# Prove if $a$ is a nonnegative real number and $n$ is a positive integer, there exists a $b \geq 0$ such that $b^n = a$

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?

$b$ 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. $$\Big(b-\frac{1}{m}\Big)^n = \sum_{k = 0}^n(-1)^k {n \choose k}b^k\frac{1}{m^{n-k}} = b^n + \sum_{k = 0}^{n-1}(-1)^k {n \choose k}b^k\frac{1}{m^{n-k}} > b^n - \sum_{k=0}^{n-1}\frac{\delta}{n} = a$$
• How does the negative sign appear in front of the summation with $\delta$ in the second to last expression? Can't the original summation be positive or negative depending on $k$ and $(-1)^k$? – mathjacks Oct 18 '14 at 15:40
• So, if I wanted to add in another step for clarity, I could have added $b^n + \sum_{k = 0}^{n-1}(-1)^k {n \choose k}b^k\frac{1}{m^{n-k}} > b^n - \sum_{k = 0}^{n-1} {n \choose k}b^k\frac{1}{m^{n-k}}$, right? Thank you for the explanation, this is very clear. – mathjacks Oct 18 '14 at 16:03