Real Addition is Well-Defined
Jump to navigation
Jump to search
Theorem
The operation of addition on the set of real numbers $\R$ is well-defined.
Proof
From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers:
- $\eqclass {\sequence {x_n} } {} \equiv \eqclass {\sequence {y_n} } {} \iff \forall \epsilon > 0: \exists n \in \N: \forall i, j > n: \size {x_i - y_j} < \epsilon$
Let $x = \eqclass {\sequence {x_n} } {}$ and $y = \eqclass {\sequence {y_n} } {}$, where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such equivalence classes.
From the definition of real addition, $x + y$ is defined as:
- $\eqclass {\sequence {x_n} } {} + \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n + y_n} } {}$.
We need to show that:
- $\sequence {x_n}, \sequence {x'_n} \in \eqclass {\sequence {x_n} } {}, \sequence {y_n}, \sequence {y'_n} \in \eqclass {\sequence {y_n} } {} \implies \sequence {x_n + y_n} = \sequence {x'_n + y'_n}$
That is:
- $\forall \epsilon > 0: \exists N: \forall i, j > N: \size {\paren {x_i + y_i} - \paren {x'_j + y'_j} } < \epsilon$
Let $\epsilon > 0$, such that:
- $\exists N_1: \forall i, j \ge N_1: \size {x_i - x'_j} < \epsilon / 2$
- $\exists N_2: \forall i, j \ge N_2: \size {y_i - y'_j} < \epsilon / 2$
Now let $N = \max \set {N_1, N_2}$.
Then we have:
- $\forall i, j \ge N: \size {x_i - x'_j} + \size {y_i - y'_j} < \epsilon$
But:
\(\ds \epsilon\) | \(>\) | \(\ds \size {x_i - x'_j} + \size {y_i - y'_j}\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds \size {\paren {x_i - x'_j} + \paren {y_i - y'_j} }\) | Triangle Inequality | |||||||||||
\(\ds \) | \(=\) | \(\ds \size {\paren {x_i + y_i} - \paren {x'_j + y'_j} }\) |
So:
- $\forall i, j \ge N: \size {\paren {x_i + y_i} - \paren {x'_j + y'_j} } < \epsilon$
Hence the result.
$\blacksquare$