Proof: Algebra 16
Let's prove the following theorem:
(a ⋅ (s / 2)) / s = a / 2
Proof:
# | Claim | Reason |
---|---|---|
1 | s / 2 = s ⋅ (1 / 2) | s / 2 = s ⋅ (1 / 2) |
2 | a ⋅ (s / 2) = a ⋅ (s ⋅ (1 / 2)) | if s / 2 = s ⋅ (1 / 2), then a ⋅ (s / 2) = a ⋅ (s ⋅ (1 / 2)) |
3 | a ⋅ (s ⋅ (1 / 2)) = (a ⋅ s) ⋅ (1 / 2) | a ⋅ (s ⋅ (1 / 2)) = (a ⋅ s) ⋅ (1 / 2) |
4 | a ⋅ (s / 2) = (a ⋅ s) ⋅ (1 / 2) | if a ⋅ (s / 2) = a ⋅ (s ⋅ (1 / 2)) and a ⋅ (s ⋅ (1 / 2)) = (a ⋅ s) ⋅ (1 / 2), then a ⋅ (s / 2) = (a ⋅ s) ⋅ (1 / 2) |
5 | (a ⋅ (s / 2)) / s = ((a ⋅ s) ⋅ (1 / 2)) / s | if a ⋅ (s / 2) = (a ⋅ s) ⋅ (1 / 2), then (a ⋅ (s / 2)) / s = ((a ⋅ s) ⋅ (1 / 2)) / s |
6 | ((a ⋅ s) ⋅ (1 / 2)) / s = ((a ⋅ s) ⋅ (1 / 2)) ⋅ (1 / s) | ((a ⋅ s) ⋅ (1 / 2)) / s = ((a ⋅ s) ⋅ (1 / 2)) ⋅ (1 / s) |
7 | ((a ⋅ s) ⋅ (1 / 2)) ⋅ (1 / s) = ((a ⋅ s) ⋅ (1 / s)) ⋅ (1 / 2) | ((a ⋅ s) ⋅ (1 / 2)) ⋅ (1 / s) = ((a ⋅ s) ⋅ (1 / s)) ⋅ (1 / 2) |
8 | (a ⋅ s) ⋅ (1 / s) = a ⋅ 1 | (a ⋅ s) ⋅ (1 / s) = a ⋅ 1 |
9 | a ⋅ 1 = a | a ⋅ 1 = a |
10 | (a ⋅ s) ⋅ (1 / s) = a | if (a ⋅ s) ⋅ (1 / s) = a ⋅ 1 and a ⋅ 1 = a, then (a ⋅ s) ⋅ (1 / s) = a |
11 | ((a ⋅ s) ⋅ (1 / s)) ⋅ (1 / 2) = a ⋅ (1 / 2) | if (a ⋅ s) ⋅ (1 / s) = a, then ((a ⋅ s) ⋅ (1 / s)) ⋅ (1 / 2) = a ⋅ (1 / 2) |
12 | ((a ⋅ s) ⋅ (1 / 2)) ⋅ (1 / s) = a ⋅ (1 / 2) | if ((a ⋅ s) ⋅ (1 / 2)) ⋅ (1 / s) = ((a ⋅ s) ⋅ (1 / s)) ⋅ (1 / 2) and ((a ⋅ s) ⋅ (1 / s)) ⋅ (1 / 2) = a ⋅ (1 / 2), then ((a ⋅ s) ⋅ (1 / 2)) ⋅ (1 / s) = a ⋅ (1 / 2) |
13 | ((a ⋅ s) ⋅ (1 / 2)) / s = a ⋅ (1 / 2) | if ((a ⋅ s) ⋅ (1 / 2)) / s = ((a ⋅ s) ⋅ (1 / 2)) ⋅ (1 / s) and ((a ⋅ s) ⋅ (1 / 2)) ⋅ (1 / s) = a ⋅ (1 / 2), then ((a ⋅ s) ⋅ (1 / 2)) / s = a ⋅ (1 / 2) |
14 | (a ⋅ (s / 2)) / s = a ⋅ (1 / 2) | if (a ⋅ (s / 2)) / s = ((a ⋅ s) ⋅ (1 / 2)) / s and ((a ⋅ s) ⋅ (1 / 2)) / s = a ⋅ (1 / 2), then (a ⋅ (s / 2)) / s = a ⋅ (1 / 2) |
15 | a ⋅ (1 / 2) = a / 2 | a ⋅ (1 / 2) = a / 2 |
16 | (a ⋅ (s / 2)) / s = a / 2 | if (a ⋅ (s / 2)) / s = a ⋅ (1 / 2) and a ⋅ (1 / 2) = a / 2, then (a ⋅ (s / 2)) / s = a / 2 |
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