Solution to Assignment 3 of Discrete Mathematics. It includes induction, proof by contrapositive, proof by contradiction.

1. Assignment 3
Deadline: April 26 (Thursday Class), April 27(Friday Class)
Harshit Kumar
April 25, 2012

2. Q1. Write the following sets?
1. x ∈ Z | x = y 2 f or some integer y ≤ 3 - {0, 1, 4, 9, 16}
2. x ∈ Z | x2 = y f or some integer y ≤ 3 - {0, 1, −1}
Q2. Suppose the universal set U = {−1, 0, 1, 2}, A = {0, 1, 2} and B = {−1, 2}. What are the following sets?
1. A ∩ B - {2}
2. A ∪ B - {−1, 0, 1, 2}
3. Ac - {−1}
4. A − B - {0, 1}
5. A × B - {(0, −1)(0, 2)(1, −1)(1, 2)(2, −1)(2, 2)}
6. P (B) - {} - {∅, {−1}, {2}, {−1, 2}}
Q3. Prove the following by Induction
1. If a is odd and b is odd, then a ∗ b is odd.
2. Any integer i > 1 is divisible by p, where p is a prime number.
r n+1 −1
3. r = 1, ∀n ≥ 1, 1 + r + ..... + rn = r−1
2
n∗(n+1)
4. ∀n ≥ 1, 13 + 23 + ..... + n3 = 2
5. ∀n ≥ 1, 22n − 1isdivisibleby3
6. ∀n ≥ 2, n3 − nisdivisibleby6
7. ∀n ≥ 3, 2n + 1 < 2n
1 1 1 √
8. ∀n ≥ 2, √
1
+ √
2
+ .... + √
n
> n
Ans. Find the attached pdf file induction.pdf
Q4. Prove the following by Contradiction
1. There exists no integers x and y such that 18x + 6y = 1
2. If x, y ∈ Z, then x2 − 4y − 3 = 0.
Ans 4 (i)
The contradiction would be, let there exist integers x and y such that
=⇒ 18x + 6y = 1.
=⇒ 1 = 2 × (9x + 3y)
=⇒ 1 is even, and hence a contradiction. We can thus arrive at the conclusion that our assumption , there exists integers x and y such that 18x + 6y = 1,
is wrong.
=⇒ there exists no integers x and y such that 18x + 6y = 1
4(ii) The contradiction would be, let there exist x, y ∈ Z such that
x2 − 4y − 3 = 0.
=⇒ x2 = 4y + 3.
=⇒ x2 = 2 ∗ 2y + 2 + 1
=⇒ x2 = 2(2y + 1) + 1
=⇒ x2 is odd
=⇒ x is odd
since x is odd, =⇒ x = 2n + 1, for some integer n
Substitute x = 2n + 1 in the original equation x2 − 4y − 3 = 0, we get
(2n + 1)2 − 4y − 3 = 0
=⇒ 4n2 + 1 + 4n − 4y − 3 = 0
=⇒ 4n2 + 4n − 4y = 2
=⇒ 2n2 + 2n − 2y = 1
=⇒ 2(n2 + n − 1) = 1
=⇒ 1 is even. This is a contradiction. We can thus arrive at the conclusion that our assumption, there exists x, y ∈ Z such that x2 − 4y − 3 = 0, is wrong.
implies there exists x, y ∈ Z such that x2 − 4y − 3 = 0.
Q5. Prove the following by Contrapositive
1. ∀n ∈ Z, if nk is even, then n is even.
2. ∀x, y ∈ Z, if x2 (y 2 − 2y) is odd, then x and y are odd.
3. ∀x ∈ R, if x2 + 5x < 0, then x < 0
4. If n is odd, then (n2 − 1) is divisible by 8.
5. If n ∈ N and 2n − 1 is prime, then n is prime.
6. ∀x, y ∈ Z and n ∈ N, if x3 ≡ y 3 (mod n)
Ans
1

3. 2