Proof: Inequality Problem 4

Let's prove the following theorem:

For clarity, here is the condition in fraction notation:

$\frac{x+\mathrm{10}}{-5}>3$

The inequality properties allow us to multiply -5 to both sides, but we need to flip the operator to the less-than sign (<).

Then this equation becomes:

x + 10 < 15

Next, we can add -10 to both sides to reach the conclusion.

Proof:

Proof Table

#	Claim	Reason
1	(x + 10) / (-5) = (x + 10) ⋅ (1 / (-5))	(x + 10) / (-5) = (x + 10) ⋅ (1 / (-5)) (Division is Multiplication by Inverse)
2	(x + 10) ⋅ (1 / (-5)) > 3	if (x + 10) / (-5) > 3 and (x + 10) / (-5) = (x + 10) ⋅ (1 / (-5)), then (x + 10) ⋅ (1 / (-5)) > 3 (Transitive Property of Inequality 4)
3	-5 < 0	-5 < 0 (Fixed Value Statement)
4	((x + 10) ⋅ (1 / (-5))) ⋅ (-5) < 3 ⋅ (-5)	if (x + 10) ⋅ (1 / (-5)) > 3 and -5 < 0, then ((x + 10) ⋅ (1 / (-5))) ⋅ (-5) < 3 ⋅ (-5) (Inequality Change)
5	3 ⋅ (-5) = -15	3 ⋅ (-5) = -15 (Fixed Value Statement)
6	(1 / (-5)) ⋅ (-5) = 1	(1 / (-5)) ⋅ (-5) = 1 (Fixed Value Statement)
7	((x + 10) ⋅ (1 / (-5))) ⋅ (-5) = (x + 10) ⋅ 1	if (1 / (-5)) ⋅ (-5) = 1, then ((x + 10) ⋅ (1 / (-5))) ⋅ (-5) = (x + 10) ⋅ 1 (Apply Associative Multiply)
8	(x + 10) ⋅ 1 < -15	if ((x + 10) ⋅ (1 / (-5))) ⋅ (-5) < 3 ⋅ (-5) and ((x + 10) ⋅ (1 / (-5))) ⋅ (-5) = (x + 10) ⋅ 1 and 3 ⋅ (-5) = -15, then (x + 10) ⋅ 1 < -15 (Double Transitive Property Less)
9	x + 10 < -15	if (x + 10) ⋅ 1 < -15, then x + 10 < -15 (Multiply by 1 Less)
10	(x + 10) + (-10) < (-15) + (-10)	if x + 10 < -15, then (x + 10) + (-10) < (-15) + (-10) (Inequality Addition Property)
11	(-15) + (-10) = -25	(-15) + (-10) = -25 (Fixed Value Statement)
12	10 + (-10) = 0	10 + (-10) = 0 (Fixed Value Statement)
13	(x + 10) + (-10) = x + 0	if 10 + (-10) = 0, then (x + 10) + (-10) = x + 0 (Apply Associative Add)
14	x + 0 < -25	if (x + 10) + (-10) < (-15) + (-10) and (x + 10) + (-10) = x + 0 and (-15) + (-10) = -25, then x + 0 < -25 (Double Transitive Property Less)
15	x < -25	if x + 0 < -25, then x < -25 (Add by 0 Less)

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