(related to Proposition: Multiplication of Real Numbers Is Cancellative)
Because the multiplication of real numbers is commutative, it is without loss of generality sufficient to show the right cancellation property, i.e. \[x\cdot z=y\cdot z\Leftrightarrow x=y,~~~~~~(x,y,z\in\mathbb R,~z\neq 0).\]
By definition of real numbers, the numbers \(x,y,z\) are some equivalence classes \[\begin{array}{rcl}x&:=&(x_n)_{n\in\mathbb N}+I,\\y&:=&(y_n)_{n\in\mathbb N}+I,\\z&:=&(z_n)_{n\in\mathbb N}+I\\\end{array}\]
for some rational Cauchy sequences \((x_n)_{n\in\mathbb N}\), \((y_n)_{n\in\mathbb N}\), and \((z_n)_{n\in\mathbb N}\) denote elements of the set \(M\) of all rational Cauchy sequences, which represent the real numbers \(x\) and \(y\), while \(I\) denotes the set of all rational sequences, which converge to \(0\). Moreover, since \(z\neq 0\) by hypothesis, the rational Cauchy sequence \((z_n)_{n\in\mathbb N}\) is not convergent to 0, i.e. \((z_n)_{n\in\mathbb N}\not\in I\).
By definition of multiplying real numbers, we have \[\begin{array}{rcl} x\cdot z&=&(x_n\cdot z_n)_{n\in\mathbb N}+I,\\ y\cdot z&=&(y_n\cdot z_n)_{n\in\mathbb N}+I. \end{array}\]
By hypothesis, \((z_n)_{n\in\mathbb N}\not\in I\). Therefore, there exist \(N\in\mathbb N\) such that \(|z_n|>0\) for all \(n > N\). For such indices it follows from the cancellation property of multiplying rational Cauchy sequences that
\[\begin{array}{rcll} x\cdot z=y\cdot z&\Leftrightarrow&(x_n\cdot z_n)_{n > N}+I=(y_n\cdot z_n)_{n > N}+I&\text{by definition of multiplying real numbers}\\ &\Leftrightarrow& (x_n)_{n > N}+I=(y_n)_{n > N}+I&\text{because multiplication of rational Cauchy sequences is cancellative}\\ &\Leftrightarrow& x=y&\text{by definition of real numbers} \end{array} \]
Altogether, we have shown that the addition of real numbers is cancellative \[ x\cdot z=y\cdot z\Rightarrow x=y,~~~~~~(x,y,z\in\mathbb R,~z\neq 0),\] and its conversion \[x=y\Rightarrow x\cdot z=y\cdot z,~~~~~~(x,y,z\in\mathbb R,~z\neq 0).\]
