# 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.

• $a + \pi^{-n},\, n \geqslant n_0$. Alternatively, $(a,b)$ is uncountable, but there are only countably many algebraic numbers. – Daniel Fischer Mar 9 '16 at 15:01
• @DanielFischer Could that argument be modified to show that there is an uncountable number of them? – Farewell Mar 9 '16 at 15:03
• The second argument shows that there are $2^{\aleph_0}$ transcendental numbers between any two algebraic numbers. – Daniel Fischer Mar 9 '16 at 15:04
• @DanielFischer That thing with $\pi^{-n}$ is easy enough way, you can post it as an answer. – Farewell Mar 9 '16 at 15:05

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.