• Share
  • Email
  • Embed
  • Like
  • Save
  • Private Content
Solution3
 

Solution3

on

  • 785 views

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

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

Statistics

Views

Total Views
785
Views on SlideShare
785
Embed Views
0

Actions

Likes
1
Downloads
12
Comments
0

0 Embeds 0

No embeds

Accessibility

Categories

Upload Details

Uploaded via as Adobe PDF

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

    Solution3 Solution3 Document Transcript

    • Assignment 3Deadline: April 26 (Thursday Class), April 27(Friday Class) Harshit Kumar April 25, 2012
    • 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 > nAns. Find the attached pdf file induction.pdfQ4. 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 = 14(ii) The contradiction would be, let there exist x, y ∈ Z such thatx2 − 4y − 3 = 0.=⇒ x2 = 4y + 3.=⇒ x2 = 2 ∗ 2y + 2 + 1=⇒ x2 = 2(2y + 1) + 1=⇒ x2 is odd=⇒ x is oddsince x is odd, =⇒ x = 2n + 1, for some integer nSubstitute 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
    • 2