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:
\(\ds \eqclass {\sequence {1_n} } {} \times \eqclass {\sequence {x_n} } {}\) | \(=\) | \(\ds \eqclass {\sequence {1_n \times x_n} } {}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \eqclass {\sequence {1 \times x_n} } {}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \eqclass {\sequence {x_n} } {}\) |
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
- 1964: Iain T. Adamson: Introduction to Field Theory ... (previous) ... (next): Chapter $\text {I}$: Elementary Definitions: $\S 1$. Rings and Fields
- 1973: G. Stephenson: Mathematical Methods for Science Students (2nd ed.) ... (previous) ... (next): Chapter $1$: Real Numbers and Functions of a Real Variable: $1.2$ Operations with Real Numbers: $(7)$