Download Salem State College MAT 420 Worksheet 4

Salem State College
MAT 420
Worksheet 4 - Solutions
L. Pedro Poitevin
October 9, 2007
1. Abe Lincoln said, “You can fool all of the people some of the time, and you can fool some of the
people all of the time, but you can’t fool all of the people all of the time.” Write this sentence in
logical notation, and find its negation.
Solution. Let P be the set of people and T the set of units of time. Let F (p, t) denote the proposition
“You can fool person p at time t.” Then Lincoln’s sentence is
(∃t ∈ T )(∀p ∈ P )(F (p, t)) ∧ (∃p ∈ P )(∀t ∈ T )(F (p, t)) ∧ ¬(∀p ∈ P )(∀t ∈ T )(F (p, t)).
Its negation is
(∀t ∈ T )(∃p ∈ P )(¬F (p, t)) ∨ (∀p ∈ P )(∃t ∈ T )(¬F (p, t)) ∨ (∀p ∈ P )(∀t ∈ T )(F (p, t)).
We interpret this sentence as saying the following: “There are people you can never fool, or all of
the people enjoy some moments at which you can’t fool them, or you can fool all of the people all
of the time.”
2. Consider the equation x4 y + ay + x = 0.
(a) Show that the following statement is false. “For all a, x ∈ R, there is a unique y such that
x4 y + ay + x = 0.
Solution. A counterexample to this statement is the pair (a, x) = (0, 0). For this example, all
y ∈ R satisfy the equation.
(b) Find the set of real numbers a such that the following statement is true. “For all x ∈ R, there
is a unique y such that x4 y + ay + x = 0.”
Solution. The solution set is the set of all positive real numbers. Suppose the sentence is true
for a. Then the equation must have a unique solution y when x = 0. Thus ay = 0 must have
a unique solution; this requires that a 6= 0. Also, if a < 0, then x = −(−a)1/4 is a choice of x
for which the equation has no solution.
If a > 0, then for every x ∈ R we can solve the equation for y to obtain y = −f racxx4 + a.
This computes a unique value for y that makes the equation true. Thus the most general
condition on a that makes the sentence true is a > 0.
3. Let f be a real-valued function on a set S. In order to prove that the minimum value in the image
of f is β, two statements must be proved. Express each of these statements using quantifiers.
Solution. To prove that β = min{f (x) : x ∈ S} it must be shown that (∀x ∈ S)(f (x) ≥ β) and
(∃x ∈ S)(f (x) = β).
4. Let Tn be the set
o of ordered pairs of positive real numbers. Define f : T → R by f (x, y) =
1
1
max x, y, x + y . Determine the minimum value in the image of f . (?)
√
√
Solution. The minimum
is 2. We prove that f (x,
always, and √
that this value is achieved.
√
√ y) ≥ 2 √
If max{x, y} ≥ 2, then f (x, y) ≥ max{x, y} ≥ 2. If x < 2 and y < 2, then f (x, y) ≥ x1 + y1 ≥
√
√
√
√2 =
2.
Finally,
when
x
=
y
=
2,
we
have
f
(x,
y)
=
2.
2
5. Let f and g be functions from R to R. Determine which statements below are true. If true, provide
a proof. If false, provide a counterexample.
(a) If f and g are bounded, then f + g is bounded.
Solution. This is a true statement. By the definition of bounded function, there exist positive
constants M1 , M2 ∈ R such that for all x ∈ R, |f (x)| ≤ M1 and |g(x)| ≤ M2 . The constant
M := M1 + M2 works to show that f + g is bounded, because applying the triangle inequality
yields, for arbitrary x ∈ R,
|(f + g)(x)| = |f (x) + g(x)| ≤ |f (x)| + |g(x)| ≤ M1 + M2 = M .
(b) If f and g are bounded, then f g is bounded.
Solution. This is a true statement. By the definition of bounded function, there exist positive
constants M1 , M2 ∈ R such that for all x ∈ R, |f (x)| ≤ M1 and |g(x)| ≤ M2 . The constant
M := M1 M2 works to show that f g is bounded, because for arbitrary x ∈ R,
|(f g)(x)| = |f (x)g(x)| = |f (x)||g(x)| ≤ M1 M2 = M .
(c) If f + g is bounded, then f and g are bounded.
Solution. This statement is false. The functions f, g defined by f (x) = x and g(x) = −x
provide a counterexample. Here f and g have unbounded image, but f (x) + g(x) = 0, which
is a constant, hence bounded, function.
(d) If f g is bounded, then f and g are bounded.
Solution. This statement is false. Consider the functions f, g defined by f (x) = x and g(x) =
1/x for x 6= 0 and g(0) = 0. Here f and g have unbounded image, but f (x)g(x) is clearly
bounded.
(e) If both f + g and f g are bounded, then f and g are bounded.
Solution. This is a true statement. We are given M, N ∈ R such that for all x ∈ R, |f (x) +
g(x)| ≤ M and |f (x)g(x)| ≤ N . We proceed to show that f 2 and g 2 are bounded. We have
|f (x)2 + g(x)2 | = |(f (x) + g(x))2 − 2f (x)g(x)|
≤ |(f (x) + g(x))2 | + 2|f (x)g(x)| ≤ M 2 + 2N .
Since f (x)2 and g(x)2 are both
nonnegative, we have √
that f (x)2 and g(x)2 are bounded by
√
f (x)2 + g(x)2 . Thus |f (x)| ≤ M 2 + 2N and |g(x)| ≤ M 2 + 2N .
6. For S in the domain of a function f , let f (S) = {f (x) : x ∈ S}. Let C and D be subsets of the
domain of f .
(a) Prove that f (C ∩ D) ⊆ f (C) ∩ f (D).
Solution. Let f (x) ∈ f (C ∩ D). Then x ∈ C ∩ D, and so x ∈ C and x ∈ D. Since x ∈ C,
we have that f (x) ∈ f (C), and since x ∈ D, we have that f (x) ∈ f (D). Therefore f (x) ∈
f (C) ∩ f (D).
(b) Give an example where equality does not hold.
Solution. Consider A = {−1, 1}, B = {1}, and f (−1) = f (1) = 1. Let C = {−1} and
D = {1}. Now C ∩ D and f (C ∩ D) are empty, but 1 ∈ f (C) ∩ f (D).