Real Analysis Help

sumpm1 · 2009-02-20 04:53:24

Hi I need some more help in Real Analysis. We are still covering infimum and supremum of a set, as well as the Completeness Axiom, Archimedean Property, and the following theorems.

Completeness Axiom: Each nonempty set of real numbers that is bounded above has a supremum.

Theorem: Between any two distinct real numbers there is a rational number and an irrational number.

Archimedean Property of the Real Numbers: If

and are real numbers, then there exists a positive integer such that .

Theorem (Follows from Archimedean): The following statements are equivalent.
1. If

and are real numbers, then there exists a positive integer such that .
2. The set of positive integers is not bounded above.
3. For each real number , there exists an integer such that .
4. For each real number , there exists a positive integer such that .

I have these two questions:

1) Prove that a nonempty finite set contains its infimum.

2) Prove each of the following results - give a direct proof of each one - without the use of the Completeness Axiom. These results could be called the Archimedean Property of the rational numbers.

a) If

and are positive rational numbers, then there exists a positive integer such that .

b) For each positive integer

, there exists a rational number such that .

c) For each rational number

, there exists an integer such that .

d) For each positive rational number

, there is a positive integer such that .

Any help on #1 would be greatly appreciated.

Thanks

Last edited by sumpm1 (2009-02-20 05:44:07)

JaneFairfax · 2009-02-20 08:32:05

mathsyperson · 2009-02-20 09:29:25

I want to point out that the Archmides Principle actually says that given a real x, some natural n exists such that n > x.

The way you have it, choosing 0=a<b would make it false.

JaneFairfax · 2009-02-20 12:25:44

JaneFairfax · 2009-02-20 13:15:50

sumpm1 wrote:

4. For each real number
, there exists a positive integer
such that
.

sumpm1 · 2009-02-21 11:24:42

mathsyperson wrote:

I want to point out that the Archimedes Principle actually says that given a real x, some natural n exists such that n > x.
The way you have it, choosing 0=a<b would make it false.

JaneFairfax · 2009-02-21 11:28:27

sumpm1 · 2009-02-22 07:11:43

I know they look easy, but I was wondering what the beginning of one of the questions from #2 would look like, any one of them would do. Thank you.

Ricky · 2009-02-22 12:12:48

For 2a, expand on what it means for a and b to be rational numbers. Then multiply each side of the equation by a number "big enough" to make everything integers. Assuming you're allowed to use the archimedian property of the integers, i.e. natural numbers are not bounded, then the proposition is proven.

Math Is Fun Forum

#1 2009-02-20 04:53:24

Real Analysis Help

#2 2009-02-20 08:32:05

Re: Real Analysis Help

#3 2009-02-20 09:29:25

Re: Real Analysis Help

#4 2009-02-20 12:25:44

Re: Real Analysis Help

#5 2009-02-20 13:15:50

Re: Real Analysis Help

#6 2009-02-21 11:24:42

Re: Real Analysis Help

#7 2009-02-21 11:28:27

Re: Real Analysis Help

#8 2009-02-22 07:11:43

Re: Real Analysis Help

#9 2009-02-22 12:12:48

Re: Real Analysis Help

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