A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Use mathematical induction to prove that the statement is true for every positive integer n. 2 is a factor of n2 - n + 2 I really need help on this one

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Loser66

  2. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Have you done mathematical induction before, like -Base case -hypothesis for case n -proof for case n+1

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i has not mate

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so hows do i do that thingy majig your talking about

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    are you there my fellow good sir

  6. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, I was looking for a good reference for you. The following link shows us how to do mathematical inductions, with examples. Example 4 is particularly similar to your problem. http://www.analyzemath.com/math_induction/mathematical_induction.html

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok thnx mate i will chat you up if i have any problems

  8. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Good, way to go! :)

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok cheerio

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Statement P (n) is defined by n 3 + 2 n is divisible by 3 STEP 1: We first show that p (1) is true. Let n = 1 and calculate n 3 + 2n 1 3 + 2(1) = 3 3 is divisible by 3 hence p (1) is true. STEP 2: We now assume that p (k) is true k 3 + 2 k is divisible by 3 is equivalent to k 3 + 2 k = 3 M , where M is a positive integer. We now consider the algebraic expression (k + 1) 3 + 2 (k + 1); expand it and group like terms (k + 1) 3 + 2 (k + 1) = k 3 + 3 k 2 + 5 k + 3 = [ k 3 + 2 k] + [3 k 2 + 3 k + 3] = 3 M + 3 [ k 2 + k + 1 ] = 3 [ M + k 2 + k + 1 ] Hence (k + 1) 3 + 2 (k + 1) is also divisible by 3 and therefore statement P(k + 1) is true.

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok so mate i need more explanation on this it so short worded and down righ confusing

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @mathmate could you help me for a sec mate

  13. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The first step is to establish that the statement you want to prove is true for some n. So P(1)=n^3+2n=1^3+2(1)=3 is divisible by three. In principle, if we have time to show this for all integers, we're done, but we cannot, because n goes to infinity So the next best thing is to show that, if 3|P(n), then 3|P(n+1). 3|x reads 3 divides x. Why ? Because if the hypothesis 3|P(n) is true, then 3|P(n+1) is true, then 3|P(n+2) is true, and so on. Once we have achieved that, we go back to say that 3|P(1), therefore 3|P(2), 3|P(3)... without having to do ALL of the integers.

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok mate so i get that so how di i do part two of my fiding my answer mate

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you there mate

  16. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Ashleyisakitty

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @sky425

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @leon549

  19. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @MrNood

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    anybody of service here mate

  21. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @helpppppppppp

  22. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    f n is even, then n = 2m for some integer m. then substitute n^2 - n + 2 = (2m)^2 - (2m) + 2 =4m^2 - 2m + 2 = 2 ( 2m^2 - m + 1) . and this is a factor of 2 by definition, since it is 2 times some integer , if n is odd , then n = 2r+1 for some integer r n^2 - n + 2 = (2r + 1) ^2 - (2r + 1) + 2 = 4r^2 + 4r + 1 - 2r -1 + 2 by foiling and distributing = 4r^2 + 2r + 2 = 2 ( 2r^2 + r + 1) . and this is a factor of 2 again. we are done :) ok i tried my best and this is what i got is it true

  23. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    This is a correct direct proof using "proof by cases". The question asks for proof by mathematical induction, which involves the three steps: 1. Proof a base case (say, n=1). 2. Inductive hypothesis: assume statement is true for case n=k. 3. Inductive proof and conclusion: prove that statement is true for case n=k+1. Done. The steps are extremely similar to example 4 that you have studied. Just have to replace the statement with yours. 1. base case : 2 is a factor of f(n)=n2 - n + 2 when n=1. Well, n^2-n+2 = f(1)= 1^2-1+2 = 2, so 2 is a factor of 2, hence statement is true for n=1 (base case). 2. Inductive hypothesis: Assume that (or IF) statement is true for case n=k: 2 is a factor of n2 - n + 2 for n=k, therefore 2 is a factor of f(k)= k^2-k+2, or f(k)=2m, where m\(\in\)Z 3. Work on case n=k+1 f(k+1)=(k+1)^2-(k+1)+2 = k^2+2k+1-k-1+2 =(k^2-k+2) +2k =(2m)+2k =2(m+1), hence 2 is a factor of f(k+1), given that 2 is a factor of f(k). we have just shown that 2 is a factor of f(k+1), which completes the proof by mathematical induction that 2 is a factor of f(n) for all n\(\in\)Z+.

  24. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.