Proof: Adding Fractions

If two fractions share a denominator, we can add them by adding the numerators. For example,

$\frac{1}{6}+\frac{3}{6}=\frac{4}{6}$

Let's prove the following theorem:

We can use the distributive property to claim that:

a ⋅ (1 / c) + b ⋅ (1 / c) = (a + b) ⋅ (1 / c)

Then we can reach the conclusion by simplifying this equation using the following property:

a ⋅ (1 / b) = a / b

Proof:

Proof Table

#	Claim	Reason
1	(a ⋅ (1 / c)) + (b ⋅ (1 / c)) = (a + b) ⋅ (1 / c)	(a ⋅ (1 / c)) + (b ⋅ (1 / c)) = (a + b) ⋅ (1 / c) (distributive_property_5)
2	a / c = a ⋅ (1 / c)	a / c = a ⋅ (1 / c) (Division is Multiplication by Inverse)
3	b / c = b ⋅ (1 / c)	b / c = b ⋅ (1 / c) (Division is Multiplication by Inverse)
4	(a ⋅ (1 / c)) + (b ⋅ (1 / c)) = (a / c) + (b / c)	if a / c = a ⋅ (1 / c) and b / c = b ⋅ (1 / c), then (a ⋅ (1 / c)) + (b ⋅ (1 / c)) = (a / c) + (b / c) (Add Equal Terms)
5	(a + b) ⋅ (1 / c) = (a + b) / c	(a + b) ⋅ (1 / c) = (a + b) / c (division_theorem)
6	(a / c) + (b / c) = (a + b) / c	if (a ⋅ (1 / c)) + (b ⋅ (1 / c)) = (a + b) ⋅ (1 / c) and (a ⋅ (1 / c)) + (b ⋅ (1 / c)) = (a / c) + (b / c) and (a + b) ⋅ (1 / c) = (a + b) / c, then (a / c) + (b / c) = (a + b) / c (Transitive Property Application 2)

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