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

$x≮x$ and $x≯x$

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 $y <x.$

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

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

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 — $>,$ $≤,$ and $≥.$

- If $x>y$ and $y>z,$ then $x>z.$
- If $x≤y$ and $y≤z,$ then $x≤z.$
- If $x≥y$ and $y≥z,$ then $x≥z.$

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 $x+z<y+z.$

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

$x<y⇔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

$(y+z)−(x+z)>0$

$(y+z)−(x+z)>0⇕x+z<y+z $

Finally, because $x<y,$ the property is obtained. If $x<y,$ then $x+z<y+z.$

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 $x−z<y−z.$

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

$x<y⇔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

$(y−z)−(x−z)>0$

$(y−z)−(x−z)>0⇕x−z<y−z $

Finally, because $x<y,$ the property has been proven. If $x<y,$ then $x−z<y−z.$

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 $z =0.$ Then, the equivalent inequalities can be written depending on the sign of $z.$

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

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 $xy>0.$
- 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 $xz<yz.$
- If $x<y$ and $z<0,$ then $xz>yz.$

Each conditional statement will be analyzed separately.

$x<y⇔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 $y−x>0z(y− andz>0⇓x)>0 $

Now, the second part of this conditional statement can be rewritten using the Distributive Property. $z(y−x)>0⇔zy−zx>0 $

From the first property, it can be said that $zy−zx>0$ if and only if $zx<zy.$ Additionally, because $x<y,$ the conditional statement has been proven. If $x<y$ and $z>0,$ then $zx<zy.$

$x<y⇔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

$(-a)b=-ab$

$-zy−(-zx)>0$

SubNeg

$a−(-b)=a+b$

$-zy+zx>0$

AddIneq

$LHS+zy>RHS+zy$

$zx>zy$

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

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 $z =0.$ Then, the equivalent inequalities can be written depending on the sign of $z.$

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

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

Each case will be analyzed separately.

$x<y⇔y−x>0 $

Furthermore, because $z>0,$ from the second property, it can be stated that $y−x$ divided by $z$ is also $y−x>0zy−x andz>0⇓>0 $

Now, the second part of this conditional statement can be rewritten. $zy−x >0⇔zy −zx >0 $

By using the first property, it can be said that $zx $ is If $x<y$ and $z>0,$ then $zx <zy .$

$x<y⇔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-zy−x and-z>0⇓>0 $

Now, the second part of this statement can be rewritten.
$-zy−x >0$

Simplify

MoveNegDenomToNum

Put minus sign in numerator

$z-(y−x) >0$

DistrNegSignSwap

$-(b−a)=a−b$

$zx−y >0$

WriteDiffFrac

Write as a difference of fractions

$zx −zy >0$

AddIneq

$LHS+zy >RHS+zy $

$zx >zy $

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