# How do we prove the existence of uncountably many transcendental numbers?

I know how to prove the countability of sets using equivalence relations to other sets, but I'm not sure how to go about proving the uncountability of the transcendental numbers (i.e., numbers that are not algebraic).

If a number $t$ is algebraic, it is the root of some polynomial with integer coefficients. There are only countably many such polynomials (each having a finite number of roots), so there are only countably many such $t$. Since there are uncountably many real (or complex) numbers, and only countably many of them are algebraic, uncountably many of them must be transcendental.

• I understand this answer, but is it acceptable to just assume that reals are uncountable? – user10507 May 5 '11 at 6:36
• Acceptable to whom? If you want a proof that the reals are uncountable, just ask. – Gerry Myerson May 5 '11 at 6:38
• @user 10507. I don't know what you have or have not seen in class; I was assuming that you already knew that R was uncountable (this is probably the first result a student sees in class after learning the definition of an uncountable set). If you havn't seen this proof, it fairly simple. See Cantor's Diagonal Argument for instance. I don't know if it is possible to prove that the trancedental numbers are uncountable without first knowing that R is uncountable. – user3180 May 5 '11 at 6:52
• the numbers $\sum a_n10^{n!}$ with $a_n$ zero or one, are transcendental (provided infinitely many of the $a_n$ are one) and are uncountable in number. Both facts can be proved without knowing $\bf R$ is uncountable. – Gerry Myerson May 5 '11 at 12:46
• @RossMillikan, OP wanted a proof that there are uncountably many transcendentals. Any proof of that is more-or-less the same as proving $\bf R$ uncountable, no? – Gerry Myerson Dec 4 '11 at 5:33

If you accept the premise that $|\mathbb R| = |P(\mathbb N)|$ then you know it is uncountable, due to Cantor's theorem.

Take all polynomials in rational numbers, each have only finitely many roots in $\mathbb R$.

Since the set of polynomials is equivalent to $\bigcup_{n\in\mathbb N} \mathbb Q^n$ (the union of polynomials from all degrees), which is countable you have only countably many possibly roots for all rational polynomials and thus only countably many algebraic numbers.

Now consider $\mathbb R\setminus A$ where $A$ is the set of all algebraic numbers, if it was not uncountable then $\mathbb R = \mathbb R\setminus A\cup A$ which was a union of two countable sets, which then would be countable.

• @augurar: Please don't make these sort of very minor edits to very old posts. – Asaf Karagila Feb 28 '14 at 5:31
• @Shuchang: Please don't approve such minor edits. Especially when the post is nearly three years old. – Asaf Karagila Feb 28 '14 at 5:32
• Why not? The question doesn't "expire". I was looking for an answer, read this one, and didn't know why the first statement was true. So I looked it up and added that detail. The answer is now more useful. – augurar Mar 3 '14 at 2:59
• I guess we'll have to agree to disagree. I can't see any reason not to make improvements of this sort if I happen to come across them, regardless of the age of the question. – augurar Mar 3 '14 at 3:06
• Okay, I have posted a question on the meta. – augurar Mar 3 '14 at 3:21

There is a more number-theoretic way of looking at the matter, one that harks back to the work of Liouville.

He showed, among other things, that a number whose decimal expansion consists of $0$s and $1$s, such that the number of $0$s between consecutive $1$s grows faster than any polynomial, must be transcendental.

The standard example is the Liouville Constant, with the $n$-th gap being of length $n!$.

It is easy to produce continuum many transcendental numbers of the Liouville type.

The reason the result is worth mentioning is that it arises from examination of the approximability of algebraic numbers by rationals, a subject with a long and glorious history.

The required approximability result has a pretty quick proof that uses only high-school level mathematics.

• I am sure that one can think of a very very complicated forcing argument requiring very large cardinals in order to prove that :-) – Asaf Karagila May 5 '11 at 9:01