Theorem:- Let a be any real number and b any +ve real number. Then there exists a
positive integer n such that
nb>a
Proof:- Given that aϵ R & bϵ R+
So there are two possible cases.
Case 1:-
When a≤0
In
this case the relation nb>a is always true because the value of nb is always
+ve.
Case 2:- When a>0
Let
us that there exists no +ve integar such that nb>a
Then
we have nb≤a ⩝ n∈N
It means
that a be the upper bound of set S which is given by
S= {b,2b,3b,………..} = {nb:n∈N}.
But by
completeness property of real numbers, S must have the supremum K(say).
Then nb≤K ⩝ n∈N
& also (n+1)b≤K ⩝ n∈N
Because if n∈N
⇨ n+1∈N
⇨ nb+b≤K
⩝ n∈N
⇨ nb≤K-b ⩝ n∈N
It means that (K-b) be the upper bound of S. So this shows that there is a upper bound
of set S which is less than the supremum of S.
But this is contradiction because
supremum of a set is the least upper bound.
Hence our assumption is
wrong & so ther must be exist an +ve integar n such that
nb>a
By- Vinay Mishra
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