In an induction step after having establised the base case, May one prove $p(k+1)\implies p(k+2)$ instead of $p(k)\implies p(k+1)$?

Question

I picked up on the fact that a proof I was working on could be more easily made: it would involve less lines, If we proved as our induction step $p(k+1)\implies p(k+2)$. I stay on top of the multiple variations of the principle of mathematical induction so I was wondering whether this could be one of those or not.

Whether you call the integer $k$ or $k+1$ is irrelevant. What's important is that the inductive step progress from a positive integer to the next integer. — Matthew Leingang, Jul 07 '21 at 15:33

score 1 · Answer 1 · answered Jul 07 '21 at 15:30

1

Put $m=k+1$ to see that $m+1=k+2$

So, $p(k+1)\implies p(k+2)$ is equivalent to $p(m)\implies p(m+1)$ which is equivalent to $p(k)\implies p(k+1)$

So, you can of course go for $p(k+1)\implies p(k+2)$

answered Jul 07 '21 at 15:30

Sayan Dutta

8,831

In an induction step after having establised the base case, May one prove $p(k+1)\implies p(k+2)$ instead of $p(k)\implies p(k+1)$?

1 Answers1