2

The exercise goes like this:

Prove that the statement P(n)

$n^2 + 3*n + 1$ is even

always fails.

My question is if it is sufficient to show that the base case fails for some of the first terms, or is that too trivial and I have to show that it fails when $n$ is odd and when it's even.

Thank you in prior

Allorja
  • 303
  • If you want to prove that the statement always fails, then you should prove (by induction) that $n^2+3n+1$ is odd, for all $n$, and the hint given in the answers should help you with this. – user505379 Oct 05 '18 at 20:32
  • Thaaanks. It is sooo obvious but I just couldn't think of it that way. I proved that it fails whenever n is odd and even, but the way you suggested is way more convenient. All the best – Allorja Oct 05 '18 at 20:38

2 Answers2

1

No it is not enough to prove that it fails just for some small (base) cases. You have to do also IS.

By IH the $P(n)$ is odd, now

$$P(n+1)= (n+1)^2+3(n+1)+1 = (n^2+3n+1)+2n+4 = P(n)+2(n+2)$$

nonuser
  • 90,026
  • This was another part of the exercise where I had to show P(n) implies P(n+1) , but anyway thank you for your trying. – Allorja Oct 05 '18 at 20:34
0

You don't need to show that base case fails for some cases because there is only one base case in this question and that is for $n = 0$ or $n = 1$ or it depends on whatever restriction $n$ has (for instance $n \in \mathbb{N}$). And even if it is too trivial, for the sake of completeness, you should show that the first statement $P(1)$ fails in this case. That's how induction starts.

ArsenBerk
  • 13,211