Edit made on May 01, 2009 by GuestEditor at 13:18:35

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WW

HEADERS_END

Proof by induction is a proof made by ~~proving~~ first assuming that a statement is true for a general case (i.e. when EQN:n=k ), then proving that it still holds true for the next case (i.e. when EQN:n=k+1 ) and then proving that it is true for the first (base) case; if ~~all three of these are true,~~ the statement /does/ hold for /both/ the base case /and/ the inductive step ( EQN:n=k+1 ), then, by induction, the statement must be true.

!! METHOD

To prove that something is true for all integers EQN:n{\ge}r ~~, first prove~~ :

* assume that it is true for EQN:n=k ~~, then~~

* prove that it ~~is~~ remains true for EQN:n=k+1 ~~, and finally~~

* prove that it is true for EQN:n=r.

~~----~~ !! Examples

#Hello# #Josh#

case n = k is not proved but assumed.

The proof consists of two steps:

* The basis (base case):

** Show that the statement holds when n = 0 (or n = 1 ...)

* The inductive step:

** show that if the statement holds for It would be nice to have some ~~n, then the statement also holds when n + 1 is substituted for n. ~~

Hope this helps small, clean examples here. Not too many, not too much.