(related to Proposition: Multiplying Negative and Positive Real Numbers)
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:
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.
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.
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.
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.
