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)
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)??
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
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?
Yes it does ,what you have done is correct.
But we have : 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 :
.
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?
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 => .
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 or
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.