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Rule

Addition Property of Inequality

Adding the same number to both sides of an inequality generates an equivalent inequality. This equivalent inequality will have the same solution set and the inequality sign remains the same. Let

x,

y,

and

z

be real numbers such that

x < y .

Then, the following conditional statement holds true.

If $x < y,$ then $x + z < y + z .$

This property holds for the other types of inequalities.

Proof

Addition Property of Inequality

The case when

x < y

will be proven. The remaining cases can be proven similarly. Before starting the proof, the following biconditional statement needs to be considered.

x < y \Leftrightarrow y - x > 0

Now, the Identity Property of Addition can be applied to the second part of the statement.

y - x > 0

IdPropAdd

Identity Property of Addition

y - x + 0 > 0

Rewrite $0$ as $z - z$

y - x + z - z > 0

CommutativePropAdd

Commutative Property of Addition

y + z - x - z > 0

$- a - b = - (a + b)$

y + z - (x + z) > 0

AddPar

Add parentheses

(y + z) - (x + z) > 0

Using the biconditional statement, the last inequality can be rewritten.

(y + z) - (x + z) > 0 ⇕ x + z < y + z

Finally, because

x < y,

the property is obtained.

If $x < y,$ then $x + z < y + z .$

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