 1, a contradiction. Applying Intermediate Value Theorem to a = N1 ,
b = N2 and y ∈ (y − 1, y + 1) ⊆ (f (N1 ), f (N2 )), the conclusion follows,
∃x ∈ (N1 , N2 ) ⊆ R, such that f (x) = y.
Problem 7: Let I denote the irrational numbers. Define the function
f :I×I→R×R
by
f (x, y) = x + y, x2 + y 2 .
(i) Does there exist (x, y) ∈ I × I such that f (x, y) ∈ I × {1}?
(ii) Does there exist (x, y) ∈ I × I such that f (x, y) = Q × I?
Solution:
√
√
(i) Yes. Let x = y = 22 , then f (x, y) = (x+y, x2 +y 2 ) = ( 2, 1) ∈ I×1.
√
√
(ii) Yes.
Let x = 2, y = 1 − 2, then f (x, y) = (x + y, x2 + y 2 ) =
√
(1, 5 − 2 2) ∈ Q × I.
Problem 8: Given n ∈ N, let Nn = {1, 2, . . . , n} = {a ∈ Z | 1 ≤ a ≤ n}.
Let X be a finite set. The number of elements in X, called the cardinality of
X, is denoted by |X|. We have |X| = n if and only if there exists a bijection
f : Nn → X. Answer correctly the following (no need to prove anything).
(i) If X ⊆ Y , then how are |X| and |Y | related?
Ans: |X| ≤ |Y |.
(ii) If f : A → B is an injection, then how are |A| and |B| related?
Ans: |A| ≤ |B|.
(iii) If g : C → D is a surjection, then how are |C| and |D| related?
Ans: |C| ≥ |D|.
(iv) If h : E → F is a bijection, then how are |E| and |F | related?
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Ans: |E| = |F |.
Problem 9: Do Problem 18 on p. 118.
Solution: Let z ∈ Z be an arbitrary element, by definition of surjection, ∃y ∈ Y , such that g(y) = z, and ∃x ∈ X, such that f (x) = y. It
follows that g ◦ f (x) = g(f (x)) = g(y) = z, and hence g ◦ f is surjective.
Problem 10: Do Problem 20 on p. 118.
Solution:
→
−
(i) Let y ∈ f (A1 ). By definition, ∃x ∈ A1 , such that f (x) = y. It then
→
−
→
−
→
−
follows that x ∈ A1 ⊆ A2 and y = f (x) ∈ f (A2 ). Thus f (A1 ) ⊆ f (A2 ),
this completes the proof.
An counterexample for the converse: Let f : R → R be a constant
→
−
function, i.e.f (x) ≡ 0. Let A1 = [0, 1], A2 = [3, 4]. Obviously, f (A1 ) =
→
−
→
−
→
−
{0} = f (A2 ), thus f (A1 ) ⊆ f (A2 ). However A1 ⊆ A2 is not true.
A simple condition on f to make the converse come true is that f is
injective:
→
−
→
−
Suppose f (A1 ) ⊆ f (A2 ) and f is injective, let x ∈ A1 , then f (x) ∈
→
−
→
−
f (A1 ) ⊆ f (A2 ). Thus ∃x1 ∈ A2 such that f (x1 ) = f (x). By injectivity,
x = x1 ∈ A2 , it follows that A1 ⊆ A2 .
→
−
(ii) Let y ∈ f (A1 ∩ A2 ), then ∃x ∈ A1 ∩ A2 , such that f (x) = y. It
→
−
→
−
follows from x ∈ A1 and x ∈ A2 that f (x) ∈ f (A1 ) and f (x) ∈ f (A2 ),
→
−
→
−
→
−
→
−
→
−
i.e.y ∈ f (A1 ) ∩ f (A2 ). Therefore, f (A1 ∩ A2 ) ⊆ f (A1 ) ∩ f (A2 ).
→
−
→
−
→
−
Take the example in (i). f (A1 ∩ A2 ) = f (∅) = ∅, while f (A1 ) ∩
→
−
f (A2 ) = {0} ∩ {0} = {0}. In this case, the equality does not hold.
→
−
(iii) y ∈ f (A1 ∪ A2 ) ⇔ ∃x ∈ A1 ∪ A2 , f (x) = y ⇔ ∃x ∈ A1 , f (x) = y or
→
−
→
−
→
−
∃x ∈ A2 , f (x) = y ⇔ y ∈ f (A1 ) or f (A2 ) ⇔ y ∈ f (A1 ∪ A2 ).
Remark: The original problems #11 and #12 on the inclusion-exclusion
principle for 3 sets, have been moved to the 4th HW assignment.
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