# e as a Dedekind cut

• Feb 13th 2011, 07:17 PM
magus
e as a Dedekind cut
I'm just learning about Dedekind cuts and I've been shown how the $\displaystyle \sqrt{2}$ is can be a cut and can infer how all algebraic numbers have cuts.

A question popped into my mind that I can't seem to get is how do you get transcendential numbers like $\displaystyle \pi$ and $\displaystyle e$ as cuts. Could someone give me a hint as how to prove this?
• Feb 13th 2011, 07:51 PM
theodds
I would think you could take

$\displaystyle e = \left\{x \in \mathbb Q: x < \sum_{i = 1} ^ n \frac{1}{i!}, \ \mbox{for some } n\right\}$

just as an example.
• Feb 14th 2011, 03:41 AM
HallsofIvy
Have you proved that the union of two cuts is a cut? Is so, since any real number can be written as a decimal, Let $\displaystyle U_1= \{x| x\in \mathbb Q, x< 3.1\}$, $\displaystyle U_2= \{x| x\in \mathbb Q, x< 3.14\}$, $\displaystyle U_3= \{x| x\in \mathbb Q, x< 3.141\}$, etc. where $\displaystyle U_n$ is the set of all rational numbers less than $\displaystyle \pi$ written to n decimal places. The Dedekind cut corresponding to $\displaystyle \pi$ is the union of all those sets.

Since every terminating decimal is a rational number, given any real number, a, and any positive integer, n, "a to n decimal places" is a rational number and so has a "rational cut" corresponding to it. The cut representing that real number is the union of those rational cuts for all n.
• Feb 14th 2011, 08:38 PM
magus
I have not proved yet that the union of two cuts is a cut but I imagine it goes something like this.

Using the order of cuts it is reasonable to assume that either one cut will encompass rationals that the other does not or the two cuts are exactly equal in which case the union is equal to each set. Since each dedekind cut will share some elements and their intersection will leave values from the cut of greater order.

Since the union of two will encompass both the lower portion of the cut and include the members left in the difference the union must simply be the cut of higher order which is a cut as required.

Now as long as I proved that right I can take the converging decimal expansions and say that they converge (is it ok to say that here?) to a cut which corresponds to the needed value.

One question remains though. Since I can take infinite decimal expansions and let them converge on some set can I say that by this all transcendentals are cuts since they are all representable like this. Do I have to prove that all transcendentals are representable by decimal expansions or does this prove that?