0

If you want to do induction on a parameter $a \in \mathbb{Z}$. In the base case, you prove it for $a=0$. And then you suppose it is true for $n$ and prove it for $n+1$ and $n-1$? Is that correct?

Or you have to do induction for all $a\geq 0$ and then for all $a \leq 0$?

bob
  • 1,256
  • Are you saying that you need to prove some statement is true for all integers (not just non-negative ones)? In any case, how you use induction will depend on the problem you are trying to solve. What are you trying to prove? – Luftbahnfahrer Mar 28 '17 at 17:04
  • Yes indeed for ALL integers. It's something about Fibonacci numbers (for all $a,m,n \in \mathbb{Z}$): \begin{equation} F_{2mn+a} \equiv (-1)^{m(n+1)} F_a \mod L_n \end{equation}. And I want to do induction on $m$ for abritrary but fixed $n,a \in \mathbb{Z}$ – bob Mar 28 '17 at 17:08
  • Your suggested method would certainly suffice. Prove that it is true when $m=0$. Let $r$ be an integer and suppose it is true for $m=r$, then prove that it is true when $m=r+1$ and $m=r-1$. Then the statement will be proven true for all integers by induction. – Luftbahnfahrer Mar 28 '17 at 17:43
  • Okay, thanks a lot! – bob Mar 28 '17 at 18:21

0 Answers0