# Archimedean property

• Feb 8th 2010, 07:28 PM
harish21
Archimedean property
The archimedean property is when these two equivalent properties hold:
- for every positive number c, there is a natural number n such that n>c
- for any positive number e, there is a natural number n such that 1/n < e

How do we prove that the Archimedean Property is the outcome of the statement that for any real number c, there is an integer in the interval [c,c+1)
• Feb 8th 2010, 07:30 PM
Drexel28
Quote:

Originally Posted by harish21
The archimedean property is when these two equivalent properties hold:
- for every positive number c, there is a natural number n such that n>c
- for any positive number e, there is a natural number n such that 1/n < e

How do we prove that the Archimedean Property is the outcome of the statement that for any real number c, there is an integer in the interval [c,c+1)

What's the actual question? Also, the most common form of the Archimedean principle I have seen (besides your second form) is that given any $\displaystyle 0<x<y$ there exists some $\displaystyle n\in\mathbb{N}$ such that $\displaystyle y<nx$.
• Feb 8th 2010, 07:38 PM
harish21
Quote:

Originally Posted by Drexel28
What's the actual question? Also, the most common form of the Archimedean principle I have seen (besides your second form) is that given any $\displaystyle 0<x<y$ there exists some $\displaystyle n\in\mathbb{N}$ such that $\displaystyle y<nx$.

The first three lines are how my book defines the Archimedean property. the second part is the real question which is confusing me!

How do we prove that the Archimedean Property is the outcome of the statement that for any real number c, there is an integer in the interval [c,c+1)??
• Feb 8th 2010, 07:58 PM
Drexel28
Quote:

Originally Posted by harish21
The first three lines are how my book defines the Archimedean property. the second part is the real question which is confusing me!

How do we prove that the Archimedean Property is the outcome of the statement that for any real number c, there is an integer in the interval [c,c+1)??

What I'm saying, are you asking to prove that

Quote:

for any real number c, there is an integer in the interval [c,c+1)??
implies or is implied by the Archimedean principle?
• Feb 8th 2010, 08:09 PM
harish21
Quote:

Originally Posted by Drexel28
What I'm saying, are you asking to prove that

implies or is implied by the Archimedean principle?

I am trying to prove that the statement "for any real number c, there is an integer in the interval [c,c+1) " IMPLIES the archimedian property!

• Feb 8th 2010, 08:16 PM
Drexel28
Quote:

Originally Posted by harish21
I am trying to prove that the statement "for any real number c, there is an integer in the interval [c,c+1) " IMPLIES the archimedian property!

Ok. So where have you started from? I need a little effort here!
• Feb 8th 2010, 08:30 PM
harish21
Quote:

Originally Posted by Drexel28
Ok. So where have you started from? I need a little effort here!

So if we assume an integer k, then according to the statement,

c<= k< c+1.

So for any value of c, this inequality holds true.

there is a natural number less than c+1, which is greater than k!

So doesn't this imply the 1st Archimedean property (out of the two properties that I have stated at the beginning of this question) ? I am not sure if what I have done is correct
• Feb 9th 2010, 02:35 AM
HallsofIvy
Not immediately but almost. Your statement of the Archimedian property was that given any positive c, there exist n larger than c while your given property says that n is larger than or equal to c. You have to handle the case that c= n< c+1. Are there any integers larger than n?
• Feb 9th 2010, 01:36 PM
xalk
Quote:

Originally Posted by harish21
So if we assume an integer k, then according to the statement,

c<= k< c+1.

So for any value of c, this inequality holds true.

there is a natural number less than c+1, which is greater than k!

So doesn't this imply the 1st Archimedean property (out of the two properties that I have stated at the beginning of this question) ? I am not sure if what I have done is correct

Yes it does ,what you have done is correct.

But we have : $\displaystyle c\leq k\leq c+1$ and you want n>c

So take n=k+1 ,and since k+1>k ,then n>c.

But the difficult part is ,when we are asked to prove that the Archimedean implies that:

for any real No c there exists a natural No k such that :

$\displaystyle c\leq k\leq c+1$.

Can you prove that??
• Feb 9th 2010, 02:23 PM
harish21
Quote:

Originally Posted by xalk
Yes it does ,what you have done is correct.

But we have : $\displaystyle c\leq k\leq c+1$ and you want n>c

So take n=k+1 ,and since k+1>k ,then n>c.

But the difficult part is ,when we are asked to prove that the Archimedean implies that:

for any real No c there exists a natural No k such that :

$\displaystyle c\leq k\leq c+1$.

Can you prove that??

According to the question, that for any real number c, there is an integer in the interval [c,c+1). Here c+1 is NOT included in the interval!!

But you have stated that c<=k<=c+1. I don't think both c and c+1 are included in the interval.

I have done the proof that there is exactly one integer in the interval [c,c+1).

But the one you are asking for is the interval [c,c+1]

Am I right?
• Feb 9th 2010, 04:08 PM
xalk
Quote:

Originally Posted by harish21
According to the question, that for any real number c, there is an integer in the interval [c,c+1). Here c+1 is NOT included in the interval!!

But you have stated that c<=k<=c+1. I don't think both c and c+1 are included in the interval.

I have done the proof that there is exactly one integer in the interval [c,c+1).

But the one you are asking for is the interval [c,c+1]

Am I right?

Yes you are right ,in the oppening post it was [c,c+1) insteadof [c,c+1] that i assumed .But it is not important,because [c,c+1) => [c,c+1].

Generaly for any two Nos x<y => $\displaystyle x\leq y$.

The two theorems are :

1) for any real c>0 there exists a natural No n such that n>c

2) for any real c there exists a natural No n such that $\displaystyle c\leq n<c+1$ or $\displaystyle c\leq n\leq c+1$

Now you have proved that (2) implies (1) (in the case where c>0) .

But you have not proved that (1) implies (2),unless you have done this proof at home.
• Feb 9th 2010, 04:34 PM
Plato
Quote:

Originally Posted by xalk
But you have not proved that (1) implies (2),unless you have done this proof at home.

xalk, do you realize that this question did not ask for that implication?
The question as posted is:
"How do we prove that the Archimedean Property is the outcome of the statement that for any real number c, there is an integer in the interval [c,c+1)"?
• Feb 9th 2010, 05:13 PM
xalk
Quote:

Originally Posted by harish21

I have done the proof that there is exactly one integer in the interval [c,c+1).

Quote:

Originally Posted by xalk

The two theorems are :

1) for any real c>0 there exists a natural No n such that n>c

2) for any real c there exists a natural No n such that $\displaystyle c\leq n<c+1$ or $\displaystyle c\leq n\leq c+1$

Now you have proved that (2) implies (1) (in the case where c>0) .

But you have not proved that (1) implies (2),unless you have done this proof at home.

Quote:

Originally Posted by Plato
xalk, do you realize that this question did not ask for that implication?
The question as posted is:
"How do we prove that the Archimedean Property is the outcome of the statement that for any real number c, there is an integer in the interval [c,c+1)"?