Proof: Algebra 17b

Let's prove the following theorem:

if not (s = 0), then (s / 2) / s = 1 / 2

Proof:

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Given

1	not (s = 0)

Proof Table

#	Claim	Reason
1	s / 2 = s ⋅ (1 / 2)	s / 2 = s ⋅ (1 / 2) (Division is Multiplication by Inverse)
2	(s / 2) / s = (s ⋅ (1 / 2)) / s	if s / 2 = s ⋅ (1 / 2), then (s / 2) / s = (s ⋅ (1 / 2)) / s (Division Property of Equality)
3	(s ⋅ (1 / 2)) / s = (s ⋅ (1 / 2)) ⋅ (1 / s)	(s ⋅ (1 / 2)) / s = (s ⋅ (1 / 2)) ⋅ (1 / s) (Division is Multiplication by Inverse)
4	(s ⋅ (1 / 2)) ⋅ (1 / s) = (s ⋅ (1 / s)) ⋅ (1 / 2)	(s ⋅ (1 / 2)) ⋅ (1 / s) = (s ⋅ (1 / s)) ⋅ (1 / 2) (Swap Terms 2 and 3)
5	s ⋅ (1 / s) = 1	if not (s = 0), then s ⋅ (1 / s) = 1 (Algebra 17)
6	(s ⋅ (1 / s)) ⋅ (1 / 2) = 1 ⋅ (1 / 2)	if s ⋅ (1 / s) = 1, then (s ⋅ (1 / s)) ⋅ (1 / 2) = 1 ⋅ (1 / 2) (Multiplicative Property of Equality)
7	1 ⋅ (1 / 2) = 1 / 2	1 ⋅ (1 / 2) = 1 / 2 (multiplicative_identity_3)
8	(s ⋅ (1 / s)) ⋅ (1 / 2) = 1 / 2	if (s ⋅ (1 / s)) ⋅ (1 / 2) = 1 ⋅ (1 / 2) and 1 ⋅ (1 / 2) = 1 / 2, then (s ⋅ (1 / s)) ⋅ (1 / 2) = 1 / 2 (Transitive Property of Equality)
9	(s ⋅ (1 / 2)) ⋅ (1 / s) = 1 / 2	if (s ⋅ (1 / 2)) ⋅ (1 / s) = (s ⋅ (1 / s)) ⋅ (1 / 2) and (s ⋅ (1 / s)) ⋅ (1 / 2) = 1 / 2, then (s ⋅ (1 / 2)) ⋅ (1 / s) = 1 / 2 (Transitive Property of Equality)
10	(s ⋅ (1 / 2)) / s = 1 / 2	if (s ⋅ (1 / 2)) / s = (s ⋅ (1 / 2)) ⋅ (1 / s) and (s ⋅ (1 / 2)) ⋅ (1 / s) = 1 / 2, then (s ⋅ (1 / 2)) / s = 1 / 2 (Transitive Property of Equality)
11	(s / 2) / s = 1 / 2	if (s / 2) / s = (s ⋅ (1 / 2)) / s and (s ⋅ (1 / 2)) / s = 1 / 2, then (s / 2) / s = 1 / 2 (Transitive Property of Equality)

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