[proofplan] We argue by contradiction. If the transcendental numbers $\mathbb{R} \setminus \mathbb{A}$ were [countable](/page/Countable%20Set), then together with the countability of the algebraic numbers $\mathbb{A}$, the [real line](/page/Real%20Numbers) would be a union of two countable sets. The [Countable Union of Countable Sets](/theorems/755) would then imply that $\mathbb{R}$ is countable, contradicting the [Uncountability of the Reals](/theorems/758). [/proofplan] [step:Assume the transcendental numbers are countable] Let $\mathbb{A} \subset \mathbb{R}$ denote the [set](/page/Set) of algebraic numbers. Suppose, for contradiction, that the set $\mathbb{R} \setminus \mathbb{A}$ is countable. [/step] [step:Combine the countability of algebraic and transcendental numbers] By the [Countability of Algebraic Numbers](/theorems/757), the set $\mathbb{A}$ is countable. By the contradiction assumption, the set $\mathbb{R} \setminus \mathbb{A}$ is countable. Since both sets are countable, the finite-indexed family $(E_i)_{i \in \{1,2\}}$ defined by \begin{align*} E_1 &= \mathbb{A}, & E_2 &= \mathbb{R} \setminus \mathbb{A} \end{align*} is a countable family of countable sets. Hence the Countable Union of Countable Sets applies and gives that $E_1 \cup E_2$ is countable. [guided] We need to convert the two pieces of $\mathbb{R}$ into a form where the Countable Union of Countable Sets applies. The theorem requires a countable indexed family of countable sets. Define the finite-indexed family $(E_i)_{i \in \{1,2\}}$ by \begin{align*} E_1 &= \mathbb{A}, & E_2 &= \mathbb{R} \setminus \mathbb{A}. \end{align*} The index set $\{1,2\}$ is finite, hence countable. The set $E_1 = \mathbb{A}$ is countable by the Countability of Algebraic Numbers. The set $E_2 = \mathbb{R} \setminus \mathbb{A}$ is countable by the contradiction assumption. Therefore $(E_i)_{i \in \{1,2\}}$ is a countable family of countable sets, so the Countable Union of Countable Sets gives that \begin{align*} E_1 \cup E_2 \end{align*} is countable. [/guided] [/step] [step:Identify the union with the real line and obtain the contradiction] For every real number $x \in \mathbb{R}$, either $x \in \mathbb{A}$ or $x \notin \mathbb{A}$. Hence \begin{align*} \mathbb{R} = \mathbb{A} \cup (\mathbb{R} \setminus \mathbb{A}) = E_1 \cup E_2. \end{align*} The preceding step therefore implies that $\mathbb{R}$ is countable. This contradicts the Uncountability of the Reals, which states that $\mathbb{R}$ is uncountable. [/step] [step:Conclude that transcendental numbers are uncountable] The contradiction came from the assumption that $\mathbb{R} \setminus \mathbb{A}$ is countable. Therefore $\mathbb{R} \setminus \mathbb{A}$ is uncountable. Since every uncountable set is nonempty, transcendental numbers exist. [/step]