If $a$ is a nonnegative real number and $n$ is a positive integer, there exists a real number $b\geq 0$ such that $b^{n}=a$ . 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
 A: 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.
A: 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
$$b^n-\left(b-\frac{1}{m}\right)^n \lt \sum_{k=1}^{n} \frac{\delta}{n}= \delta.$$
