Mathematical Deduction

Date: 07/22/97 at 12:32:32
From: Cat Fyle
Subject: Modern Algebra

I have a test tomorrow in Modern Algebra, but the test is going to 
consist of complex numbers. I have not had complex numbers, so
I have no idea where to begin. 

Here is one of the questions:

Let E be a subset of P such that i) 1 is in E ii) whenever n is in E, 
also n+1 is in E. Then E = P.

Prove the above statement.  This is the principle of Mathematical 
Induction for well-ordered positive integers: Let P denote the set of 
all positive integers such that S <= P and S <> P, then there exists 
an integer m in S such that m <= n for all n in S.

I would appreciate any help you could provide.

Thank you very much, CAT

Date: 07/22/97 at 12:58:26
From: Doctor Rob
Subject: Re: Modern Algebra

Assume that E <> P.  Let S be the set of integers in P but not in E.
Apply the Principle of Mathematical Induction.  Then show that the m 
that it produces leads to a contradiction, that there is an element of 
E which is larger than m by showing m+1 is in E.  Then the assumption 
that E <> P must be wrong, so E = P.

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/