MATH 311-01
Problem of the Day
1. Prove that there are infinitely many prime numbers congruent to 5 modulo 6.
Proof. Let us suppose counterfactually that there are only finitely many primes congruent to
5 modulo 6. One of those is 5; let us denote the list of others as p1 , p2 , . . . , pn (we know this
list is nonempty since 11 is prime). Now, let P = 6p1 p2 p3 p4 · · · pn + 5. By construction P ≡ 5
(mod 6); by the Fundamental Theorem of Arithmetic we may consider P ’s decomposition into
primes P = q1 q2 q3 · · · qm . Since the only product which yields 5 modulo 6 is 1 · 5 ≡ 5 (mod 6),
we know at least one of the qi is congruent to 5, and must thus be equal to either 5 or some
pj . We divide into cases and show a contradiction in each case:
Case I: qi = 5. Then 5 | P , and we know that 5 | 5, so 5 | P − 5, which is equal to the product
p1 p2 p3 · · · pn Since 5 is prime, it must divide one of these factors, so there is some j such that
5 | pj ; but since all the pj are primes not equal to 5, this is imposssible.
Case II: qi = pj for some j. Then pj | P , and we know that pj | p1 p2 · · · pn , so pj |
P − p1 p2 · · · pn , which is equal to 5. Thus pj | 5; but since pj is a prime not equal to 5, this is
imposssible.
It is also possible to avoid the need for a casewise argument with a different choice of P :
Alternative proof. Let us suppose counterfactually that there are only finitely many primes
congruent to 5 modulo 6. Let us denote the list of such numbers as p1 , p2 , . . . , pn (we know
this list is nonempty since 5 is prime). Now, let P = 6p1 p2 p3 p4 · · · pn − 1. By construction
P ≡ −1 ≡ 5 (mod 6); by the Fundamental Theorem of Arithmetic we may consider P ’s
decomposition into primes P = q1 q2 q3 · · · qm . Since the only product which yields 5 modulo 6
is 1 · 5 ≡ 5 (mod 6), we know at least one of the qi is congruent to 5, and must thus be equal to
some pj . Thus pj | P and pj | 6p1 p2 · · · pn , so pj divides their difference, which is 1. However,
the only divisors of 1 are 1 and −1, neither of which are prime., leading to a contradiction.
2. (a) Prove that for any sets A and B, P(A ∩ B) = P(A) ∩ P(B).
Proof. To prove equality of two sets, we prove that any element of one of the sets must
be an element of the other, and vice versa.
Consider any S ∈ P(A ∩ B); we shall seek to show that this S must also be an element of
P(A) ∩ P(B). Then by the defintion of a power set, S ⊆ A ∩ B. Since A ∩ B is known to
be a subset of both A and B, transitivity of subset inclusion yields that S ⊆ A and S ⊆ B.
Thus S ∈ P(A) and S ∈ P(B), so by the defintion of intersection, S ∈ P(A) ∩ P(B).
Conversely, let us consider any T ∈ P(A) ∩ P(B) and show that it is an element of
P(A ∩ B). By the defintion of an intersection, T must be an element of both P(A) and
P(B). Thus T ⊆ A and T ⊆ B. Thus each element of T is an element of both A and of
B, and is thus an element of A ∩ B; hence T ⊆ A ∩ B, so by defintition, T ∈ P(A ∩ B).
(b) Prove that for sets A and B, P(A ∪ B) = P(A) ∪ P(B) if and only if A ⊆ B or B ⊆ A.
Proof. Let us start by proving that if A ⊆ B or B ⊆ A, then P(A ∪ B) = P(A) ∩ P(B).
WLOG we may assume the specific premise that A ⊆ B. Then A ∪ B = B. Also, any
element S of P(A) is described by the relation S ⊆ A ⊆ B, so S is also an element of P(B).
Thus P(A) ⊆ P(B), so P(A) ∪ P(B) = P(B). Thus, P(A ∪ B) = P(B) = P(A) ∪ P(B).
Conversely, let us show that if A * B and B * A, then P(A∪B) = P(A)∩P(B). In order
for A to not be a subset of B, there must be some x ∈ A which is not in B; likewise, for
Page 1 of 2
due December 4, 2013
MATH 311-01
Problem of the Day
B to not be a subset of A, there must be some y ∈ B which is not in A. Since x ∈ A ∪ B
and y ∈ A ∪ B, {x, y} ⊆ A ∪ B and thus {x, y} ∈ P(A ∪ B). However, since x ∈
/ B,
{x, y} ∈
/ P(B), and likewise since y ∈
/ A, {x, y} ∈
/ P(A), so {x, y} ∈
/ P(A) ∪ P(B). Since
some element of P(A ∪ B) is not in P(A) ∪ P(B), e know these two sets are nonequal.
Page 2 of 2
due December 4, 2013