Homework 1
ˆ Please submit your answers to all questions.
ˆ We will mark your answers to 3 questions.
ˆ We will provide you with full solutions to all questions.
1. Write the following sentences in symbolic logic notation and state whether
they are true or false, or open sentences (you do not need to prove your
answer). Make sure to note which statements/open sentences are denoted
with which letter.
Example: The sentence, “The car is red and blue but not green” can be
written as (P ∧ Q) ∧ (∼ R), (or (P ∧ Q) ∧ (¬R)), where P : “The car is
red”, Q: “The car is blue”, and R: “The car is green”. (You could also write
P (car), Q(car), and R(car) to reflect what variable the truth values of P , Q,
& R depend on.) Since the truth value of the original sentence depends on
the car, it is an open sentence, not a statement.
(a)
(b)
(c)
(d)
2 is even and 11 is prime.
If n is a multiple of 7 and 4, then it is a multiple of 14.
5 ≤ x ≤ 17.
A real number x is less than −3 or greater than 3 if its square is greater
than 9.
Solution:
(a) 2 is even and 11 is prime.
This statement can be written as P ∧ Q for
P : “2 is even”, and
Q: “11 is prime”.
The statement is true.
(b) If n is a multiple of 7 and 4, then it is a multiple of 14.
This statement can be written as (P (n) ∧ Q(n)) =⇒ R(n) for
P (n): “n is a multiple of 7”,
Q(n): “n is a multple of 4”, and
R(n): “n is a multiple of 14”.
The statement is true.
Homework 1
(c) 5 ≤ x ≤ 17.
This statement can be written as P (x) ∧ Q(x) for
P (x): “x ≥ 5”, and
Q(x): “x ≤ 17”.
It is an open sentence.
(d) A real number x is less than −3 or greater than 3 if its square is
greater than 9.
This statement can be written as R(x) =⇒ (P (x) ∨ Q(x)) for
P (x): “x is less than −3”,
Q(x): “x greater than 3”, and
R(x): “x2 is greater than 9”.
The statement is true.
2. Prove that the product of two odd numbers is odd.
Solution: Assume n, m are odd integers. Then we know n = 2k + 1 and
m = 2ℓ + 1 for some k, ℓ ∈ Z. Thus,
nm = (2k + 1)(2ℓ + 1) = 2(2kℓ + k + ℓ) + 1.
Therefore, since 2kℓ + k + ℓ ∈ Z for k, ℓ ∈ Z, we see nm is odd.
3. Let n ∈ Z. Prove that if 3 | n − 2 then 3 ∤ n2 + 2n + 8.
Solution: Assume that 3 | n − 2 and hence n − 2 = 3k, for some integer
k. Then n = 3k + 2. So,
n2 +2n+8 = (3k+2)2 +2(3k+2)+8 = 9k 2 +18k+16 = 3(3k 2 +6k+5)+1.
Since (3k 2 + 6k + 5) ∈ Z it follows that 3 ∤ n2 + 2n + 8.
4. Prove that: If −3 < x < 1, then x2 + 2x − 3 < 0.
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Homework 1
Solution: Since −3 < x < 1, we have x + 3 > 0 and x − 1 < 0. So
x2 + 2x − 3 = (x + 3)(x − 1) < 0.
5. Let n, a, b, c, d ∈ Z with n > 0. Prove: if n|a and n|c, then n|(ab + cd + ac).
Solution: Since n|a and n|c, we can write a = nj and c = nk for some
j, k ∈ Z.
Then ab + cd + ac = njb + nkd + n2 jk = n(jb + kd + njk).
Since (jb + kd + njk) ∈ Z, we have n | (ab + cd + ac).
6. Let a, b ∈ R. Prove that
1
1
ab − b2 ≤ a2
2
2
Solution: We know that (a−b)2 ≥ 0 because it is a squared real number.
Hence a2 − 2ab + b2 ≥ 0 so a2 + b2 ≥ 2ab and so ab − 21 b2 ≤ 12 a2 .
7. Let c ∈ Z. Prove that if 7c + 11 is odd then 9c + 17 is odd.
Solution: Assume that 7c + 11 is odd. Hence 7c + 11 = 2k + 1 for some
k ∈ Z.
Add 2c + 6 to both sides to get 9c + 17 = 2k + 2c + 7 = 2(k + c + 3) + 1.
Since (k + c + 3) ∈ Z it follows that 9c + 17 is odd.
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