5
$\begingroup$

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:

proof

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

Improve this question
$\endgroup$
2
$\begingroup$

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

Improve this answer
$\endgroup$
  • $\begingroup$ 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 $(-1)^k$? $\endgroup$ – mathjacks Oct 18 '14 at 15:40
  • $\begingroup$ I just replaced all positive signs with negatives, so sum became even less, that's why we have strict equality. $\endgroup$ – Jihad Oct 18 '14 at 15:53
  • $\begingroup$ 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. $\endgroup$ – mathjacks Oct 18 '14 at 16:03
  • $\begingroup$ @flapjackery exactly! $\endgroup$ – Jihad Oct 18 '14 at 16:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.