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Transcript
```Math 347 F1, Spring 2010
Midterm Exam 1 Solutions
Prof. A.J. Hildebrand
Midterm Exam 1 Solutions, Comments, and Feedback
• Grading: Each problem was worth 20 points, for a total of 100 points. In multipart
problems, each part was worth 5 points, with the exception of Problems 4 and 5, which
had a breakdown of 15 points for the proof (part (a)) and 5 for the counterexample
(part (b)). For exam statistics and a letter grade correspondence see the course
webpage.
• Solutions: Solutions, along with some remarks about common errors, are attached.
• “Only if ”: “P only if Q” means P ⇒ Q. In the combination “P if and only if Q”
(P ⇔ Q) the “if” part (“P if Q”) stands for the implication P ⇐ Q, while the “only
if” part (“P only if Q”) stands for the reverse implication, P ⇒ Q. See Problem 1 of
the logic worksheet for various ways of expressing an implication in English.
• Negation of implications: The negation of an implication P ⇒ Q is not an
implication and cannot be expressed as such, but an “and” statement: “P and not
Q” (P ∧ ¬Q).
• Implications involving variables: In statements such as “f (x) < f (y) whenever
x < y” or “if n is odd, then n is prime” the variables (x, y, and n) are understood to be
arbitrary elements of the underlying universe, i.e., in the sense of a “for all” quantifier.
When negating such statements, this quantifier turns into an existential quantifier,
and it is absolutely crucial to include this quantifier in the negated statement: “there
exist x, y ∈ R such that x < y and f (x) ≥ f (y)” and “there exists n ∈ N such that
n is odd and not prime”. Without such a quantifier the negated statements would
not make sense, and if interpreted in the “for all” sense, they would be completely
wrong. (Compare “all odd integers are not prime” to “there exists an odd integer
that is not prime”.) See Problem 3 of the logic worksheet for further examples and
• Negation of “and” statements. When negating an “and” statement, “and” turns
into “or”. For example, the negation of “pass midterm and pass final” is “fail midterm
or fail final”.
• Not decreasing versus non-decreasing. As pointed out on the logic worksheet (see Problems 4(c)–(d)), these two terms have completely different meanings.
The first, “not decreasing”, is the negation of decreasing, while the second (“nondecreasing”) is a variant of increasing (with the strict inequality < replaced by ≤).
• Counterexamples: A counterexample most be completely specified. For example,
to show that a relation such as (∗) f (C ∩ D) = f (C) ∩ f (D) is not true requires
specifying not only the sets C, D, but also the function f . Specifying the sets C and
D alone is not enough and does not guarantee that (∗) is false. (In fact, given any
sets C and D, there is a function f for which (∗) is true ...)
1
Math 347 F1, Spring 2010
Midterm Exam 1 Solutions
Prof. A.J. Hildebrand
1. Definitions: For the following questions an answer is sufficient: Just state the requested definition, theorem, formulas, etc. Be sure to use correct notation, include
any necessary quantifiers in the appropriate order, and connecting words (e.g., “such
that”) if necessary. Definitions should be stated in English, i.e., not using logical
symbols (∀, ∃, ¬, =⇒ , etc.).
(a) A function f from R to R is not decreasing if . . .
Solution: “(∃x, y ∈ R)((x < y) ∧ (f (x) ≤ f (y)))”
“There exist real numbers x < y such that f (x) ≤ f (y).”
(b) The graph of a function f : A → B is defined as the set . . .
Solution:
a ∈ A}.
{(a, b) ∈ A × B : b = f (a)} or, equivalently, {(a, f (a)) :
(c) Given a function f : A → B, the image of a set C ⊆ A is the set . . .
Solution:
{b ∈ B : b = f (a) for some a ∈ C }, or, equivalently,
{f (a) : a ∈ C}.
(d) Let A = {1, 2}, and B = {2, 3, 4}. Write down explicitly (i.e., with all elements
explicitly listed) the following sets. Be sure to use correct set-theoretic notation.
(i) A − B = . . . {1}
(ii) A × B = . . . {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4)}
(iii) The power set of A: P (A) = . . . {∅, {1}, {2}, {1, 2}}
2. Conditionals: Consider the following statement (where the underlying universe is
the set N of natural numbers): “If n is odd, then n is prime.”
(a) Write the contrapositive of this statement.
Solution:
“If n is composite, then n is even.”
(b) Write the converse of this statement.
Solution:
“If n is prime, then n is odd.”
(c) Rewrite the statement in the form “... only if ...”.
Solution: “n is odd only if n is prime.”
Remark: See Problem 1 on the Logic Worksheet: “A only if B” means
the same as A =⇒ B, which in turn means the same as “if A, then B”.
The above statement is of the latter form, with A being “n odd” and B
being “n prime”.]
(d) Write the negation of this statement without using words of negation. (You
can use “composite” as a synonym for “not prime” and “even” for “not odd”.)
Solution: “There exists an odd number that is composite.”
Remark: Here the quantifier “there exists” is crucial and cannot be
omitted.
3. Negations: Negate the following statements without using words of negation. Write
your negation first in symbolic form (i.e., using logical symbols ∀, ∃, =⇒ , lnot, ∧,
etc.), then as an English sentence (without using logical symbols). (Here f is assumed
to be a function from R to R.)
2
Math 347 F1, Spring 2010
Midterm Exam 1 Solutions
Prof. A.J. Hildebrand
(a) f (x, y) 6= 0 whenever x 6= 0 and y 6= 0.
Solution: [cf. 3(a), Logic worksheet]
Symbolic notation: (∀x, y ∈ R)[x 6= 0 ∧ y 6= 0 =⇒ f (x, y) 6= 0]
Negation: (∃x, y ∈ R)[x 6= 0 ∧ y 6= 0 ∧ f (x, y) = 0]
“There exist x, y ∈ R − {0} such that f (x, y) = 0.”
Or:
“There exist x, y ∈ R such that x 6= 0, y 6= 0, and f (x, y) = 0.”
(b) For every > 0 there exists δ > 0 such that |f (x)| < whenever |x| < δ.
Solution: [cf. 5(a), Logic worksheet]
Symbolic notation: (∀ > 0)(∃δ > 0)[|x| < δ =⇒ |f (x)| < ]
(∀ > 0)(∃δ > 0)(∀x ∈ R)[|x| < δ =⇒ |f (x)| < ]
Negation: (∃ > 0)(∀δ > 0)(∃x ∈ R)[|x| < δ ∧ |f (x)| ≥ ]
“There exists > 0 such that for all δ > 0 there exists x ∈ R such that
|x)| < δ and |f (x)| ≥ .”
Note: Note the implied quantifier “for all x ∈ R” (cf. the remarks for
Problems 5(b)(d) of the Logic worksheet).
(c) There exists M ∈ R such that for all x ∈ R there exists y > x such that
f (y) ≤ M .
Solution: [cf. 4(e) on the Logic worksheet]
Symbolic notation: (∃M ∈ R)(∀x ∈ R)(∃y > x)(f (y) ≤ M ).
Negation: (∀M ∈ R)(∃x ∈ R)(∀y > x)(f (y) > M ).
“For all M ∈ R there exists x ∈ R such that for all y > x we have
f (y) > M .”
(d) To pass this class requires passing the midterm and the final. (Hint: To avoid
words of negation, use the word fail as the opposite of “pass”).
Solution:
Negation: “You can pass this class, but fail the midterm or the final.”
Remark: Note the negation changes the “and” in “pass midterm and
pass final” to an “or”.
4. Let f be a function from R to R and let g be defined by g(x) = ef (x) for all x ∈ R.
(a) Prove that if f is bounded, then g is bounded. (You may use the standard
properties of the exponential function.)
(b) Give an example showing that converse is not true. (No proof is required for
this part. Just give a counterexample.
Solution:
(a) Suppose f is bounded. By the definition of a bounded
function, this means that there is a constant M ∈ R such that
|f (x)| ≤ M
for all x ∈ R.
Hence
|g(x)| = |ef (x) | ≤ e|f (x)| ≤ eM
for all x ∈ R,
since |ey | ≤ e|y| for any real y, and since the exponential function is increasing and thus preserves inequalities. Thus, g is bounded, with bound
M 0 = eM .
3
Math 347 F1, Spring 2010
Midterm Exam 1 Solutions
Prof. A.J. Hildebrand
(b) The converse is not true, as shown by the functions f (x) = −x2 , g(x) =
2
ef (x) = e−x . With this choice, f (x) is unbounded, but g(x) is between 0
and 1 and hence bounded.
5. Let f : A → B be a function, and let C and D be subsets of A.
(a) Give a careful, step-by-step proof of the relation f (C ∩ D) ⊆ f (C) ∩ f (D).
Your write-up must include all necessary steps, with appropriate justifications
(e.g., “by the def. of ...”), in the correct logical order, use proper notation and
terminology, and include any necessary quantifiers and connecting words (e.g.,
“therefore”, “such that”).
(b) Give an example showing that equality does not hold. (No formal proof required
for this part. A brief explanation is enough.)
Solution: [cf. Problem 1.50 from HW 2] (a) To show that f (C ∩ D) ⊆
f (C) ∩ f (D), we need to show that every element in the set on the left is
also an element in the set on the right.
• Suppose y ∈ f (C ∩ D).
• Then there exists x ∈ C ∩ D such that f (x) = y, by the definition of
an image set,
• Since x ∈ C ∩ D, we have x ∈ C and x ∈ D, by the definition of an
intersection.
• Since x ∈ C, we have f (x) ∈ f (C), by the definition of an image set,
and hence y ∈ f (C).
• Similarly, since x ∈ D, we get y = f (x) ∈ f (D).
• Since y ∈ f (C) and y ∈ f (D), we have y ∈ f (C) ∩ f (D), by the
definition of an intersection.
(b) A counterexample is given by taking f (x) = |x| (as a function from R
to R), C = {x ∈ R : x > 0}, D = {x ∈ R : x < 0}. Then C ∩ D = ∅, so
f (C ∩ D) = ∅, but f (C) and f (D) are both equal to the set of positive real
numbers, and hence f (C) ∩ f (D) is also equal to the set of positive reals.
4
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