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

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 $x + z < y + z .$

Rule

Subtraction Property of Inequality

Subtracting the same number from both sides of an inequality produces an equivalent inequality. The solution set and inequality sign of this equivalent inequality does not change. 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

Subtraction Property of Inequality

The case when

x < y

will be proven. The other cases can be proven using a similar reasoning. Consider the biconditional statement before beginning the proof.

x < y \Leftrightarrow y - x > 0

This property can be proven using the Additive Inverse of

z,

which is

- z .

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 - (- z)) - x > 0

RemovePar

Remove parentheses

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

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

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

CommutativePropAdd

Commutative Property of Addition

y - z - (x - z) > 0

AddPar

Add parentheses

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

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

(y - z) - (x - z) > 0 ⇕ x - z < y - z

Finally, because

x < y,

the property has been proven.

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

Rule

Multiplication Property of Inequality

Multiplying both sides of an inequality by a nonzero real number $z$ produces an equivalent inequality. The following conditions about $z$ need to be considered when applying this property.

Positive $z$	If $z$ is positive, the inequality sign remains the same.
Negative $z$	If $z$ is negative, the inequality sign needs to be reversed to produce an equivalent inequality.

For example, let $x,$ $y,$ and $z$ be real numbers such that $x < y$ and $z \neq = 0 .$ Then, the equivalent inequalities can be written depending on the sign of $z .$

If $x < y$ and $z > 0,$ then $x z < y z .$
If $x < y$ and $z < 0,$ then $x z > y z .$

This property holds for the other types of inequalities.

Proof

Multiplication Property of Inequality

The case when $x < y$ will be proven. The remaining cases can be proven following a similar reasoning. Before starting the proof, the following properties of real numbers need to be considered.

$x < y$ if and only if $y - x > 0 .$
If $x$ and $y$ are positive, then $x y > 0 .$
If $z$ is negative, then $- z$ is positive.

Using these properties, the following conditional statements can be proven.

If $x < y$ and $z > 0,$ then $x z < y z .$
If $x < y$ and $z < 0,$ then $x z > y z .$

Each conditional statement will be analyzed separately.

When $z$ Is Greater Than $0$

It is given that

x < y,

then using the first property, it is known that

y - x

is greater than

0 .

x < y \Leftrightarrow y - x > 0

Furthermore, because

z > 0,

from the second property, it can be stated that the product of

z

and

y - x

is also greater than

0 .

y - x > 0 z (y - and z > 0 ⇓ x) > 0

Now, the second part of this conditional statement can be rewritten using the Distributive Property.

z (y - x) > 0 \Leftrightarrow z y - z x > 0

From the first property, it can be said that

z y - z x > 0

if and only if

z x < z y .

Additionally, because

x < y,

the conditional statement has been proven.

x < y

and

z > 0,

then

z x < z y .

When $z$ Is Less Than $0$

Again, because

x < y,

the following statement is valid.

x < y \Leftrightarrow y - x > 0

Additionally, since

z < 0,

from the third property it follows that

- z

is positive. Moreover, the product of

- z

and

y - z

will be positive.

y - x > 0 - z (y - and - z > 0 ⇓ x) > 0

Now,

- z

can be distributed in the second part of the statement.

- z (y - x) > 0

▼

Simplify

Distr

Distribute $(- z)$

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

MultNegPos

$(- a) b = - a b$

- z y - (- z x) > 0

SubNeg

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

- z y + z x > 0

AddIneq

$LHS + z y > RHS + z y$

z x > z y

Finally, because

x < y,

the property has been proven.

If $x < y$ and $z < 0,$ then $z x > z y .$

Rule

Division Property of Inequality

Dividing both sides of an inequality by a nonzero real number $z$ produces an equivalent inequality. However, the following conditions need to be considered.

Positive $z$	If $z$ is positive, the inequality sign remains the same.
Negative $z$	If $z$ is negative, the inequality sign needs to be reversed to produce an equivalent inequality.

For example, let $x,$ $y,$ and $z$ be real numbers such that $x < y$ and $z \neq = 0 .$ Then, the equivalent inequalities can be written depending on the sign of $z .$

If $x < y$ and $z > 0,$ then $\frac{x}{z} < \frac{y}{z} .$
If $x < y$ and $z < 0,$ then $\frac{x}{z} > \frac{y}{z} .$

This property holds for the other types of inequalities.

Proof

Division Property of Inequality

The case when $x < y$ will be proven. The remaining cases can be proven following a similar reasoning. Before starting the proof, the following properties of real numbers need to be considered.

$x < y$ if and only if $y - x$ is positive.
If $x$ and $z$ are positive, then $\frac{x}{z}$ is also positive.
If $z$ is negative, then $- z$ is positive.

Using these properties, the following conditional statements can be proven.

If $x < y$ and $z > 0,$ then $\frac{x}{z} < \frac{y}{z} .$
If $x < y$ and $z < 0,$ then $\frac{x}{z} > \frac{y}{z} .$

Each case will be analyzed separately.

$z > 0$

It is given that

x < y,

then using the first property, it is known that

y - x

is greater than

0 .

x < y \Leftrightarrow y - x > 0

Furthermore, because

z > 0,

from the second property, it can be stated that

y - x

divided by

z

is also greater than

0 .

y - x > 0 \frac{y - x}{z} and z > 0 ⇓ > 0

Now, the second part of this conditional statement can be rewritten.

\frac{y - x}{z} > 0 \Leftrightarrow \frac{y}{z} - \frac{x}{z} > 0

By using the first property, it can be said that

\frac{x}{z}

is less than

\frac{y}{z} .

Additionally, because

x < y,

the property has been proven.

If $x < y$ and $z > 0,$ then $\frac{x}{z} < \frac{y}{z} .$

$z < 0$

Again, because

x < y,

the following statement is valid.

x < y \Leftrightarrow y - x > 0

Additionally, since

z < 0,

from the third property, it follows that

- z

is positive. Moreover, the quotient of

y - x

and

- z

will be positive.

y - x > 0 \frac{y - x}{- z} and - z > 0 ⇓ > 0

Now, the second part of this statement can be rewritten.

\frac{y - x}{- z} > 0

▼

Simplify

MoveNegDenomToNum

Put minus sign in numerator

\frac{- ( y - x )}{z} > 0

DistrNegSignSwap

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

\frac{x - y}{z} > 0

WriteDiffFrac

Write as a difference of fractions

\frac{x}{z} - \frac{y}{z} > 0

AddIneq

$LHS + \frac{y}{z} > RHS + \frac{y}{z}$

\frac{x}{z} > \frac{y}{z}

Finally, because

x < y,

the property has been obtained.

If $x < y$ and $z < 0,$ then $\frac{x}{z} > \frac{y}{z} .$

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Proof

Proof

Proof

When z Is Greater Than 0

When z Is Less Than 0

Proof

z>0

z<0

When $z$ Is Greater Than $0$

When $z$ Is Less Than $0$

$z > 0$

$z < 0$