Re: Mathematical induction
- From: "[Mr.] Lynn Kurtz" <kurtz@xxxxxxxxxxxxxxx>
- Date: Mon, 04 Feb 2008 04:48:31 GMT
On Sun, 03 Feb 2008 21:32:52 -0500, quasi <quasi@xxxxxxxx> wrote:
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
Did you guys ever wind up simplifying an expression like
- ( a +- b) to (-a -+ b) and turn to your class and tell them that is
the suminrosulp sign? Great fun and I think you are obviously ready
for it.
--Lynn
.
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