Quantcast
Channel: Active questions tagged real-analysis - Mathematics Stack Exchange
Viewing all articles
Browse latest Browse all 9155

How to showcase an inequality with a Dedekind cut of $\sqrt2$

$
0
0

I've been reading through Foundations of Analysis by Taylor, and I'm stumped on one part of the Dedekind cut section for formulating the reals. Right now, I'm on the section where it talks about a Dedekind cut with cut number $\sqrt2$.

There's a point where it states that if $\frac{n}{m}$ is any positive element of $L$, then we can always choose a larger rational number which still has a square less than 2 as follows: $\frac{kn+1}{km} > \frac{n}{m}$ for every $k \in \mathbb N$ and

$\left(\frac{kn+1}{km}\right)^2 = (\frac{n}{m})^2 + \frac{1}{km}(\frac{2n}{m} + \frac{1}{km})$.

The text states that by choosing $k$ large enough, we can make the second term on the right less than $2 - \left(\frac{n}{m}\right)^2$, which would imply $\left(\frac{kn+1}{km}\right)^2 < 2$

However, I'm a bit confused on this part. How exactly does $\frac{1}{km}(\frac{2n}{m} + \frac{1}{km})$ become less than $2 - \left(\frac{n}{m}\right)^2$?

Any help would be appreciated.


Viewing all articles
Browse latest Browse all 9155

Trending Articles



<script src="https://jsc.adskeeper.com/r/s/rssing.com.1596347.js" async> </script>