Algebra I / Chapter 4: Fractions / Fractions

Proof: Fraction Multiplication

Let's prove the following theorem:

if the following are true:

not (b = 0)
not (d = 0)
not (b ⋅ d = 0)

then (a / b) ⋅ (c / d) = (a ⋅ c) / (b ⋅ d)

Proof:

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Given

1	not (b = 0)
2	not (d = 0)
3	not (b ⋅ d = 0)

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) (Multiply 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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