Let a b R Suppose that a 0 Prove that there is some n N suc

Let a, b R. Suppose that a > 0. Prove that there is some n N such that b [-na,na]

Solution

Proof: May assume b >0 (if b <0, argue with -b on similar lines).

So we are given a>0, b>0 and have to show the existence of a natural number n such that na>b.

Suppose there does not exist such n.

Then the set {m: m<b/a} is the set N and is bounded from above (by b/a).

But N is not a bounded subset of R (For proof ,see *)

This contradiction proves the result.

* N is not a bounded subset of R

Proof: If it were bounded , there exists a least upper bound , say M.

By definition of M, there exists an integer k such that k>M-1. so k+1>M, which is absurd.

Hence * is true.