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:
Given
1 | not (b = 0) |
---|---|
2 | not (d = 0) |
3 | not (b ⋅ d = 0) |
# | 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) |
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)) |
3 | (a ⋅ c) ⋅ (1 / (b ⋅ d)) = (a ⋅ c) / (b ⋅ d) | (a ⋅ c) ⋅ (1 / (b ⋅ d)) = (a ⋅ c) / (b ⋅ d) |
4 | (a ⋅ c) ⋅ ((1 / b) ⋅ (1 / d)) = (a / b) ⋅ (c / d) | (a ⋅ c) ⋅ ((1 / b) ⋅ (1 / d)) = (a / b) ⋅ (c / d) |
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) |
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