TITLE: Prove if $a$ is a nonnegative real number and $n$ is a positive integer, there exists a $b \geq 0$ such that $b^n = a$
QUESTION [5 upvotes]: This is Theorem 7.5 in Foundations of Mathematical Analysis by Johnsonbaugh and Pfaffenberger.
At the end of the proof in the book, we want to show by contradiction that $b^n < a$ and $b^n > a$ are not true. The proof in the book excludes the part for $b^n > a$. The text suggest it's to be proved in a similar manner to $b^n<a$, but I can't seem to figure it out how to set up the variables to prove $b^n > a$ is not true.
Here is the proof in the book:

My problem lies in "Similarly, one shows that $a < b^n$ is false...". How is this done?

REPLY [2 votes]: $b$ is a least upper bound. Just make the same inequalities for $(b-\frac{1}{m})^n$. Then you will get that $(b-\frac{1}{m})^n > a$ then $b$ is not a least upper bound.
$$
\Big(b-\frac{1}{m}\Big)^n = \sum_{k = 0}^n(-1)^k {n \choose k}b^k\frac{1}{m^{n-k}}
=
b^n + \sum_{k = 0}^{n-1}(-1)^k {n \choose k}b^k\frac{1}{m^{n-k}}
>
b^n - \sum_{k=0}^{n-1}\frac{\delta}{n} = a
$$