Real Multiplication Identity is One

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Theorem

The identity element of real number multiplication is the real number $1$:

$\exists 1 \in \R: \forall a \in \R_{\ne 0}: a \times 1 = a = 1 \times a$


Corollary

$\forall x \in \R_{\ne 0}: x \times y = x \implies y = 1$


Proof

From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers.


Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}$, where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such equivalence classes.

From the definition of real multiplication, $x \times y$ is defined as:

$\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n \times y_n} } {}$


Let $\sequence {1_n}$ be such that $\forall i: 1_n = 1$.

Then we have:

\(\displaystyle \eqclass {\sequence {1_n} } {} \times \eqclass {\sequence {x_n} } {}\) \(=\) \(\displaystyle \eqclass {\sequence {1_n \times x_n} } {}\)
\(\displaystyle \) \(=\) \(\displaystyle \eqclass {\sequence {1 \times x_n} } {}\)
\(\displaystyle \) \(=\) \(\displaystyle \eqclass {\sequence {x_n} } {}\)


Similarly for $\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {1_n} } {}$.


So the identity element of $\struct {\R_{\ne 0}, \times}$ is the real number $1$.

$\blacksquare$


Sources