Archimedean Property of Real Numbers
​
The order – completeness property of real number has important consequence. The most important of them is the following Archimedean property of real numbers.
​
Theorem 1 : Let a be any real number and b any positive real number. Then there exists a positive integer n such that
nb > a.
​
Proof: If a ≤ 0, the theorem is obvious because then for every positive integer n we have nb > a.
​
[ n >0, b >0 ⇒ nb > 0 and so if a ≤ 0, then we have nb > 0 ] . So now let a > 0.
In this case assume that there exists no positive integer n such that nb >a.
​
Then we have nb ≤ a ∀ n ∈ N.
Its means that a is an upper bound of the non – empty subset S of R given by
S = { b, 2b, 3b, …………….}= { nb: n ∈ N}.
​
Therefore by the completeness property of R, S must have the supremum, say s.
​
Then nb ≤ s ∀ n ∈ N and so
(n + 1)b ≤ s ∀ n ∈ N
nb + b ≤ s ∀ n ∈ N
i.e. nb ≤ s – b ∀ n ∈ N.
s – b is an upper bound of S.
since b > 0. Therefore s-b < s.
​
Its means that we have an upper bound for S which is less than the supremum s of S. But this contradicts the property of the supremum.
​
Hence our initial assumption is wrong and so there must exists some positive integer n such that
nb > a.
​
​