 Hermit Notebook

# How do we know the formula for fraction multiplication is right ?

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## The question

We all learnt in college this formula:

$\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$

Where $a, b, c$ and $d$ are integers and $b$ and $d$ are different from zero, or in other words: $a, c \in \mathbb{N}$ and $b, d \in \mathbb{N}^*$

When we apply it, it just seems to work. But how are we sure the formula is always true ?

## The prerequisites

We will take for granted the properties of the operations on integers, especially for multiplication:

1. The associativity of multiplication:

$a \times b \times c \;\ = \;\ (a \times b) \times c \;\ = \;\ a \times (b \times c)$

1. The commutativity of multiplication

$a \times b = b \times a$

1. The equal priorities of multiplication and division, which means the order in which we do multiplication and division does not change the result:

$a \times b \div c = (a \times b) \div c = a \times (b \div c)$

1. The neutrality of $1$ for multiplication:

$\forall a \in \mathbb{N}^*, \;\ a \times 1 = a$

1. A number divided by itself equals $1$

$\forall a \ne 0, \quad a \div a = 1$

We can try to use simple (?) algebra to transform $\frac{a}{b} \times \frac{c}{d}$ into $\frac{a \times c}{b \times d}$ and thus be sure of our formula.

As in many algebra transformations, we will introduce a neutral expression, which means an expression that evaluates to the neutral element (here $1$) for the operator (here $\times$) and thus will let the value of the formula unchanged. We use property (5) from the prerequisites to introduce a handy neutral expression:

$\frac{a}{b} \times \frac{c}{d} \ = \ \frac{a}{b} \times \frac{c}{d} \;\ \times \;\ \left((b \times d) \ \div \ (b \times d) \right)$

Now we use properties (1), (2) and (3) from the prerequisites to rewrite the equation above:

$\frac{a}{b} \times \frac{c}{d} \ = \ \left(\frac{a}{b} \times b\right) \times \left(\frac{c}{d} \times d \right) \div \left( b \times d \right)$

Dividing by a number then multiplying by that same number does not change the initial number, so $\frac{a}{b} \times b \ = \ a$ and $\frac{c}{d} \times d \ = \ c$. So we now have:

$\frac{a}{b} \times \frac{c}{d} \ = \ a \times c \div \left( b \times d \right)$

Using associativity, we write:

$\frac{a}{b} \times \frac{c}{d} \ = \ \left(a \times c\right) \div \left( b \times d \right)$

Which is the same as:

$\boxed{ \frac{a}{b} \times \frac{c}{d} \ = \ \frac{a \times c}{b \times d} }$

Yay ! Here we have it ! Q.E.D - Quod Erat Demonstrandum !

## A postcriptum

Notice that $a \div b$ is not exactly the same mathematical object as $\frac{a}{b}$. The first is a binary operator applied to 2 integers and the second is a number (a rational number). They are equal, meaning they can be interchanged in an expression, because they eventually represent the same value.

See you soon !
Keep learning !

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