# Proof

According to the definition of negative and positive real numbers, we can represent a real number $$x$$ in three ways: $x=(x_n)_{n\in\mathbb N} + I:\begin{cases}\text{there is } N\text{ such that } x_n > 0\text{ for all } n > N&\Longleftrightarrow \text{ if }x\text{ is a positive real number}\\ \text{for all rational } \epsilon > 0 \text{ there is }N\text{ such that }-\epsilon < x_n < \epsilon\text{ for all }n > N&\Longleftrightarrow \text{ if }x\text{ equals 0}\\ \text{there is } N\text{ such that } x_n < 0\text{ for all } n > N&\Longleftrightarrow \text{ if }x\text{ is a negative real number} \end{cases}$

Due to the definition of multiplying real numbers, we have for two real numbers $$x=(x_n)_{n\in\mathbb N} + I$$ and $$y=(y_n)_{n\in\mathbb N} + I$$:

$\begin{array}{ccl} x\cdot y:=(x_n\cdot y_n)_{n\in\mathbb N} + I.\quad\quad ( * ) \end{array}$

Because the multiplication of real numbers is commutative, it is sufficient to prove the following four cases, applying the rules of multiplying negative and positive rational numbers:

### $$(1)$$ A positive real number times a positive real number gives a positive real number.

Assume $$x > 0$$ and $$y > 0$$. Then there exist indices $$N_x,N_y$$ such that $$x_n > 0$$ for all $$n > N_x$$ and $$y_n > 0$$ for all $$n > N_y$$. Set $$N=\max(N_x,N_y)$$, then $$x_n\cdot y_n > 0$$ for all $$n > N$$, and $$xy$$ proves to be a positive real number.

### $$(2)$$ A negative real number times a positive real number gives a negative real number.

Assume $$x < 0$$ and $$y > 0$$. Then there exist indices $$N_x,N_y$$ such that $$x_n < 0$$ for all $$n > N_x$$ and $$y_n > 0$$ for all $$n > N_y$$. Set $$N=\max(N_x,N_y)$$, then $$x_n\cdot y_n < 0$$ for all $$n > N$$, and $$xy$$ proves to be a negative real number.

### $$(3)$$ A negative real number times a negative real number gives a positive real number.

Assume $$x < 0$$ and $$y < 0$$. Therefore, there exist indices $$N_x,N_y$$ such that $$x_n < 0$$ for all $$n > N_x$$ and $$y_n < 0$$ for all $$n > N_y$$. Set $$N=\max(N_x,N_y)$$, then $$x_n\cdot y_n > 0$$ for all $$n > N$$, and $$xy$$ is a positive real number.

### $$(4)$$ Zero times any real number gives zero.

Assume $$x = 0$$ and $$y$$ be any real number. Note that, like in any rational Cauchy sequence, the sequence members in $$(y_n)_{n\in\mathbb N}$$ must be bounded. This means that there exists a rational constant $$c\in\mathbb Q$$ such that $$|y_n| < c$$ for all $$n\in\mathbb N$$. Moreover, for any rational $$\epsilon > 0$$, there exists an index $$N(\epsilon,c)$$ such that $\frac{-\epsilon}c < x_n < \frac{\epsilon}c$ for all $$n > N(\epsilon,c)$$. Then $-\epsilon < x_n\cdot y_n < \epsilon$ for all $$n > N(\epsilon,c)$$. This means that the product $$xy$$ equals zero.

Github: ### References

#### Bibliography

1. Piotrowski, Andreas: Own Research, 2014