Proof: Algebra 6
Let's prove the following theorem:
(a / b) ⋅ d = (d / b) ⋅ a
Proof:
| # | Claim | Reason |
|---|---|---|
| 1 | (a / b) ⋅ d = d ⋅ (a / b) | (a / b) ⋅ d = d ⋅ (a / b) |
| 2 | d ⋅ (a / b) = (d ⋅ a) / b | d ⋅ (a / b) = (d ⋅ a) / b |
| 3 | (d ⋅ a) / b = (d ⋅ a) ⋅ (1 / b) | (d ⋅ a) / b = (d ⋅ a) ⋅ (1 / b) |
| 4 | (d ⋅ a) ⋅ (1 / b) = (d ⋅ (1 / b)) ⋅ a | (d ⋅ a) ⋅ (1 / b) = (d ⋅ (1 / b)) ⋅ a |
| 5 | d ⋅ (1 / b) = d / b | d ⋅ (1 / b) = d / b |
| 6 | (d ⋅ (1 / b)) ⋅ a = (d / b) ⋅ a | if d ⋅ (1 / b) = d / b, then (d ⋅ (1 / b)) ⋅ a = (d / b) ⋅ a |
| 7 | (d ⋅ a) ⋅ (1 / b) = (d / b) ⋅ a | if (d ⋅ (1 / b)) ⋅ a = (d / b) ⋅ a and (d ⋅ a) ⋅ (1 / b) = (d ⋅ (1 / b)) ⋅ a, then (d ⋅ a) ⋅ (1 / b) = (d / b) ⋅ a |
| 8 | (d ⋅ a) / b = (d / b) ⋅ a | if (d ⋅ a) ⋅ (1 / b) = (d / b) ⋅ a and (d ⋅ a) / b = (d ⋅ a) ⋅ (1 / b), then (d ⋅ a) / b = (d / b) ⋅ a |
| 9 | d ⋅ (a / b) = (d / b) ⋅ a | if (d ⋅ a) / b = (d / b) ⋅ a and d ⋅ (a / b) = (d ⋅ a) / b, then d ⋅ (a / b) = (d / b) ⋅ a |
| 10 | (a / b) ⋅ d = (d / b) ⋅ a | if d ⋅ (a / b) = (d / b) ⋅ a and (a / b) ⋅ d = d ⋅ (a / b), then (a / b) ⋅ d = (d / b) ⋅ a |
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