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?


$1 > \sqrt(2) > 0$ 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.

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  • $\begingroup$ 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

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