Properties of Inequalities

Reading and Combining Inequalities

In the previous tutorial, we treated the order as part of the structure of an ordered field and established how strict inequalities behave under translation and multiplication. This tutorial develops further properties that follow from those rules. These properties are useful in their own right, and they are essential when rearranging inequalities: the sign of the number used to multiply or divide both sides determines whether the comparison keeps its direction or reverses.

Recall that $a\leq b$ means $a<b$ or $a=b$. Thus, a non-strict inequality allows equality, while a strict inequality does not. In an ordered field $F$, trichotomy says exactly one of $a<b$, $a=b$, or $b<a$ holds. In particular, any two elements can be compared using $\leq$, and if both $a\leq b$ and $b\leq a$, then they must be equal.

Definition (Non-strict order). For $a,b\in F$, write $a\leq b$ if $a<b$ or $a=b$. Write $a\geq b$ if $b\leq a$. The symbols $<$ and $\leq$ therefore express different claims: $a<b$ excludes equality, whereas $a\leq b$ permits it.

The distinction matters in chains. If $a\leq b$ and $b\leq c$, then $a\leq c$. If at least one comparison in the chain is strict, then the resulting comparison is strict. For example, $a\leq b<c$ implies $a<c$, and $a<b\leq c$ also implies $a<c$. These facts let us combine bounds without accidentally losing information about whether equality is possible.

Theorem (Chaining strict and non-strict inequalities). Let $a,b,c\in F$.

If $a\leq b$ and $b\leq c$, then $a\leq c$.
If $a\leq b$ and $b<c$, or if $a<b$ and $b\leq c$, then $a<c$.
If $a\leq b$ and $b\leq a$, then $a=b$.

Proof. For the first claim, $a\leq b$ means either $a<b$ or $a=b$, and $b\leq c$ means either $b<c$ or $b=c$. If $a=b$, then $a\leq c$ follows directly from $b\leq c$. If $a<b$, then either $b=c$, giving $a<c$, or $b<c$, giving $a<c$ by transitivity of the strict order. In every case $a\leq c$.

For the second claim, suppose first that $a\leq b$ and $b<c$. If $a=b$, then $a<c$ follows by substitution. If $a<b$, transitivity of the strict order gives $a<c$. Now suppose $a<b$ and $b\leq c$. If $b=c$, substitution gives $a<c$; if $b<c$, strict transitivity gives $a<c$.

For the final claim, trichotomy gives exactly one of $a<b$, $a=b$, or $b<a$. The first alternative contradicts $b\leq a$, and the third contradicts $a\leq b$. Therefore $a=b$. $\square$

Worked Example: Combining Bounds

Suppose $x\leq 4$ and $4<9$. The mixed-chain property gives $x<9$. The conclusion is strict even though the first comparison permits equality: if $x=4$, then $x=4<9$, and if $x<4$, strict transitivity gives $x<9$.

By contrast, if $x\leq 4$ and $4\leq 9$, the chain only guarantees $x\leq 9$. The general conclusion is non-strict, although a particular value of $x$ might make the comparison strict.

Negating Both Sides Reverses the Order

Negation reverses an inequality. This can be seen from the translation rule in the previous tutorial: adding the same element to both sides preserves a strict inequality. If $a<b$, adding $-a-b$ to each side gives $-b<-a$. The same reversal holds for non-strict inequalities, including the equality case.

Theorem (Order reversal under negation). For $a,b\in F$, $$ a\leq b \quad\Longleftrightarrow\quad -b\leq -a. $$ In particular, $a<b$ if and only if $-b<-a$.

Proof. First suppose $a\leq b$. If $a=b$, then $-b=-a$, so $-b\leq -a$. If $a<b$, translation of $a<b$ by $-a-b$ gives

$$ a+(-a-b)<b+(-a-b), $$

which simplifies, using the field laws for additive inverses, to $-b<-a$. Thus $a\leq b$ implies $-b\leq -a$. Applying the same implication to $-b\leq -a$ gives $-(-a)\leq-(-b)$, or $a\leq b$, because $-(-x)=x$. This proves the equivalence. The strict version follows by the same translation argument in each direction. $\square$

Worked Example: Reversing a Comparison by Negation

Since $-8<-3$, order reversal under negation gives $3<8$. The calculation uses the original endpoints in reverse order:

-(-3)<-(-8) \qquad\Longleftrightarrow\qquad 3<8.

For a non-strict comparison, if $u\leq 6$, then negating both sides gives $-6\leq -u$. If $u=6$, both sides of the new comparison are equal; if $u<6$, the new comparison is strict.

A common error is to negate both sides while keeping the endpoints in the same order. From $a<b$, the correct result is $-b<-a$, not $-a<-b$. The direction reverses because $-b$ is the left-hand side after negation, and $-a$ is the right-hand side.

Multiplication and Division Depend on the Sign

The multiplication-and-order theorem from the previous tutorial states that multiplying a strict inequality by a positive element preserves its direction, while multiplying by a negative element reverses it. The same rules apply to non-strict inequalities: equality remains equality, while a strict comparison follows the corresponding strict rule. Division is multiplication by a reciprocal, so it follows the same sign test. In particular, division is only defined when the divisor is nonzero.

Theorem (Multiplying a non-strict inequality). Suppose $a\leq b$ and $c\in F$.

If $c>0$, then $ac\leq bc$.
If $c<0$, then $bc\leq ac$.
If $c=0$, then $ac=bc$.

If $c\ne0$, each implication can also be reversed: the resulting comparison determines the original one, with the same sign-dependent direction.

Proof. Suppose first that $c>0$. If $a=b$, then $ac=bc$. If $a<b$, the multiplication-and-order theorem gives $ac<bc$. Either way $ac\leq bc$. If $c<0$, equality again gives $ac=bc$; if $a<b$, the same theorem gives $bc<ac$. Thus $bc\leq ac$. If $c=0$, the field law $x0=0$ gives $ac=0=bc$.

Now suppose $c\ne0$. Its reciprocal $c^{-1}$ exists. If $c>0$, then $c^{-1}>0$, so applying the positive case to $ac\leq bc$ and multiplying by $c^{-1}$ recovers $a\leq b$. If $c<0$, then $c^{-1}<0$. Multiplying $bc\leq ac$ by $c^{-1}<0$ reverses the inequality, giving $(ac)c^{-1}\leq(bc)c^{-1}$, hence $a\leq b$. More directly, from $bc\leq ac$, multiplication by $c^{-1}<0$ gives

a=(ac)c^{-1}\leq (bc)c^{-1}=b.

So $a\leq b$. This establishes the reverse implication in the negative case as well. $\square$

The theorem also gives the usual division rules. When $c>0$, dividing $a\leq b$ by $c$ yields $a/c\leq b/c$. When $c<0$, it yields $b/c\leq a/c$. The condition $c\ne0$ is necessary: division by zero is undefined, and multiplying by zero erases the comparison because both products become zero.

Worked Example: Multiplying by a Negative Number

Start with $-2\leq 5$ and multiply both sides by $-3$. Since $-3<0$, the order reverses:

5(-3)\leq (-2)(-3), \qquad\text{so}\qquad -15\leq 6.

The calculation confirms the direction: $-15$ is less than $6$. Keeping the original direction would incorrectly claim $6\leq-15$.

Worked Example: Dividing by a Negative Number

Suppose $12< -4x$. Divide both sides by $-4$, which is negative, so the inequality reverses:

\frac{12}{-4}>\frac{-4x}{-4}, \qquad\text{hence}\qquad -3>x.

Equivalently, $x<-3$. To check the direction, take $x=-4$, which satisfies the conclusion: $12<-4(-4)=16$. At $x=-2$, the conclusion fails, and so does the original inequality, since $12<8$ is false.

Comparing Reciprocals

The previous tutorial established that if $0<x<y$, then $0<y^{-1}<x^{-1}$. This positive case does not extend as one universal rule to every pair of nonzero numbers. If two numbers have opposite signs, their reciprocals keep those signs, so the comparison between them is preserved. If they have the same sign, taking reciprocals reverses their order.

Theorem (Reciprocal comparisons by sign). Let $x<y$, where $x\ne0$ and $y\ne0$.

If $x$ and $y$ have the same sign, then $y^{-1}<x^{-1}$.
If $x<0<y$, then $x^{-1}<y^{-1}$.

Proof. If $0<x<y$, the first conclusion is the reciprocals-reverse-positive-order theorem from the previous tutorial. If $x<y<0$, then $0<-y<-x$: the last two inequalities follow by reversing $x<y$ under negation, and by $y<0$. Applying the positive reciprocal theorem to $0<-y<-x$ gives

(-x)^{-1}<(-y)^{-1}.

Since $(-x)^{-1}=-x^{-1}$ and $(-y)^{-1}=-y^{-1}$, negating this strict inequality reverses it and yields $y^{-1}<x^{-1}$. Finally, if $x<0<y$, then $x^{-1}<0$ and $0<y^{-1}$, because reciprocals have the same sign as their nonzero inputs. Transitivity gives $x^{-1}<y^{-1}$. $\square$

Worked Example: Reciprocals of Two Negative Numbers

Since $-6<-3<0$, the numbers have the same sign, so their reciprocals reverse order:

(-3)^{-1}<(-6)^{-1}, \qquad\text{that is}\qquad -\frac{1}{3}<-\frac{1}{6}.

The comparison is correct because $-1/3$ is more negative than $-1/6$. For opposite signs, $-2<4$ instead gives $(-2)^{-1}<4^{-1}$, or $-1/2<1/4$: the order is preserved.

A Reliable Way to Use Inequality Rules

When transforming an inequality, first identify the operation and the sign of any factor involved. Translation by the same element preserves an inequality, while multiplication by a positive factor preserves its direction and multiplication by a negative factor reverses it. Negating both sides is the special case of multiplying by $-1$. For reciprocals, check whether the numbers are positive, negative, or of opposite signs before deciding how their order changes.

These rules apply in any ordered field, not only in $\mathbb{R}$. They follow from the order and field properties developed earlier, so the same reasoning works whenever those properties are available. The essential caution is that the sign condition is part of the rule: without knowing whether a factor is positive, negative, or zero, one cannot determine the effect of multiplication on the comparison.

Key takeaway. Strict and non-strict inequalities can be combined using order transitivity. Negation reverses their direction; multiplication by a positive number preserves direction, and multiplication by a negative number reverses it. Division follows the same rules and requires a nonzero divisor.

Check Your Understanding

Use the order properties in this tutorial to answer the following questions.

If $p\leq q$ and $q<r$, what strict or non-strict comparison follows between $p$ and $r$?
If $m<n$, what inequality results after negating both sides?
What happens to $a\leq b$ when both sides are multiplied by a negative number?
Why is multiplying an inequality by zero not reversible?
If $x<y<0$, how do $x^{-1}$ and $y^{-1}$ compare?
When dividing an inequality by a number, what information must be checked before deciding whether its direction changes?

What You'll Learn