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Question: If $a$ is a nonnegative real number and $n$ is a positive integer, there exists a real number $b\geq 0$ such that $b^{n}=a$ .
The book gives the proof: Let $X= \{ x \in \mathfrak{R} | x \geq 0 \; and\ \; x^{n} \leq a \}$

Assuming that the set is bounded above(can be shown), assume that the LUB is b. Suppose that $b^{n} <a$, and let $\delta = a- b^{n}$. Choose positive integers $ m_{0},...m_{n-1} $ such that $ { n \choose k } b^{k} \frac{1}{m_{k}^{n-k}} < \frac{\delta}{n}, k=0,...n-1$ Let $ m= max \{ m_{0},...m_{n-1}\}. $ Then

$$ ( b+ \frac{1}{m}) ^{n} = \sum_{k=0}^{n} { n \choose k} b^{k} \frac{1}{m^{n-k}} = \sum_{k=0}^{n-1} { n \choose k } b^{k} \frac{1}{m^{n-k}} + b^{n} < \sum_{k=0}^{n-1} \frac{\delta}{n} + b^{n} = \delta+ b^{n} = a$$

Thus $ b+ \frac{1}{m} \in X$ but $b< b +\frac{1}{m}$, which is impossible; thus $a \leq b$ I am having trouble showing that $a < b^{n}$ is false. The books says that it is proves similarly though. Thanks

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Hint, let $X= \{ x \in \mathfrak{R} | x \geq 0 \; \mathrm{and}\ \; x^n \geq a \}$. –  peoplepower Dec 22 '12 at 3:02
    
If we consider this new set, won't we have to discard of the $b$ we are trying to prove this fact about? I am not sure how the proof above could be tweaked to involve that set. –  Jmaff Dec 22 '12 at 3:22
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Yes, you find $b'$. Then since every element of the old $X$ is less than or equal to every element of the new $X$, $b\leq b'$. Hence $a\leq b\leq b'\leq a$. –  peoplepower Dec 22 '12 at 3:45
    
Thanks. Then to use this...I take the greatest lower bound of the set you described b'(which is $\leq$ a) and then show the b must be less than this b'. I show this must be true by assuming that $b'< b$ which implies that for some $x$ in the set $X$ $b'<x$ and thus $ (b')^{n} \leq x^{n} \leq a$ which contradicts our saying that $b'$ is the G.L.B of the new X. I think this works? –  Jmaff Dec 22 '12 at 4:32
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The proof in the other direction uses essentially the same estimate. Suppose that $b^n-a=\delta$. Choose $m$ exactly as in the OP. Then $\left(b-\dfrac{1}{m}\right)^n \lt b^n$. We will show that $\left(b-\dfrac{1}{m}\right)^n \gt a$. This implies that $b-\dfrac{1}{m}$ is an upper bound for $X$, contradicting the fact that $b$ is the least upper bound of $X$.

For the estimate, use the fact that $$ \left( b- \frac{1}{m}\right) ^{n} = b^n +\sum_{k=1}^{n} { n \choose k } b^{n-k} \frac{(-1)^{k}}{m^{k}}.$$ Using the Triangle Inequality, we conclude that $$b^n-\left(b-\frac{1}{m}\right)^n \lt \sum_{k=1}^{n} \frac{\delta}{n}= \delta.$$

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A different proof can be done by finding a sequence of numbers $b_m$ such that $b_m^n \to a$ and using the completeness of the reals.

Possible ways to generate the $b_m$ are binary search (choose $b_0^n < a$ and $b_1^n > a$ and successively bisect, choosing the interval that contains $a$) and Newton's iteration (I think the iteration will converge from any starting point).

I don't know whether the equivalence of Dedekind (cut) and Cauchy (sequential) completeness has been proved yet for the OP.

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