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} , 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