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Reference

Properties of Inequalities

Rule

Anti Reflexive Property of Inequality

A real number can never be less than or greater than itself.

$x ≮ x$ and $x ≯ x$

This property is an axiom. Therefore, it can be accepted as true without proof.

Rule

Anti Symmetric Property of Inequality

For any two real numbers $x$ and $y,$ if $x$ is less than $y,$ then $y$ cannot be less than $x .$

If $x < y,$ then $y \neq < x .$

Alternatively, if $x$ is greater than $y,$ then $y$ cannot be greater than $x .$

If $x > y,$ then $y \neq > x .$

This property is an axiom. Therefore, it can be accepted as true without proof.

Rule

Transitive Property of Inequality

Let $x,$ $y,$ and $z$ be real numbers. If $x$ is less than $y$ and $y$ is less than $z,$ then $x$ is less than $z .$

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

This property also applies to other types of inequalities — $>,$ $\leq,$ and $\geq .$

If $x > y$ and $y > z,$ then $x > z .$
If $x \leq y$ and $y \leq z,$ then $x \leq z .$
If $x \geq y$ and $y \geq z,$ then $x \geq z .$

Since this property is an axiom, it does not need proof to be accepted as true.

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

\IdPropAdd

y - x + 0 > 0

Rewrite $0$ as $z - z$

y - x + z - z > 0

CommutativePropAdd

\CommutativePropAdd

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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Proof