Is there an easy enough way to show that between two algebraic numbers there is an infinite number of transcendental numbers? We know that between two different rational numbers there is an infinite number of irrational numbers and that between two different irrational numbers there is an infinite number of rational numbers.
With the help of the fact that all rational numbers are algebraic and of the fact that rational numbers are dense enough on the real line we could show that between two different transcendental numbers there is an infinite number of algebraic numbers.

But how to show that between two different algebraic numbers there is an infinite number of transcendental numbers?

I would like to see the proof which is as simple as possible.
Thank you.
 A: We can explicitly give infinitely many transcendental numbers between any two algebraic numbers. Let the algebraic numbers be $a,b$ and suppose $a < b$. Choose $n_0 \in \mathbb{N}$ so large that $\pi^{-n_0} < b-a$, then $\{a + \pi^{-n} : n \geqslant n_0\}$ is an infinite family of transcendental numbers between $a$ and $b$.
Since there are only countably many algebraic numbers, and the interval $(a,b)$ has cardinality $2^{\aleph_0}$, there are in fact $2^{\aleph_0}$ transcendental numbers between any two algebraic numbers.
A: I don't think this needs a very deep explanation.   If you agree there are infinitely many transcendental numbers in any finite interval (say between 0 and 1), then you can easily (linearly) scale this between any two other numbers, be they algebraic or not, resulting in an infinite number of remapped transcendentals in the new interval.   Without getting too rigorous, it should be pretty obvious that the remapped numbers are still transcendentals because they are of the form (algebraic + (algebraic) scalar * algebraic * transcendental = transcendental).   You can't get the transcendental out of the transcendental by multiplying by an algebraic.
