Proof: Fraction Multiplication

Let's prove the following theorem:

For clarity, here is the conclusion in fraction notation:

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

This theorem states that when we are multiplying fractions, the numerator is the product of numerators and the denominator is the product of denominators.

Before you read the proof, we encourage you to try to prove this theorem on your own.

This proof begins with the following theorem:

$\frac{1}{b}\cdot \frac{1}{d}=\frac{1}{b\cdot d}$

The proof multiplies both sides of this equation by a ⋅ c. It then shows that the left side is equal to

$\frac{a}{b}\cdot \frac{c}{d}$

and the right side is equal to

$\frac{a\cdot c}{b\cdot d}$

Proof:

Proof Table

#	Claim	Reason
1	(1 / b) ⋅ (1 / d) = 1 / (b ⋅ d)	if not (b = 0) and not (d = 0) and not (b ⋅ d = 0), then (1 / b) ⋅ (1 / d) = 1 / (b ⋅ d) (Multiplying Denominators)
2	(a ⋅ c) ⋅ ((1 / b) ⋅ (1 / d)) = (a ⋅ c) ⋅ (1 / (b ⋅ d))	if (1 / b) ⋅ (1 / d) = 1 / (b ⋅ d), then (a ⋅ c) ⋅ ((1 / b) ⋅ (1 / d)) = (a ⋅ c) ⋅ (1 / (b ⋅ d)) (Substitution)
3	(a ⋅ c) ⋅ (1 / (b ⋅ d)) = (a ⋅ c) / (b ⋅ d)	(a ⋅ c) ⋅ (1 / (b ⋅ d)) = (a ⋅ c) / (b ⋅ d) (division_theorem)
4	(a ⋅ c) ⋅ ((1 / b) ⋅ (1 / d)) = (a / b) ⋅ (c / d)	(a ⋅ c) ⋅ ((1 / b) ⋅ (1 / d)) = (a / b) ⋅ (c / d) (simplify_product_2)
5	(a / b) ⋅ (c / d) = (a ⋅ c) / (b ⋅ d)	if (a ⋅ c) ⋅ ((1 / b) ⋅ (1 / d)) = (a ⋅ c) ⋅ (1 / (b ⋅ d)) and (a ⋅ c) ⋅ ((1 / b) ⋅ (1 / d)) = (a / b) ⋅ (c / d) and (a ⋅ c) ⋅ (1 / (b ⋅ d)) = (a ⋅ c) / (b ⋅ d), then (a / b) ⋅ (c / d) = (a ⋅ c) / (b ⋅ d) (Transitive Property Application 2)

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