## Algebraic Numbers Are Countable

**Point of post:** In this post I will give a proof that the algebraic numbers are countable, and thus we will prove that there are non-algebraic (transcendental) numbers.

*Remark:* Personally I find my proof much simpler and more intuitive than the usual method by considering the “height” of numbers (viz. Rudin, pg. 43)

Call a number *algebraic *if their exists such that . In other words, the algebraic numbers are those which are the roots of polynomials with integer coefficients. So, formally

**Theorem:** is countably infinite.

**Proof:** For each define

Then, for each define

Then, by the Fundamental Theorem of Algebra we have that . We claim that

To see this let , then there exists a non-zero polynomial such that , but this tells us that and so evidently . Conversely, if then there exists some (we disregard the case when since these are just constants and are zero, only when the constant itself is zero) and some such that , and so by definition and since are not all zero we see that . Thus, .

Note though that since for all that for each fixed we have that

is the countable union of finite (non-empty) sets and is thus itself countable. Thus, being the countable union of such countable sets is itself countable.

From this we have the following very nice corollary:

**Corollary:** There exists uncountably real numbers which are non-algebraic (transcendental).

**Proof:** Suppose to the contrary that was countable, then since the finite union of countable sets is countable and

we see that is countable. But, this is a contradiction and so the conclusion follows.

**References:**

1. Rudin, Walter. *Principles of Mathematical Analysis*. New York: McGraw-Hill, 1976. Print.

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