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Recitation 4: Quantifiers and basic proofs
Math 299
1. Quantifiers in sentences are one of the linguistic constructs that are hard for computers
to handle in general. Here is a nice pair of example dialogues:
1. Student A: “How was the birthday party after I left?”
Student B: “It was really fun. Everybody had a drink.”
2. Student A: “How was the birthday party after I left?”
Student B: “It was really fun. Everybody watched a movie.”
Did everybody have their own drink, or did they share the same drink?
Solution: They had their own.
Did they watch the same movie or a different movie (e.g., with their smart phone)?
Solution: They all watched the same movie (this is most likely what Student B meant,
anyway).
Use quantifiers to express the statements Everybody had a drink and Everybody
watched a movie. (The order of quantifiers are important to interpret the situation
correctly.)
Solution:
∀p ∈ {people at the party} , ∃D ∈ {drinks} , p had D
∃M ∈ {movies} ∀p ∈ {people at the party} , p watched M
2. Try to rewrite each statement in one of the standard forms
∀x ∈ S, P (x);
∃x ∈ S, P (x);
∀x ∈ S ∃y ∈ T, Q(x, y);
for suitable sets S, T and open sentences P (x), Q(x, y).
(a) For any natural number n, there is a prime number m larger than n.
Solution: There are two quantifiers;
∀n ∈ N ∃m ∈ {prime numbers} , m > n
etc.
(b) There is an integer M larger than the number of contacts saved in any MSU
student’s cell phone.
Solution: There are two quantifiers;
∃M ∈ Z ∀x ∈ {MSU students’ cell phones} , M > number of contacts in x
(c) Every quadratic equation has the form ax2 +bx+c = 0, where a, b, c are complex
numbers, and every equation of the form ax2 +bx+c = 0 has a complex solution.
Solution: There are four quantifiers;
∀Q ∈ {quadratic equations} ∃a, b, c ∈ C, Q has the form ax2 + bx + c = 0
∧ ∀a, b, c ∈ C ∃x ∈ C, ax2 + bx + c = 0
3. Negate each of the statements in the previous problem. Try to express your answers
using common English.
Solution:
(a) There is a natural number n so that for all prime numbers m, n is at least as
big as m.
(b) For any integer M , there is a student cell phone with at least M contacts.
(c) Either there are quadratic equations that do not have the form ax2 + bx + c = 0
for any complex numbers a, b, c, or there are quadratic equations that do not
have any complex solutions.
(d) There are students at MSU that are less than 52 years old.
(e) There are some students who do not experience moments of frustration.
(f) I don’t know anyone who is never shy.
4. Negate the following statements.
(a) ∀x ∈ N ∃y ∈ Z, P (x, y) =⇒ Q(x, y)
Solution: ∃x ∈ N ∀y ∈ Z, P (x, y)∧ ∼ Q(x, y).
(b) (∀x ∈ N ∃y ∈ Z, P (x, y)) =⇒ Q
Solution: (∀x ∈ N ∃y ∈ Z, P (x, y)) ∧ ∼ Q
5. Express the following statement using as few English words as possible (i.e., try to
only use symbols from math +, =, x, . . . and logic ∀, ∃, ⇒, ∧, . . .):
(a) The equation x2 + x + 1 = 0 has exactly one real solution.
Solution: ∃x ∈ R, (x2 + x + 1 = 0) ∧ (y ∈ R, y 2 + y + 1 = 0 =⇒ x = y)
(b) The equation x2 + x + 1 = 0 has at least two distinct real solutions.
Solution: ∃x, y ∈ R, (x2 + x + 1 = 0) ∧ (y 2 + y + 1 = 0) ∧ (x 6= y).
6. Let S ⊆ R, and define S to be topped out if there is some B ∈ R such that for all
s∈S
s + 2 ≤ B.
(a) Is the set {1, 2, 3} topped out?
Solution: Yes. Take B = 5, then s + 2 ≤ 5 for all s ∈ {1, 2, 3}.
(b) Find a set S ⊆ R that is not topped out.
Solution: Take S = R, or S = N.
(c) Suppose S ⊆ R is a finite set. Is it possible for S to not be topped out?
Solution: Every finite set is topped out. To see this, note that since S is finite,
it contains a largest element x (so this satisfies x ≥ s for all s ∈ S). If we set
B = x + 2, and if s ∈ S, then s + 2 ≤ x + 2 = B.
(d) Suppose S ⊆ R and T ⊆ R are both topped out, and define a new set U by
U = {s + t |s ∈ S, t ∈ T } .
Is U topped out? Explain.
Solution: Yes, U is topped out. Since S, T are topped out, there are B, C ∈ R so
that s + 2 ≤ B and t + 2 ≤ C for all s ∈ S and all t ∈ T . Suppose u ∈ U . Then
we can write u = s + t for some s ∈ S and t ∈ T . This implies
u + 2 = s + t + 2 = (s + 2) + (t + 2) − 2 ≤ B + C − 2.
7. Let n, k ∈ Z. Prove the following statements
(a) If n is even, then n2 is even.
Solution: If n is even, then we can write n = 2m for some integer m. Then
n2 = (2m)n = 2(mn). Since this is divisible by 2 it follows that n2 is even.
(b) If n is even, then n3 is even.
Solution: If n is even, then we can write n = 2m for some integer m. Then
n3 = (2m)n2 = 2(mn2 ). Since this is divisible by 2 it follows that n3 is even.
(c) If n is even and k ≥ 1, then nk is even.
Solution: If n is even, then we can write n = 2m for some integer m. Then
nk = (2m)nk−1 = 2(mnk−1 ). Since this is divisible by 2 it follows that nk is
even.
In the last statement, what happens if we remove the assumption that k ≥ 1?
Solution: We needed k ≥ 1 to ensure that mnk−1 is an integer. For example, if k = 0,
then nk−1 = 1/n.
8. Let b, c, x ∈ R.
(a) Prove that x satisfies the equation x2 + bx + c = 0 if and only if x satisfies the
equation (x + b/2)2 + c = b2 /4.
Solution: If x2 +bx+c = 0, then adding b2 /4 to both sides gives x2 +bx+b2 /4+c =
b2 /4. The left-hand side can be factored to give (x + b/2)2 + c = b2 /4.
Conversely, if (x + b/2)2 + c = b2 /4, then expand out the left-hand side to get
x2 + bx + b2 /4 + c = b2 /4. Now subtract b2 /4 from both sides.
(b) Assume b2 − 4c ≥ 0. Prove
(x + b/2)2 + c = b2 /4 if
√ that x satisfies1 the equation
√
1
2
2
and only if x = 2 (−b + b − 4c) or x = 2 (−b − b − 4c).
Solution: If (x + b/2)2 + c = b2 /4, then subtracting c from both sides gives
(x + b/2)2 = b2 /4 − c. Write the right-hand side as 41 (b2 − 4c), and note that
this is non-negative by assumption.
We can thereforep
take the square root of both
p
sides to get x + b/2 = 12 b2 /4 − c or x + b/2 = − 21 b2 /4 − c. Subtracting b/2
from both sides gives the desired result.
√
√
To see the converse, simply plug x = 21 (−b + b2 − 4c) or x = 12 (−b − b2 − 4c)
into (x + b/2)2 + c. In both cases, the expression simplifies to b2 /4.
9. Let n, k ∈ Z and assume k ≥ 1. Use the contrapositive to prove that if nk is odd,
then n is odd.
Solution: The contrapositive is “If n is even, then nk is even”. Since we are also
assuming k ≥ 1, it suffices to prove the statement “if n is even and k ≥ 1, then nk is
even”. However, this was done in Problem 6 (c) above.
Quiz 4 (Solutions)
Math 299
Please answer each question in the space provided. Use complete sentences and correct
mathematical notation to write your answers. You have 20 minutes to complete this quiz.
1. (4 points) Rewrite each statement using English words instead of the symbols ∀ and
∃.
(a) ∀x ∈ Z ∃y ∈ R, x + y = 7.
Solution: For every x ∈ Z, there is some y ∈ R so that for all x + y = 7.
(b) ∃x ∈ (0, ∞) ∀n ∈ N, x ≤ 1/n.
Solution: There is some x ∈ (0, ∞) so that for all n ∈ N, x ≤ 1/n.
2. (4 points) Rewrite each statement below in the one of the forms
∀x ∈ S, P (x)
or
∃x ∈ S, P (x)
for appropriate S and P (x).
(a) Every natural number is an integer.
Solution: ∀n ∈ N, n ∈ Z.
(b) There is a real number that is both positive and negative.
Solution: ∃x ∈ R, (x > 0) ∧ (x < 0).
3. (4 points) Negate each of the following statements. Simplify your answer as much as
possible.
(a) ∀x ∈ Z ∃y ∈ R, x + y = 7.
Solution: ∃x ∈ Z ∀y ∈ R, x + y 6= 7.
(b) ∃x ∈ (0, ∞) ∀n ∈ N, x ≤ 1/n.
Solution: ∀x ∈ (0, ∞) ∃n ∈ N, x > 1/n.
4. (3 points) True or False.
(a) The following statements are equivalent:
∀x ∈ R ∃y ∈ R, x + y = 9
∃y ∈ R ∀x ∈ R, x + y = 9
Solution: False
(b) The following statements are equivalent:
∀x ∈ R ∀y ∈ Z, x + 2y = 7
∀y ∈ Z ∀x ∈ R, x + 2y = 7
Solution: True
(c) The following statements are equivalent:
∃x ∈ R ∃y ∈ Z, x2 = y 2
∃(y, x) ∈ Z × R, x2 = y 2
Solution: True