Math 3000 Homework 1 Answers
1. pg. 83, problem 24.
Prove that if a+b and ab have the same parity, then a and b are even.
Pf: By cases:
Case I: a+b and ab are even.
Since ab is even, at least one of a or b must be even. WLOG assume that a is even. Since a+b is even
and a is even, b must be even.
Case II: a+b and ab are odd.
Since a+b is odd, a and b must have different parities, but then ab would be even since one of a or b
is even. This case is thus impossible.
2. pg. 102, problem 20.
Prove that if 3x4 + 1 ≤ x7 + x3 then x > 0.
Assume that x ≤ 0. Then 3x4 + 1 ≥ 1 > 0 ≥ x7 + x3, a contradiction.
3. pg. 103, problem 38
Let A, B and C be sets. Prove that (A - B) ∪ (A - C) = A - (B ∩ C).
Proof 1:
A - (B ∩ C) = A ∩ (B ∩ C)c = A ∩ (B ∪ C ) = (A ∩ B) ∪ (A ∩ C)
= (A - B) ∪ (A - C).
Proof 2: (element chasing)
Let x ∈ (A - B) ∪ (A - C).
Case 1: x ∈ (A - B).
Then x ∈ A and x ∉ B. Si nce B ∩ C ⊆ B, x ∉ B ∩ C. Ther ef or e,
x ∈ A - (B ∩ C).
Case 2: x ∈ (A - C).
Then x ∈ A and x ∉ C. Si nce B ∩ C ⊆ C, x ∉ B ∩ C. Ther ef or e,
x ∈ A - (B ∩ C).
So i n ei t her case we have x ∈ A - (B ∩ C). Thus, (A - B)∪(A - C
A-(B ∩ C).
Now suppose t hat y
A - (B ∩ C). So, y
A and y
B ∩ C. By t he DeMor gan l aws,
y
B
C. I f y
B t hen y
A - B, and i f y
C t hen y
A - C. Hence, we have y
A - B) ∪ (A - C). Ther ef or e, A - (B ∩ C
A - B) ∪ (A - C) and we have pr oved t he
r esul t .
4. pg. 124, problem 12
Show that 1000 cannot be written as the sum of three positive integers, an even number of which
are even.
Proof: Assume 1000 = a + b + c (all positive integers).
Case 1: Two of a, b or c are even.
WLOG assume that a and b are even. Then a = 2s, b = 2t and c = 2u + 1 where s,t and u are integers.
Then a + b + c = 2s + 2t + 2u + 1 = 2(s + t + u) + 1. Since s + t + u is an integer, this sum is odd and so
cannot be 1000.
Case 2: None of a, b or c are even.
Then a = 2s +1, b = 2t + 1 and c = 2u + 1 where s, t and u are integers. Then,
a + b + c = 2s + 1 + 2t + 1 + 2u + 1 = 2(s + t + u + 1) + 1. Since s + t + u + 1 is an integer, the sum is odd
and so cannot be 1000.
5. pg. 124, problem 22
Let m be a positive integer of the form m = 2s, where s is an odd integer. Prove that there do not
exist positive integers x and y such that x2 - y2 = m.
Pf: BWOC, suppose that there exist integers x and y so that x2 - y2 = m = 2s. Thus, (x+y)(x-y) = 2s, so at
least one of the factors on the left must be even (if they are both odd their product is odd, a
contradiction). If either x + y or x - y is even, then so is the other since (x + y) = (x - y) + 2y. Since they
are both even, (x+y)(x-y) = 4t = 2s, so s = 2t contradicting the fact that s is odd. Therefore, there are no
integers x and y such that x2 - y2 = m.