Addition and Multiplication on the Rational Numbers

Patrick Chen

We may not realize it, but addition and multiplication on the integers can be seen as functions. For example, we can represent addition on the integers as the function $a: \mathbb{Z}^{2} \rightarrow \mathbb{Z}$ defined by $a(x,y) = x + y$ for all $(x,y) \in \mathbb{Z}^{2}$ . Similarly, we can represent multiplication on the integers as $m: \mathbb{Z}^{2} \rightarrow \mathbb{Z}$ defined by $m(x,y) = xy$ for all $(x,y) \in \mathbb{Z}^{2}$ . Because of this, addition and multiplication just like functions must be well-defined, meaning that the sum of two elements elements and the multiplication of two elements must be unique. In the case of integers by the way they are defined, this is pretty trivial.

However, it becomes less trivial as we move to the rational numbers $\mathbb{Q}$ . This is because every rational number can be represented infinitely many ways. For example, $\frac{1}{2}$ can be represented as $\frac{2}{4}, \frac{3}{6}, \frac{-1}{-2}$ , …. and $\frac{1}{3}$ can be represented as $\frac{2}{6}, \frac{3}{9}, \frac{-1}{-3}$ , …. since they all represent the same value. Because of this, we must ensure that the result from addition and multiplication are well defined, meaning they don’t depend on the way we represent the different values.

In this post, we will define addition and multiplication on the rational numbers and prove that they are well defined.

Recall from Class

As we have seen in lecture, we can build the rational numbers through the equivalence relation $\simeq$ on $\mathbb{Z} \times \mathbb{Z}\backslash\{0\}$ defined by $(a,b) \simeq (c,d)$ if $ad = bc$ . For example,

$(0,1) \simeq (0, 2) \simeq (0, k)$ for any $k \in \mathbb{Z}$
$(1,1) \simeq (-1, -1) \simeq (k, k)$ for any $k \in \mathbb{Z}$
$(-2,4) \simeq (4, -8) \simeq (k, -2k)$ for any $k \in \mathbb{Z}$

Thus, the quotient set $\mathbb{Z} \times \mathbb{Z}\backslash\{0\}/ \simeq$ defines the rational numbers $\mathbb{Q}$ as the equivalence classes of represent the rational numbers: $\mathbb{Q} = \mathbb{Z} \times \mathbb{Z}\backslash\{0\}/ \simeq$ .

Addition

We define addition on $\mathbb{Q}$ in terms of those on $\mathbb{Z}$ . Given, $a,b,c,d \in \mathbb{Z}$ where $b, d, \neq 0$ , then

$\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$ .

After defining addition on $\mathbb{Q}$ , we must show that addition is well-defined. The approach to show that addition is well-defined is very similar to the proof of uniqueness since we must show that there is a unique result between the sum of two rational numbers.

Proposition: Let $a,b,c,d \in \mathbb{Z}$ where $b, d, \neq 0$ , so $\frac{a}{b}, \frac{c}{d} \in \mathbb{Q}$ . The sum of $\frac{a}{b} + \frac{c}{d}$ is well-defined.

Proof. Let $a, b, c, d, e, f, g, h \in \mathbb{Z}$ where $b, d, f, h \neq 0$ be arbitrary. Suppose that $\frac{a}{b} = \frac{e}{f}$ and $\frac{c}{d} = \frac{g}{h}$ . It follows that $af = be$ and $ch =gd$ . By definition of addition, $\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$ and $\frac{e}{f} + \frac{g}{h} = \frac{eh + fg}{fh}$ .

To show that addition is well-defined, we must prove that $\frac{ad + bc}{bd} = \frac{eh + fg}{fh}$ . To do this we can prove $(fh)(ad + bc) = (bd)(eh + fg)$ instead. We know that

$(fh)(ad + bc) = fhad + fhbc = afhd + chfb = behd + gdfb = (bd)(eh + fg)$ .

Therefore, addition on the rational numbers is well-defined.

Multiplication

We define multiplication on $\mathbb{Q}$ in terms of those on $\mathbb{Z}$ . Given, $a,b,c,d \in \mathbb{Z}$ where $b, d, \neq 0$ , then

$\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}$ .

After defining multiplication on $\mathbb{Q}$ , we must show that multiplication is well-defined.

Proposition. Let $a,b,c,d \in \mathbb{Z}$ where $b, d, \neq 0$ , so $\frac{a}{b}, \frac{c}{d} \in \mathbb{Q}$ . The multiplication of $\frac{a}{b} \cdot \frac{c}{d}$ is well-defined.

Proof. Let $a, b, c, d, e, f, g, h \in \mathbb{Z}$ where $b, d, f, h \neq o$ be arbitrary. Suppose $\frac{a}{b} = \frac{e}{f}$ and $\frac{c}{d} = \frac{g}{h}$ . It follows that $af = be$ and $ch =gd$ . By definition of multiplication, $\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}$ and $\frac{e}{f} \cdot \frac{g}{h} = \frac{eg}{fh}$ .

To show that multiplication is well-defined, we must prove that $\frac{ac}{bd} = \frac{eg}{fh}$ . To do this we can prove $(ac)(fh) = (eg)(bd)$ instead. We know that