0
$\begingroup$

Assume $a$ is a dedekind cut. By definition it is nonempty so there exist $b\in a$. Similarly, we can find $p\in a$ such that $p>b$. It has no maximals element so, we can repeat this process to infinity.

Claim: There exist atleast one irrational between two rationals.

Consider $p$ and $b$ that we talked about earlier. W.l.o.g say both are positive and observe that $1>\frac{1}{\sqrt2}>0$ so we can find $n\in \mathbb{N}$ such that $p-b>\frac{1}{\sqrt2 n}>0$, by archimedean property. Now by adding b to the inequality, we get $p>\frac{1}{\sqrt2 n}+b>b$. (It is trivial that $\frac{1}{\sqrt2 n}+b$ is irrational).

As we can find infinitely many rationals that are greater than other rational in our dedekind cut $a$(because dedekind cut has no maximal element) we can find infinitely many irrationals under a cut, in our case $p$,$b$$<a$. As there exist infinitely many irrationals under a cut and they also have rational between them rationals are dense in reals. Similarly as said earlier, we have infinitely many cuts that are greater than other under a cut, irrationals are dense in reals.

So my question is that, is this proof valid? if not, then why?

$\endgroup$
0
$\begingroup$

$1 > \sqrt(2) > 0$ is not true, is it a mistype?

Anyway, it could be a lot simpler. Consider this number:

$a + \frac{(b - a)}{\sqrt(2)}$

This gives you an algebraic irrational between a and b. Use $\pi$ in place of $\sqrt(2)$ if you want a transcendental.

$\endgroup$
  • $\begingroup$ oops that should be $1>\frac{1}{\sqrt2}>0$, it's a typo my bad. I'm editing my post. In case of $1>\frac{1}{\sqrt2}>0$, is my proof correct? $\endgroup$ – dankmemer Mar 24 '17 at 22:11

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.