Thread: Real number system

How can I show that if A > 0 and n is a positive, integer there is a positive number c such that c^n = A.

Here's my start for proving that c is a positive number

Let S be the set of all positive numbers x such that x^n < A. This set is not empty(how do I show this?). It is a bounded set and has a least upper bound(difficulty on proving ) which is c.

Then I have to prove that c^n = A.

If e > 0, there is some x in S s.t. c-e < x. hence (c-e)^n < A. Now let e -> 0 then c^n <= A. If c^n < A then I have prove this is not possible

I got this far... help !!

[QUOTE=mathsohard;608199]How can I show that if A > 0 and n is a positive, integer there is a positive number c such that c^n = A.

Here's my start for proving that c is a positive number

Let S be the set of all positive numbers x such that x^n < A. This set is not empty(how do I show this?).[\quote]
If A is a positive number then so is [tex]A^{1/n} so there exist postive numbers less than . , then .

It is a bounded set and has a least upper bound(difficulty on proving )

A basic property of the real numbers: if a non-empty set of real numbers has an upper bound, it has a least upper bound.

which is c.

Then I have to prove that c^n = A.

Suppose that 0< c^n< A. What can you say about c?
Suppose that 0< A< c^n. What can you say about c?

If e > 0, there is some x in S s.t. c-e < x. hence (c-e)^n < A. Now let e -> 0 then c^n <= A. If c^n < A then I have prove this is not possible

I got this far... help !!

Good! You might look at c- e/n, say, and compare to (c- e)^n. Remember that you only have to show "<" or ">", not "=" so you have more flexibility.

form a cut.
if x^n < A, x in L
if y^n > A, y in R

c is lub of L

That's the outline. The details are less interesting.

OK, details:

if x=A/2, x^n < A
if y=2A, y^n > A