Proof: Algebra Divide
Let's prove the following theorem:
if the following are true:
- a / b = c / d
- not (a = 0)
- not (d = 0)
then d / b = c / a
Proof:
Given
| 1 | a / b = c / d |
|---|---|
| 2 | not (a = 0) |
| 3 | not (d = 0) |
| # | Claim | Reason |
|---|---|---|
| 1 | (a / b) ⋅ d = (c / d) ⋅ d | if a / b = c / d, then (a / b) ⋅ d = (c / d) ⋅ d |
| 2 | (c / d) ⋅ d = c | if not (d = 0), then (c / d) ⋅ d = c |
| 3 | (a / b) ⋅ d = c | if (a / b) ⋅ d = (c / d) ⋅ d and (c / d) ⋅ d = c, then (a / b) ⋅ d = c |
| 4 | (a / b) ⋅ d = (d / b) ⋅ a | (a / b) ⋅ d = (d / b) ⋅ a |
| 5 | (d / b) ⋅ a = c | if (a / b) ⋅ d = (d / b) ⋅ a and (a / b) ⋅ d = c, then (d / b) ⋅ a = c |
| 6 | ((d / b) ⋅ a) / a = c / a | if (d / b) ⋅ a = c, then ((d / b) ⋅ a) / a = c / a |
| 7 | ((d / b) ⋅ a) / a = d / b | if not (a = 0), then ((d / b) ⋅ a) / a = d / b |
| 8 | d / b = c / a | if ((d / b) ⋅ a) / a = d / b and ((d / b) ⋅ a) / a = c / a, then d / b = c / a |
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