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Follow the example and use the least upper bound property of the real numbers to prove that any positive real number has a cube root.

Given an example of squared roots:

Let x be a real number such that x > 0. Then there is a positive real number y such that y2 = y?y = x
Let S = {s &#1108; R: s>0 and s2<x}
The S is not empty since x/2 &#1108; S, if x<2 and 1 &#1108; S otherwise. S is also bounded above since, x+1 is an upper bound for S. Let y be the least upper bound. Thus 0<min{x/2,1}<=y
If y2 < x, let &#949; =
(y+ &#949;)2 = y2 + 2y&#949; + &#949;2 = y2 + 2y +
< y2 +
< y2 +
< y2 + (x-y2)
= x
Thus (y + &#949;) &#1108; S &#61664; y cannot be an upper bound for S and y2 !< x
For the case y2 > x, let &#949; =

(y - &#949;)2 >=x &#61664; y- &#949; is also an upper bound for S
Thus y is not the least upper bound and y2 !> x, that y2 = x

Follow the example and use the least upper bound property of the real numbers to prove that any positive real number has a cube root.

Solution Summary

An example if followed and the least upper bound property of the real numbers is used to prove that any positive real number has a cube root. The solution is detailed and well presented. The response received a rating of "5/5" from the student who originally posted the question.

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