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### 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/
```
Associated Topics:
College Modern Algebra

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