Hermit Notebook

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

Reminder: you can left-scroll and right-scroll the long math blocks

The question

We all learnt in college this formula:

ab×cd=a×cb×d\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}

Where a,b,ca, b, c and dd are integers and bb and dd are different from zero, or in other words: a,cNa, c \in \mathbb{N} and b,dNb, 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×b×c = (a×b)×c = a×(b×c)a \times b \times c \;\ = \;\ (a \times b) \times c \;\ = \;\ a \times (b \times c)

  1. The commutativity of multiplication

a×b=b×aa \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×b÷c=(a×b)÷c=a×(b÷c)a \times b \div c = (a \times b) \div c = a \times (b \div c)

  1. The neutrality of 11 for multiplication:

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

  1. A number divided by itself equals 11

a0,a÷a=1\forall a \ne 0, \quad a \div a = 1

An answer

We can try to use simple (?) algebra to transform ab×cd\frac{a}{b} \times \frac{c}{d} into a×cb×d\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 11) 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:

ab×cd = ab×cd × ((b×d) ÷ (b×d))\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:

ab×cd = (ab×b)×(cd×d)÷(b×d)\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 ab×b = a\frac{a}{b} \times b \ = \ a and cd×d = c\frac{c}{d} \times d \ = \ c. So we now have:

ab×cd = a×c÷(b×d)\frac{a}{b} \times \frac{c}{d} \ = \ a \times c \div \left( b \times d \right)

Using associativity, we write:

ab×cd = (a×c)÷(b×d)\frac{a}{b} \times \frac{c}{d} \ = \ \left(a \times c\right) \div \left( b \times d \right)

Which is the same as:

ab×cd = a×cb×d\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÷ba \div b is not exactly the same mathematical object as ab\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 !

Contents

  1. 1. The question
  2. 2. The prerequisites
  3. 3. An answer
  4. 4. A postcriptum