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Math Help - STRONG induction

  1. #1
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    STRONG induction

    i have this question that i couldnt solve its about STRONG INDUCTION i need ur help pls.

    Prove by strong induction that any natural number n greater than 7 can
    be written as a sum of 3's and 5's, being understood that no 3's or no 5's is
    also accepted, as in 12 = 3 +3 + 3 + 3.

    thank you
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  2. #2
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    ??? 9 can't. Do you mean sum or difference of 3s and 5s?
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  3. #3
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    Quote Originally Posted by Debsta View Post
    ??? 9 can't. Do you mean sum or difference of 3s and 5s?
    Why not?

    9 = 3+3+3 is an accepted form, according to his post.

    Base case: n=8. 8=5+3 so we are good to go.

    Assume that n \in \mathbb{N} can be written as a sum of 3's and 5's. We want to prove that we can also express n+1 as such a sum.

    Lemma: The implementation of n as such a sum has either at least three 3s or one 5 in it.
    Proof: If n has a 5 in its sum, we are done. Assume it doesn't, then n = 3m for some positive integer m. But we only take n > 7, therefore m\geq 3 and the lemma is correct.

    According to the inductive assumption, we know that n = 3k + 5m where k\geq 3 \ \text{or} \ m \geq 1.

    Substituting that into n+1, we get: n+1 = 3k + 5m + 1. Now, if:

    (I)  k <3:
    We know that there is at least one 5 in the sum. This gives us that: n+1 = 3k + 5(m-1) + 5 + 1 = 3k + 5(m-1) + 6 = 3(k+2) + 5(m-1) and we are done.

    (II) m < 1:
    Then, n = 3m: \ m \in \mathbb{N}, \ m\geq 3. This gives us that: n+1 = 3m+1 = 3(m-3)+3\cdot3 + 1 = 3(m-3)+10 = 3(m-3) + 2\cdot 5 and then n+1 is also a sum of 3,5s. Therefore it always is, and we are done.
    Last edited by Defunkt; November 6th 2009 at 08:56 PM.
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