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Thread: Mathetmatical Induction

  1. #1
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    Mathetmatical Induction

    (a) Let k be in the set of N. Prove that if k^2+k+5 is even, then (k+1)^2 +(k+1) +5 is even.

    Well, since I wasn't sure how to prove this by induction, I did it by contrapositive and it worked out. Is that the correct way to do it?

    (b) What is the value of k^2+k+5 when k = 7... Fairly straightforward, I got 61.

    (c) What information do (a) and (b) provide?

    Part a and c are where I am confused.

    If I did part a correctly, then k^2+k+5 was proven to be even, but part b has proved otherwise.. What gives?
    Last edited by MrJank; Sep 25th 2018 at 11:51 AM.
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  2. #2
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    Re: Mathetmatical Induction

    Quote Originally Posted by MrJank View Post
    (a) Let k be in the set of N. Prove that if k^2+k+5 is even, then (k+1)^2 +(k+1) +5 is even.

    Well, since I wasn't sure how to prove this by induction, I did it by contrapositive and it worked out. Is that the correct way to do it?
    I'm not sure what you mean by "the correct way". There is never one correct way to prove a true statement. If the statement is true for a given "k" then "(k+1)^2+ (k+1)+ 5" is not going to be even so induction is not a good way to go. Instead do the algebra: (k+1)^2+ (k+1)+ 5= k^2+ 2k+ 1+ k+ 1+ 5= (k^2+ k+ 5)+ 2k+ 2= (k^2+ k+ 5)+ 2(k+ 1) and the sum of two even numbers is even.

    (b) What is the value of k^2+k+5 when k = 7... Fairly straightforward, I got 61.
    Yes, 49+ 7+ 5= 56+ 5= 61. Pretty basic arithmetic.

    (c) What information do (a) and (b) provide?
    None! (a) says what is true if k^2+ k+ 5 is even but 61 is not even so (a) does not apply.

    Part a and c are where I am confused.

    If I did part a correctly, then k^2+k+5 is was proven to be even, but part b has proved otherwise.. What gives?
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