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#3: Post edited by user avatar r~~‭ · 2022-09-05T05:34:32Z (about 2 years ago)
Add reference
  • The dimension of $\mathbb{R}$ as a vector space over $\mathbb{Q}$ is equal to the cardinality of $\mathbb{R}$.
  • In general, the dimension and cardinality of any vector space $\mathbf{V}$ and the cardinality of its scalar field $\mathbf{K}$ will obey the following equation:
  • $$
  • \|\mathbf{V}\| = \begin{cases}
  • \|\mathbf{K}\|^{\dim(\mathbf{V})} & \text{$\dim(\mathbf{V})$ is finite} \\\\
  • \max(\|\mathbf{K}\|, \dim(\mathbf{V})) & \text{$\dim(\mathbf{V})$ is infinite}
  • \end{cases}$$
  • In the case of $\mathbb{R}$ over $\mathbb{Q}$, if the dimension of $\mathbb{R}$ were somehow finite then this equation would imply that $|\mathbb{R}|$ is equal to some finite power of $|\mathbb{Q}|$, which is false (every finite power of a countable set is countable). Since $|\mathbb{R}|$ is greater than $|\mathbb{Q}|$, the second branch of the equation implies that the dimension of $\mathbb{R}$ is equal to $|\mathbb{R}|$.
  • The dimension of $\mathbb{R}$ as a vector space over $\mathbb{Q}$ is equal to the cardinality of $\mathbb{R}$.
  • In general, the dimension and cardinality of any vector space $\mathbf{V}$ and the cardinality of its scalar field $\mathbf{K}$ will obey the following equation:
  • $$
  • \|\mathbf{V}\| = \begin{cases}
  • \|\mathbf{K}\|^{\dim(\mathbf{V})} & \text{$\dim(\mathbf{V})$ is finite} \\\\
  • \max(\|\mathbf{K}\|, \dim(\mathbf{V})) & \text{$\dim(\mathbf{V})$ is infinite}
  • \end{cases}$$
  • In the case of $\mathbb{R}$ over $\mathbb{Q}$, if the dimension of $\mathbb{R}$ were somehow finite then this equation would imply that $|\mathbb{R}|$ is equal to some finite power of $|\mathbb{Q}|$, which is false (every finite power of a countable set is countable). Since $|\mathbb{R}|$ is greater than $|\mathbb{Q}|$, the second branch of the equation implies that the dimension of $\mathbb{R}$ is equal to $|\mathbb{R}|$.
  • (For a quickie proof of the infinite-dimensional case, see <https://math.stackexchange.com/a/194287>.)
#2: Post edited by user avatar Derek Elkins‭ · 2022-09-05T05:04:16Z (about 2 years ago)
cases environment accomplishes the same output in a simpler way
  • The dimension of $\mathbb{R}$ as a vector space over $\mathbb{Q}$ is equal to the cardinality of $\mathbb{R}$.
  • In general, the dimension and cardinality of any vector space $\mathbf{V}$ and the cardinality of its scalar field $\mathbf{K}$ will obey the following equation:
  • $$
  • \|\mathbf{V}\| = \left\\{
  • \begin{array}{ll}
  • \|\mathbf{K}\|^{\dim(\mathbf{V})} & \text{$\dim(\mathbf{V})$ is finite} \\\\
  • \max(\|\mathbf{K}\|, \dim(\mathbf{V})) & \text{$\dim(\mathbf{V})$ is infinite}
  • \end{array}
  • \right.
  • $$
  • In the case of $\mathbb{R}$ over $\mathbb{Q}$, if the dimension of $\mathbb{R}$ were somehow finite then this equation would imply that $|\mathbb{R}|$ is equal to some finite power of $|\mathbb{Q}|$, which is false (every finite power of a countable set is countable). Since $|\mathbb{R}|$ is greater than $|\mathbb{Q}|$, the second branch of the equation implies that the dimension of $\mathbb{R}$ is equal to $|\mathbb{R}|$.
  • The dimension of $\mathbb{R}$ as a vector space over $\mathbb{Q}$ is equal to the cardinality of $\mathbb{R}$.
  • In general, the dimension and cardinality of any vector space $\mathbf{V}$ and the cardinality of its scalar field $\mathbf{K}$ will obey the following equation:
  • $$
  • \|\mathbf{V}\| = \begin{cases}
  • \|\mathbf{K}\|^{\dim(\mathbf{V})} & \text{$\dim(\mathbf{V})$ is finite} \\\\
  • \max(\|\mathbf{K}\|, \dim(\mathbf{V})) & \text{$\dim(\mathbf{V})$ is infinite}
  • \end{cases}$$
  • In the case of $\mathbb{R}$ over $\mathbb{Q}$, if the dimension of $\mathbb{R}$ were somehow finite then this equation would imply that $|\mathbb{R}|$ is equal to some finite power of $|\mathbb{Q}|$, which is false (every finite power of a countable set is countable). Since $|\mathbb{R}|$ is greater than $|\mathbb{Q}|$, the second branch of the equation implies that the dimension of $\mathbb{R}$ is equal to $|\mathbb{R}|$.
#1: Initial revision by user avatar r~~‭ · 2022-09-04T03:41:39Z (about 2 years ago)
The dimension of $\mathbb{R}$ as a vector space over $\mathbb{Q}$ is equal to the cardinality of $\mathbb{R}$.

In general, the dimension and cardinality of any vector space $\mathbf{V}$ and the cardinality of its scalar field $\mathbf{K}$ will obey the following equation:

$$
\|\mathbf{V}\| = \left\\{
\begin{array}{ll}
\|\mathbf{K}\|^{\dim(\mathbf{V})} & \text{$\dim(\mathbf{V})$ is finite} \\\\
\max(\|\mathbf{K}\|, \dim(\mathbf{V})) & \text{$\dim(\mathbf{V})$ is infinite}
\end{array}
\right.
$$

In the case of $\mathbb{R}$ over $\mathbb{Q}$, if the dimension of $\mathbb{R}$ were somehow finite then this equation would imply that $|\mathbb{R}|$ is equal to some finite power of $|\mathbb{Q}|$, which is false (every finite power of a countable set is countable). Since $|\mathbb{R}|$ is greater than $|\mathbb{Q}|$, the second branch of the equation implies that the dimension of $\mathbb{R}$ is equal to $|\mathbb{R}|$.