??? 9 can't. Do you mean sum or difference of 3s and 5s?
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
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 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 and the lemma is correct.
According to the inductive assumption, we know that where .
Substituting that into n+1, we get: . Now, if:
(I) :
We know that there is at least one 5 in the sum. This gives us that: and we are done.
(II) :
Then, . This gives us that: and then n+1 is also a sum of 3,5s. Therefore it always is, and we are done.