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 ⋅ a) ⋅ (1 / b) = (d ⋅ (1 / b)) ⋅ a and (d ⋅ (1 / b)) ⋅ a = (d / b) ⋅ a, then (d ⋅ a) ⋅ (1 / b) = (d / b) ⋅ a |
8 | (d ⋅ a) / b = (d / b) ⋅ a | if (d ⋅ a) / b = (d ⋅ a) ⋅ (1 / b) and (d ⋅ a) ⋅ (1 / b) = (d / b) ⋅ a, then (d ⋅ a) / b = (d / b) ⋅ a |
9 | d ⋅ (a / b) = (d / b) ⋅ a | if d ⋅ (a / b) = (d ⋅ a) / b and (d ⋅ a) / b = (d / b) ⋅ a, then d ⋅ (a / b) = (d / b) ⋅ a |
10 | (a / b) ⋅ d = (d / b) ⋅ a | if (a / b) ⋅ d = d ⋅ (a / b) and d ⋅ (a / b) = (d / b) ⋅ a, then (a / b) ⋅ d = (d / b) ⋅ a |
Comments
Please log in to add comments