SOLUTION OF HW1
MINGFENG ZHAO
February 11, 2013
1. [8 Points] Foe each of the following mathematical statements, rewrite the statement making all the
quantifiers explicit. Then form the negation of the statement. Finally recast the negation in a form
similar to the original statement.
a. Every positive integer has a unique prime factorization.
b. The only even prime is 2.
c. Multiplication of integers is associative.
d. Two points in the plane determine a line.
e. The altitudes of a triangle intersect at a point.
f. Given a line in the plane and a point not on it, there exists a unique line passing through the
given point parallel to the given line.
g. Any partitioning of the integers into a finite number of disjoint subsets has the property that
one of the subsets contains arbitrarily long arithmetic progressions.
h. If there are more letters than mailboxes, at least one mailbox must get more than one letter.
Proof. a. Rewrite: For all n ∈ N, there exists finitely many prime numbers p1 , · · · , pk with p1 < · · · <
pk , and e1 , · · · , ek ∈ N such that n = pe11 · · · · · pekk , and this factorization is unique.
Negation: There exists some n ∈ N such that either for all finitely many prime numbers p1 , · · · , pk
with p1 < · · · < pk and e1 , · · · , ek ∈ N, we have n 6= pe11 · · · · · pekk , or there exists finitely many primes
numbers p1 , · · · , pk , q1 , · · · , ql with p1 < · · · < pk and q1 < · · · < ql , and e1 , · · · , ek , f1 , · · · , fl ∈ N
such that (p1 , · · · , pk , e1 , · · · , ek ) 6= (q1 , · · · , ql , f1 , · · · , fl ) and n = pe11 · · · · · pekk = q1f1 · · · · · qlfl .
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MINGFENG ZHAO
Recast Negation: There exists some n ∈ N such that either n has no prime factorization, or n has
at least two different prime factorizations.
b. Rewrite: Let n ∈ N be an even prime number, then n = 2.
Negation: There exists some even number n ∈ N such that n 6= 2, but n is prime.
Recast Negation: There exists some even prime which is not equal to 2.
c. Rewrite: Let a, b, c ∈ Z, then (a · b) · c = a · (b · c).
Negation: There exists some a, b, c ∈ Z such that (a · b) · c 6= a · (b · c).
Recast Negation: Multiplication of integers is not associative.
d. Rewrite: Let P 6= Q ∈ R2 , then there exists a unique line ` which passes through P and Q.
Negation: There exists P 6= Q ∈ R2 such that either there are at least two lines `1 and `2 such that
both `1 and `2 pass through P and Q, or there is no line ` which pass through P and Q.
Recast Negation: Two point in the plane can not determine a line.
e. Rewrite: Let ∆ABC be a triangle, and AD, BE and CF be altitudes of ∆ABC, then there exists
a point P such that P is on all of lines AD, BE and CF .
Negation: There exists a triangle ∆ABC, let AD, BE and CF be altitudes of ∆ABC, then there
is no point P such that P is on all of lines AD, BE and CF .
Recast Negation: There exists a triangle such that the altitudes of this triangle does not intersect
at a point.
f. Rewrite: Let ` be a line and P be a point such that P ∈
/ `, then there exists a unique line `0 such
that P ∈ `0 and `0 //`.
Negation: There exists a line ` and a point P with P ∈
/ `, then either there is no line `0 such that
P ∈ `0 and `0 //`, or there are at least two lines `0 and `00 such that P ∈ `0 , P ∈ `00 and `0 //`00 //`.
Recast Negation: There exists a line and a point which is not on this line, then either there is no
line which passes through the given point and is parallel to the given line.
SOLUTION OF HW1
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g. Rewrite: For any finite partition P = {A1 , · · · , An } of Z, then there exists some i such that Ai
contains arbitrarily long arithmetic progressions.
Negation: There is a finite partition P = {A1 , · · · , An } of Z such that for all i , we have Ai does
not contain arbitrarily long arithmetic progressions.
Recast Negation: There is a finite partition of integers such that every set in the partition doest
not contain arbitrarily long arithmetic progressions.
h. Rewrite: Let m < n, put n letters into m mailboxes, then there exists some one mailbox which
has more than one letter.
Negation: Let m < n, put n letters into m mailboxes, every mailbox has at most one letter.
Recast Negation: There are more letters than mailboxes, every mailbox has at most one letter.
2. [10 Points] Each of the following true statements is in universal existential form. Write the
corresponding statement with the order of quantifiers reversed, and show why it is false.
a. Every nonzero rational number has a rational reciprocal.
b. There is no largest prime.
Proof. a. Rewrite: For all x ∈ Q\{0}, there exists some y ∈ Q such that xy = 1.
Reversed: There exists some y ∈ Q such that for all x ∈ Q\{0}, we have xy = 1.
Proof of False: Take x1 = 1, then 1 = x1 y = 1 · y = y, that is, y = 1. Take x2 = 2, then
x2 y = 2 · 1 = 2 6= 1, contradiction.
b. Rewrite: For all prime number p ∈ N, there exists a prime number q ∈ N such that p < q.
Reversed: There exists a prime number p ∈ N such that for all prime number q ∈ N, we have q ≤ p.
Proof of False: Let n = 1 · 2 · · · · p + 1, we know that there exists a prime number q ∈ N such that
q | n. By the assumption, we know that q ≤ p, which implies that q | 1, contradiction.
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MINGFENG ZHAO
3. [10 Points] Prove that every subset of N is either finite or countable.
Proof. Let A be a subset of N. If A is finite, then we are done. If A is infinite. By the well ordering
property of N, there exists a1 ∈ A such that a1 = inf a. Let A1 = A\{a1 }, since A is infinite, then
a∈A
A1 is infinite. By the well ordering property of N, there exists a2 ∈ A1 such that a2 = inf a, also we
a∈A1
know that a1 < a2 . Let A2 = A1 \{a2 }, since A1 is infinite, then A2 is infinite. By induction, we know
that there is an ordering on A, that is, elements in A can be aligned by ordering, a1 < a2 < a3 < · · · .
Hence A is countable. In summary, we know that A is either finite or countable.
4. [10 Points] Show that if a countable subset is removed from an uncountable set, the remainder is
still uncountable.
Proof. Let X be an uncountable set, A be a countable subset of X, and B = X\A. If B is uncountable,
we are done. If B is finite or countable, since A is countable, then X = A
[
B is also countable, which
contradicts with the assumption that X is uncountable. Therefore, we know that B is uncountable.
5. [10 Points] Let A be a set for which there exists a function f from A to N with the property that
for every natural number k, the subset of A given by the solutions to f (a) = k is finite. Show that A
is finite or countable.
Proof. For any k ∈ N, let
Ak = f −1 ({k}) = {a ∈ A : f (a) = k} ⊂ A.
By the assumption, we know that for any k ∈ N, Ak is a finite subset of A. By the definition of
Ak , it is easy to see that Ak
T
Al = ∅ for all k 6= l. Also we know that
A=
∞
[
k=1
Ak .
SOLUTION OF HW1
Since for all k ∈ N, Ak is finite, then A =
∞
[
5
Ak is finite or countable.
k=1
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu