Math 3210, Fall 2013
Instructor: Thomas Goller
August 26, 2013
Name:
Thomas Goller
First Week Quiz
(1) Let P and Q be statements. What are the only truth values for P and Q that make
P =⇒ Q (“If P , then Q”) false?
When P is true and Q is false.
(2) What is the negation of the statement “for all x ∈ Z, x2 > x”? Is either statement true?
The negation is “For some x ∈ Z, x2 ≤ x,” which is a true statement (take x = −1,
x = 0, or x = 1). The original statement is false, because it fails when x = −1, x = 0,
and x = 1.
Comments:
• Given any statement, it is always the case that exactly one of the following is true:
either the statement is true and its negation is false, or the statement is false and
its negation is true. Remember, the truth table for the negation of a statement P
is exactly the opposite of the truth table for P .
(3) If a, b ∈ R and a < b, give a description in set theory notation for the intervals (a, b) and
[a, b).
(a, b) = {x ∈ R : a < x < b} and [a, b) = {x ∈ R : a ≤ x < b}.
(4) Let f : R → R≥0 be the function defined by the rule f (x) = x2 .
(a) What is the domain of f ?
R.
(b) What is the range (image) of f ?
(c) Is f one-to-one (injective)?
(d) Is f onto (surjective)?
(e) What is f (−3)?
(f) What is f (∅)?
R≥0 .
No, because f (−1) = 1 = f (1).
Yes, every non-negative real number has a square root.
9.
∅.
(g) What is f ({−1, 1})?
{1}.
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(5) Prove or disprove the following claim.
Claim. Let n ∈ Z. If n is odd, then n2 is odd.
Proof. Since n is odd, there is k ∈ Z such that n = 2k + 1. Then n2 = (2k + 1)2 =
2(2k 2 + 2k) + 1, so n2 is odd.
Comments:
• A well-written proof reads like a grammatically correct English essay. In this case,
the proof reads “Since n is odd, there is an integer k such that n equals two k plus
one. Then n squared equals two k plus one squared equals two times two k squared
plus two k plus one, so n squared is odd.”)
• The proof relies heavily on the definition of odd: “An integer n is odd if there is
an integer k such that n = 2k + 1.” This definition is used at the beginning of the
proof to deduce that n = 2k + 1, and again at the end of the proof, since exhibiting
n2 as twice an integer plus one shows that n2 is odd. You don’t need to write that
2k 2 + 2k is an integer – that is an important but fairly obvious point.
(6) Prove or disprove the following claim.
Claim. Let n ∈ N. Then 12 + 22 + · · · + n2 =
n(n+1)(2n+1)
.
6
Proof. Induction on n. The base case n = 1 is true since 12 = 1 = 1·2·3
. Now suppose
6
k(k+1)(2k+1)
2
2
2
the claim is true for some n = k ∈ N, namely 1 + 2 + · · · + k =
. Then
6
k(k + 1)(2k + 1)
+ (k + 1)2
6
"
k + 1!
=
k(2k + 1) + 6(k + 1)
6
k+1
=
(2k 2 + 7k + 6)
6
(k + 1)(k + 2)(2k + 3)
=
,
6
12 + 22 + · · · + k 2 + (k + 1)2 =
so the claim is true when n = k + 1. Thus we are done by induction.
Comments:
• You can try proof by induction when a claim has one case for each natural number.
• Induction sometimes allows you to prove a claim that would be really difficult to
prove directly!
(7) What are the three most recent math classes you’ve completed?
Seminar on moduli of curves in algebraic geometry. Topics course on the minimal model
program. Seminar on differential forms. Teaching: discrete math.
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