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)
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)??
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?
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??
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?
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.