Proof: Square Root Example 2

Let's prove the following theorem:

Proof:

Proof Table

#	Claim	Reason
1	(3 / 4) ⋅ (s ⋅ s) = 3 ⋅ ((s ⋅ s) ⋅ (1 / 4))	(3 / 4) ⋅ (s ⋅ s) = 3 ⋅ ((s ⋅ s) ⋅ (1 / 4)) (reduction_property_2)
2	((3 / 4) ⋅ (s ⋅ s))(1 / 2) = (3 ⋅ ((s ⋅ s) ⋅ (1 / 4)))(1 / 2)	if (3 / 4) ⋅ (s ⋅ s) = 3 ⋅ ((s ⋅ s) ⋅ (1 / 4)), then ((3 / 4) ⋅ (s ⋅ s))(1 / 2) = (3 ⋅ ((s ⋅ s) ⋅ (1 / 4)))(1 / 2) (Substitution)
3	(3 ⋅ ((s ⋅ s) ⋅ (1 / 4)))(1 / 2) = (3(1 / 2)) ⋅ (((s ⋅ s) ⋅ (1 / 4))(1 / 2))	(3 ⋅ ((s ⋅ s) ⋅ (1 / 4)))(1 / 2) = (3(1 / 2)) ⋅ (((s ⋅ s) ⋅ (1 / 4))(1 / 2)) (Power of a Product)
4	((s ⋅ s) ⋅ (1 / 4))(1 / 2) = s / 2	((s ⋅ s) ⋅ (1 / 4))(1 / 2) = s / 2 (square_root_example)
5	(3(1 / 2)) ⋅ (((s ⋅ s) ⋅ (1 / 4))(1 / 2)) = (3(1 / 2)) ⋅ (s / 2)	if ((s ⋅ s) ⋅ (1 / 4))(1 / 2) = s / 2, then (3(1 / 2)) ⋅ (((s ⋅ s) ⋅ (1 / 4))(1 / 2)) = (3(1 / 2)) ⋅ (s / 2) (Multiplicative Property of Equality Variation 1)
6	(3 ⋅ ((s ⋅ s) ⋅ (1 / 4)))(1 / 2) = (3(1 / 2)) ⋅ (s / 2)	if (3 ⋅ ((s ⋅ s) ⋅ (1 / 4)))(1 / 2) = (3(1 / 2)) ⋅ (((s ⋅ s) ⋅ (1 / 4))(1 / 2)) and (3(1 / 2)) ⋅ (((s ⋅ s) ⋅ (1 / 4))(1 / 2)) = (3(1 / 2)) ⋅ (s / 2), then (3 ⋅ ((s ⋅ s) ⋅ (1 / 4)))(1 / 2) = (3(1 / 2)) ⋅ (s / 2) (Transitive Property of Equality)
7	((3 / 4) ⋅ (s ⋅ s))(1 / 2) = (3(1 / 2)) ⋅ (s / 2)	if ((3 / 4) ⋅ (s ⋅ s))(1 / 2) = (3 ⋅ ((s ⋅ s) ⋅ (1 / 4)))(1 / 2) and (3 ⋅ ((s ⋅ s) ⋅ (1 / 4)))(1 / 2) = (3(1 / 2)) ⋅ (s / 2), then ((3 / 4) ⋅ (s ⋅ s))(1 / 2) = (3(1 / 2)) ⋅ (s / 2) (Transitive Property of Equality)

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