Re: Mathematical induction

On Mon, 04 Feb 2008 02:05:36 GMT, Gerry Myerson
<gerry@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote:

In article <48hcq3lke29a697hpbl42686bukkmnfbk9@xxxxxxx>,
quasi <quasi@xxxxxxxx> wrote:

On Sun, 03 Feb 2008 17:39:11 -0500, quasi <quasi@xxxxxxxx> wrote:

On Sun, 3 Feb 2008 07:21:08 -0800 (PST), Ed Durrett
<edward_durrett@xxxxxxxxxxxxx> wrote:

Hello folks,

just a quick question regarding mathematical induction.

The usual way (besides the base case) is to show
n -> n + 1.

However sometimes it's hard to show this.
I came across a variation of mathematical induction
due to Cauchy. One shows
n -> 2n *and* n -> n - 1 (instead of n -> n + 1).
So in the first step we go forward and in the second
backwards. I would call this method
"forward/backward-induction".

Question: Is there a mathematical expression for
this type of induction?

It's usually called just "Backwards Induction". It's a fun method.

But it doesn't actually require "P(n) => P(2n)". You could replace
that with "P(n) => P(m) for some m > n".

Here is the usual version ...

Principle of Backwards Induction:

If P(n) is true for infinitely many positive integers n,
and if (for n >1), P(n) => P(n-1), then P(n) is true for all
positive integers n.

Sorry, it's called "Reverse Induction", not "Backwards Induction".

I suppose it could simply be called noitcudni.

q: You were able to prove it? If so, by what method?

a: noitcudni

q: Your answer came out garbled. Are you saying you couldn't prove it?

a: tievorpdididiasion

quasi
.

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