# Real Multiplication is Well-Defined

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## Theorem

The operation of multiplication on the set of real numbers $\R$ is well-defined.

## Proof

From the definition, the real numbers are the set of all equivalence classes $\left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right]$ of Cauchy sequences of rational numbers.

Let $x = \left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right], y = \left[\!\left[{\left \langle {y_n} \right \rangle}\right]\!\right]$, where $\left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right]$ and $\left[\!\left[{\left \langle {y_n} \right \rangle}\right]\!\right]$ are such equivalence classes.

From the definition of real multiplication, $x \times y$ is defined as $\left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right] \times \left[\!\left[{\left \langle {y_n} \right \rangle}\right]\!\right] = \left[\!\left[{\left \langle {x_n \times y_n} \right \rangle}\right]\!\right]$.

We need to show that:

$\left \langle {x_n} \right \rangle, \left \langle {x'_n} \right \rangle \in \left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right], \left \langle {y_n} \right \rangle, \left \langle {y'_n} \right \rangle \in \left[\!\left[{\left \langle {y_n} \right \rangle}\right]\!\right] \implies \left \langle {x_n \times y_n} \right \rangle = \left \langle {x'_n \times y'_n} \right \rangle$.

That is:

$\forall \epsilon > 0: \exists N: \forall i, j > N: \left\vert{\left({x_i \times y_i}\right) - \left({x'_j \times y'_j}\right)}\right\vert < \epsilon$.

As $\left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right]$ and $\left[\!\left[{\left \langle {y_n} \right \rangle}\right]\!\right]$ are Cauchy, they are bounded.

Let $B_x = 2 \sup \left({\left \langle {x_n} \right \rangle}\right)$ and $B_y = 2 \sup \left({\left \langle {y_n} \right \rangle}\right)$.

Let $B = \max \left\{{B_x, B_y}\right\}$.

Now let $\epsilon > 0$. Then:

• $\exists N_1: \forall i, j > N_1: \left\vert{B}\right\vert \left\vert{x_i - x'_j}\right\vert < \epsilon / 2$;
• $\exists N_2: \forall i, j > N_2: \left\vert{B}\right\vert \left\vert{y_i - y'_j}\right\vert < \epsilon / 2$.

Now let $N = \max \left\{{N_1, N_2}\right\}$.

Then we have $\forall i, j \ge N: \left\vert{B}\right\vert \left\vert{x_i - x'_j}\right\vert + \left\vert{B}\right\vert \left\vert{y_i - y'_j}\right\vert < \epsilon$.

So:

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \epsilon$$ $$>$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \left\vert{B}\right\vert \left\vert{x_i - x'_j}\right\vert + \left\vert{B}\right\vert \left\vert{y_i - y'_j}\right\vert$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\ge$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \left\vert{B \left({x_i - x'_j}\right) + B \left({y_i - y'_j}\right)}\right\vert$$ $$\displaystyle$$ $$\displaystyle$$ Triangle Inequality $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\ge$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \left\vert{\left({x_i - x'_j}\right) \left({y_i + y'_j}\right) + \left({y_i - y'_j}\right) \left({x_i + x'_j}\right)}\right\vert$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \left\vert{x_i y_i - x'_j y_i + x_i y'_j - x'_j y'_j + x_i y_i + x'_j y_i - x_i y'_j - x'_j y'_j}\right\vert$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \left\vert{2 x_i y_i - 2 x'_j y'_j}\right\vert$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \left\vert{x_i y_i - x'_j y'_j}\right\vert$$ $$\displaystyle$$ $$\displaystyle$$

Hence the result.

$\blacksquare$