Properties of Inequalities: Rules, Examples & Solved Practice

An inequality compares the sizes of two numbers. The statement $a < b$ means exactly that the difference $b - a$ is positive, which on the number line says that $a$ lies to the left of $b$ . Everything else in this section follows from that one definition.

Quick Reference

Symbol or rule	Read as	Meaning
$a < b$	" $a$ is less than $b$ "	$b - a$ is positive; $a$ lies left of $b$
$a > b$	" $a$ is greater than $b$ "	Same as $b < a$
$a \leq b$	" $a$ is less than or equal to $b$ "	$a < b$ or $a = b$
$a < b \leq c$	" $b$ is between $a$ and $c$ "	$a < b$ and $b \leq c$
Trichotomy		If $a \neq b$ then $a < b$ or $a > b$
Transitivity		$a > b$ and $b > c$ give $a > c$
Add anything		$a > b$ gives $a + c > b + c$ for every $c$
Multiply by $c > 0$		$a > b$ gives $a c > b c$ ; direction kept
Multiply by $c < 0$		$a > b$ gives $a c < b c$ ; direction flipped
Reciprocals, same sign		$0 < a < b$ or $a < b < 0$ gives $\frac{1}{a} > \frac{1}{b}$
Squares		$a \neq 0$ gives $a^{2} > 0$

Given two numbers $a$ and $b$ , we write $a < b$ ( $a$ is less than $b$ ) or equivalently $b > a$ ( $b$ is greater than $a$ ) if $b - a$ is positive. Geometrically $a < b$ means $a$ lies to the left of $b$ on the number line (see the following figure).

The symbol $a \leq b$ means either $a < b$ or $a = b$ and
$a < b \leq c$ means $a < b$ and $b \leq c$ .

Illustration for Inequalities — a < b geometrically means a lies to the left of b on the number line.

The signs $<$ and $>$ are called inequality symbols and satisfy the following properties:

If 
  a
  ≠
  b
 then 
  a
  <
  b
 or 
  a
  >
  b
.
If 
  a
  >
  b
 and 
  b
  >
  c
 then 
  a
  >
  c
.
If 
  a
  >
  b
 then 
  a
  +
  c
  >
  b
  +
  c
 (and 
  a
  −
  c
  >
  b
  −
  c
) for every 
  c
 (if we add a positive or negative number to both sides of an inequality, the direction of the inequality will be preserved).
If 
  a
  >
  b
 and 
  c
  >
  d
, then 
  a
  +
  c
  >
  b
  +
  d
 (Inequalities with the same directions can be added).
If 
  a
  >
  b
 and 
  c
  >
  0
 then 
  a
  c
  >
  b
  c
 (If we multiply or divide both sides of an inequality by a positive number, the direction of the inequality will be preserved).
If 
  a
  >
  b
 and 
  c
  <
  0
 then 
  a
  c
  <
  b
  c
 (If we multiply or divide both sides of an inequality by a negative number, we need to reverse the inequality direction).
If 
  a
 and 
  b
 are both positive or both negative and 
  a
  <
  b
 then 
  
    
      1
      a
    
  
  >
  
    
      1
      b
    
  
.
If 
  a
  <
  0
  <
  b
, then 
  
    
      1
      a
    
  
  <
  0
  <
  
    
      1
      b
    
  
.
If 
  a
  ≠
  0
, 
  
    a
    
      2
    
  
  >
  0
.

The above properties remain true, if we replace $>$ by $\geq$ and $<$ by $\leq$ .

Exercises

Exercise 1.

Using the definition given at the start of this section, decide whether each statement is true or false by working out $b - a$ .

(a) $3 < 8$ (b) $- 5 < - 9$ (c) $- 4 < 0$ (d) $7 \leq 7$

Answer

(a) True (b) False (c) True (d) True

Solution

The definition says: $a < b$ means $b - a$ is positive. So in each part we subtract the left number from the right number and look at the sign of the result.

(a) $3 < 8$ . Here $a = 3$ and $b = 8$ , so

b - a = 8 - 3 = 5.

Since $5$ is positive, the statement $3 < 8$ is true. Geometrically, $3$ lies to the left of $8$ on the number line.

(b) $- 5 < - 9$ . Here $a = - 5$ and $b = - 9$ , so

b - a = - 9 - (- 5) = - 9 + 5 = - 4.

Since $- 4$ is negative, not positive, the statement is false. In fact $- 5 > - 9$ .

This is the part where mistakes happen. Because $9$ is bigger than $5$ , it is tempting to say $- 9$ is bigger than $- 5$ . The definition settles it without any guesswork. On the number line, $- 9$ lies further to the left than $- 5$ , and further left means smaller.

Note also the double negative: subtracting $- 5$ is the same as adding $5$ .

(c) $- 4 < 0$ . Here $a = - 4$ and $b = 0$ , so

b - a = 0 - (- 4) = 0 + 4 = 4.

Since $4$ is positive, the statement is true. Every negative number is less than $0$ .

(d) $7 \leq 7$ . The section states that $a \leq b$ means either $a < b$ or $a = b$ . Only one of the two needs to hold. Here $7 < 7$ is false, since $7 - 7 = 0$ is not positive, but $7 = 7$ is certainly true. Because the two options are joined by "or", the statement $7 \leq 7$ is true.

A statement like $7 \leq 7$ often looks wrong to students at first. Remember that $\leq$ is asking a generous question — it is satisfied by equality just as happily as by strict inequality.

Exercise 2.

Suppose $x > 5$ . Use the properties of this section to complete each statement, and name the property you used.

(a) $x + 3 > \underset{―}{}$ (b) $x - 2 > \underset{―}{}$ (c) $4 x > \underset{―}{}$ (d) $- 4 x < \underset{―}{}$

Answer

(a) $8$ (b) $3$ (c) $20$ (d) $- 20$

Solution

We start from $x > 5$ each time and apply one property.

(a) Property 3 says: if $a > b$ then $a + c > b + c$ for every $c$ . Adding the same number to both sides keeps the direction of the inequality. Take $c = 3$ :

x + 3 > 5 + 3,

x + 3 > 8.

(b) Property 3 again — it covers subtraction too, since it states that $a - c > b - c$ as well. Take $c = 2$ :

x - 2 > 5 - 2,

x - 2 > 3.

(c) Property 5 says: if $a > b$ and $c > 0$ then $a c > b c$ . Multiplying both sides by a positive number keeps the direction. Here $c = 4$ , which is positive, so:

4 x > 4 \cdot 5,

4 x > 20.

(d) Property 6 says: if $a > b$ and $c < 0$ then $a c < b c$ . Multiplying both sides by a **negative** number reverses the direction. Here $c = - 4$ , which is negative, so the $>$ must flip to $<$ :

- 4 x < - 4 \cdot 5,

- 4 x < - 20.

Part (d) is the one to watch, and it is the single most common error in all work with inequalities. Compare (c) and (d): the only difference is the sign of the multiplier, and that sign decides whether the inequality symbol keeps its direction or turns around.

A quick sanity check on (d): $x > 5$ , so $x$ could be $6$ . Then $- 4 x = - 24$ , and indeed $- 24 < - 20$ . Had we carelessly written $- 4 x > - 20$ , the value $x = 6$ would have given $- 24 > - 20$ , which is false.

Exercise 3.

Given that $a > b$ and $c > d$ , explain why $a + c > b + d$ , and check the conclusion with $a = 9$ , $b = 4$ , $c = 6$ , $d = 1$ .

Solution

The explanation. This is property 4 of the section: inequalities pointing in the same direction may be added.

It is worth seeing why it is believable. We are told two things: $a$ beats $b$ , and $c$ beats $d$ . If we form the sum $a + c$ , then each piece of it is at least as good as the matching piece of $b + d$ — the first piece is bigger and the second piece is bigger too. So the total must be bigger.

We can also assemble it from property 3, which lets us add the same quantity to both sides of one inequality.

Start with $a > b$ . Add $c$ to both sides, using property 3 with the number $c$ :

a + c > b + c .

Now start with $c > d$ . Add $b$ to both sides, again by property 3:

c + b > d + b,

which we may write as

b + c > b + d,

using the commutative property of addition to reorder each side.

So far we have

a + c > b + c and b + c > b + d .

Property 2 says that if $a > b$ and $b > c$ then $a > c$ — the inequalities may be chained. Chaining our two statements through the common middle quantity $b + c$ gives

a + c > b + d,

which is what we wanted.

The check. Put $a = 9$ , $b = 4$ , $c = 6$ , $d = 1$ . First confirm the hypotheses: $9 > 4$ is true, and $6 > 1$ is true, so the property applies. Now

a + c = 9 + 6 = 15, b + d = 4 + 1 = 5,

and $15 > 5$ . The conclusion holds.

A warning about what this property does not say. Inequalities pointing the same way may be added, but they may not in general be subtracted. With the same numbers, $a - c = 9 - 6 = 3$ and $b - d = 4 - 1 = 3$ , so $a - c > b - d$ is false here — the two are equal. Property 4 is about addition only.

Exercise 4.

Property 7 says that if $a$ and $b$ are both positive or both negative and $a < b$ , then $\frac{1}{a} > \frac{1}{b}$ .

(a) Check this with $a = 2$ and $b = 5$ . (b) Check it with $a = - 5$ and $b = - 2$ .

Answer

(a) $\frac{1}{2} > \frac{1}{5}$ (b) $- \frac{1}{5} > - \frac{1}{2}$

Solution

Before applying any property, check that its conditions are met. Property 7 needs two things: that $a$ and $b$ have the same sign, and that $a < b$ .

Part (a): $a = 2$ , $b = 5$ .

Conditions: both numbers are positive, so the sign condition holds. And $a < b$ means $2 < 5$ , which is true since $5 - 2 = 3$ is positive. So the property applies.

It predicts $\frac{1}{a} > \frac{1}{b}$ , that is,

\frac{1}{2} > \frac{1}{5} .

To confirm, write both as decimals: $\frac{1}{2} = 0.5$ and $\frac{1}{5} = 0.2$ , and $0.5 > 0.2$ . The prediction is correct.

Notice what happened: the inequality reversed when we took reciprocals. This makes sense — cutting a cake into $5$ pieces gives smaller pieces than cutting it into $2$ .

Part (b): $a = - 5$ , $b = - 2$ .

Conditions: both numbers are negative, so the sign condition holds. And $a < b$ means $- 5 < - 2$ , which is true since $- 2 - (- 5) = - 2 + 5 = 3$ is positive. So the property applies again.

It predicts $\frac{1}{a} > \frac{1}{b}$ , that is,

\frac{1}{- 5} > \frac{1}{- 2},

or in tidier form

- \frac{1}{5} > - \frac{1}{2} .

To confirm: $- \frac{1}{5} = - 0.2$ and $- \frac{1}{2} = - 0.5$ , and $- 0.2 > - 0.5$ because $- 0.2 - (- 0.5) = 0.3$ is positive. The prediction is correct.

Why the same-sign condition matters. Suppose we ignored it and took $a = - 2$ , $b = 3$ . Then $a < b$ holds, but the numbers have opposite signs. The property would predict $- \frac{1}{2} > \frac{1}{3}$ , which is plainly false, since a negative number cannot exceed a positive one. Property 8 is the one that covers this mixed case: when $a < 0 < b$ , it tells us $\frac{1}{a} < 0 < \frac{1}{b}$ .

Exercise 5.

Two friends are comparing bank balances. Ana's balance is $a$ dollars and Bruno's is $b$ dollars, and Ana has less money than Bruno.

(a) Write this as an inequality. (b) Each is charged the same $$ 15$ monthly fee. Write an inequality comparing their new balances and name the property used. (c) Instead, each balance is tripled. Compare the new balances and name the property used.

Answer

(a) $a < b$ (b) $a - 15 < b - 15$ , by property 3 (c) $3 a < 3 b$ , by property 5

Solution

(a) "Ana has less money than Bruno" says Ana's amount is the smaller one, so

a < b .

(b) Both balances drop by the same amount, $15$ dollars, so the new balances are $a - 15$ and $b - 15$ .

Property 3 says that if $a > b$ then $a - c > b - c$ for every $c$ — subtracting the same number from both sides leaves the direction of the inequality alone. Writing our inequality in the form $b > a$ and subtracting $15$ from both sides gives $b - 15 > a - 15$ , that is,

a - 15 < b - 15.

So Ana still has less than Bruno. That fits common sense: charging both people the same fee does not change who is ahead. Note that the gap between them is also unchanged.

(c) Now the new balances are $3 a$ and $3 b$ .

Property 5 says that if $a > b$ and $c > 0$ then $a c > b c$ — multiplying both sides by a positive number leaves the direction alone. Here the multiplier is $3$ , which is positive, so

3 a < 3 b .

Ana still has less. Here, unlike part (b), the gap between them has grown: it is now three times what it was.

A remark on why property 6 does not appear. Property 6 would reverse the inequality, but it applies only when the multiplier is negative. Multiplying a balance by $- 4$ has no sensible meaning here, so that case does not arise. In real problems the multiplier is usually a positive quantity such as a rate, a count, or a scale factor, which is why the reversal rule catches people by surprise when it finally does turn up.

Exercise 6.

A student is told that $- 3 x > 12$ and concludes that $x > - 4$ . Find the error, and give the correct conclusion.

Answer

The correct conclusion is $x < - 4$ .

Solution

Finding the error. To get from $- 3 x > 12$ to a statement about $x$ , both sides must be divided by $- 3$ . The student did the arithmetic correctly — $12$ divided by $- 3$ really is $- 4$ — but kept the symbol $>$ pointing the same way.

That is the mistake. Property 6 of this section says: if $a > b$ and $c < 0$ , then $a c < b c$ . Because $- 3$ is negative, dividing by it reverses the direction of the inequality. The student used the rule for positive multipliers, property 5, in a situation where property 6 applies.

The correct conclusion. Starting from

- 3 x > 12

and dividing both sides by $- 3$ , a negative number, we must flip $>$ to $<$ :

x < - 4.

Checking that this is right. The value of a check here is that it settles the matter beyond argument. Try a number that satisfies our answer, say $x = - 10$ , which does satisfy $x < - 4$ :

- 3 \cdot (- 10) = 30,

and $30 > 12$ is true. Good — the original inequality holds.

Now try a number that satisfies the student's answer but not ours, say $x = 0$ , which satisfies $x > - 4$ :

- 3 \cdot 0 = 0,

and $0 > 12$ is false. So the student's conclusion admits values that do not work, confirming it is wrong.

How to avoid this in future. Every time you multiply or divide an inequality, stop and ask one question: is the number I am multiplying or dividing by positive or negative? If positive, the symbol stays as it is; if negative, it turns around. Adding and subtracting, by contrast, never require a flip — property 3 has no sign condition attached to $c$ at all.

Exercise 7.

Is the following statement always true, sometimes true, or never true? "If $a > b$ , then $a^{2} > b^{2}$ ." Justify your answer.

Answer

Sometimes true.

Solution

The answer is sometimes. We need one example where the conclusion holds and one where it fails.

A case where it holds. Take $a = 5$ and $b = 3$ . The hypothesis is satisfied, since $5 > 3$ . And

a^{2} = 25, b^{2} = 9,

so $a^{2} > b^{2}$ is true here.

A case where it fails. Take $a = 1$ and $b = - 6$ . The hypothesis is still satisfied, since $1 > - 6$ : indeed $1 - (- 6) = 7$ is positive. But now

a^{2} = 1, b^{2} = 36,

and $1 > 36$ is false. So the conclusion fails.

Since both outcomes occur, the statement is sometimes true.

Why it breaks down. Squaring means multiplying a number by itself, and property 5 only lets us multiply both sides of an inequality by a positive number. Going from $a > b$ to $a^{2} > b^{2}$ would require multiplying the left side by $a$ and the right side by $b$ — two different multipliers, and possibly negative ones at that. No property in this section permits that.

Property 9 is the relevant warning: it says $a^{2} > 0$ whenever $a \neq 0$ . Squaring destroys the information about sign, turning every non-zero number, however negative, into a positive one. That is exactly what went wrong in the second example, where the very negative $b = - 6$ produced the large positive $b^{2} = 36$ .

When is it safe? If we also know that $b > 0$ — so that $a$ and $b$ are both positive — the statement does become true. In the first example both numbers were positive and it worked. The moral is to check for hidden sign conditions before applying a step to an inequality.

Exercise 8.

Suppose $a > b$ . Prove that $- a < - b$ , using only the properties listed in this section. Then explain the result in terms of the number line.

Solution

The proof. We are given

a > b,

and we want to reach $- a < - b$ .

Multiplying by $- 1$ is what turns $a$ into $- a$ and $b$ into $- b$ . Property 6 covers exactly this: if $a > b$ and $c < 0$ , then $a c < b c$ .

Check the condition first. We are taking $c = - 1$ , and $- 1 < 0$ , so $c$ is negative and property 6 applies.

Applying it with $c = - 1$ :

a \cdot (- 1) < b \cdot (- 1) .

Writing each side more simply,

- a < - b,

which is what we wanted to prove. Notice that the inequality symbol turned around, from $>$ in the hypothesis to $<$ in the conclusion; that reversal is the whole content of property 6.

A check with numbers. Take $a = 8$ and $b = 3$ , so the hypothesis $8 > 3$ holds. The conclusion claims $- 8 < - 3$ . Testing it against the definition, $- 3 - (- 8) = - 3 + 8 = 5$ , which is positive, so $- 8 < - 3$ is indeed true.

The number line picture. The section states that $a < b$ means $a$ lies to the left of $b$ . Attaching a minus sign to a number reflects its point across the origin: whatever is a certain distance to the right of $0$ moves to the same distance to the left, and the other way round.

Reflection turns the line back to front. So if $a$ started out to the right of $b$ — which is what $a > b$ says — then after reflecting, $- a$ ends up to the left of $- b$ . And lying to the left is exactly $- a < - b$ .

This picture is a good one to keep in mind, because it explains in a single image why so many inequality steps involving negatives require a flip. It is also worth noting what the reflection does not change: the distance between the two points stays the same, which is why $a - b$ and $(- b) - (- a)$ come to the same number.

Frequently Asked Questions

What does

a < b

actually mean?

It means the difference $b - a$ is a positive number. That single condition is the definition, and every property in this section is derived from it. Geometrically it says $a$ sits to the left of $b$ on the number line. To test a claim such as $- 5 < - 9$ , do not rely on the feel of the numbers; compute $b - a = - 9 - (- 5) = - 4$ , which is negative, so the claim is false.

What is the difference between

<

and

\leq

The strict symbol $<$ excludes equality, while $\leq$ allows it: $a \leq b$ is true when $a < b$ or when $a = b$ . So $7 < 7$ is false but $7 \leq 7$ is true. All nine properties listed in this section stay valid when $<$ and $>$ are replaced throughout by $\leq$ and $\geq$ , which is why the two families of statements are usually proved only once.

Why does multiplying by a negative number flip the inequality?

Multiplying by a negative number reflects the number line across the origin, and reflection reverses left and right. Whatever was farther to the right ends up farther to the left. In symbols, if $a > b$ then $a - b$ is positive, and multiplying a positive number by a negative $c$ gives a negative result $c (a - b) = c a - c b$ , so $c a < c b$ . Dividing by a negative number is multiplying by its reciprocal, which is also negative, so division flips the sign for the same reason.

Can two inequalities be added? Can they be multiplied or subtracted?

Adding is safe when the directions match: from

a > b and c > d

it follows that

a + c > b + d .

Subtracting in the same way is not safe. From $5 > 2$ and $10 > 1$ subtracting gives $5 - 10 > 2 - 1$ , that is $- 5 > 1$ , which is false.

Multiplying two inequalities is safe only when all four quantities are positive; with negatives allowed, $- 1 > - 2$ and $- 1 > - 3$ multiply to the false statement $1 > 6$ .

Why does taking reciprocals reverse an inequality?

Because on the positive side of the line, the larger a number is, the smaller a share of $1$ it represents. If $0 < a < b$ then $\frac{1}{a} > \frac{1}{b}$ ; for instance $2 < 5$ but $\frac{1}{2} > \frac{1}{5}$ . The same reversal holds when both numbers are negative, as in $- 5 < - 2$ giving $- \frac{1}{5} > - \frac{1}{2}$ .

The rule needs $a$ and $b$ to have the same sign. If $a < 0 < b$ , then $\frac{1}{a}$ is negative and $\frac{1}{b}$ is positive, so the reciprocals keep the original order instead of reversing it.

Why is

a^{2} > 0

for every nonzero

a

If $a > 0$ , multiplying the inequality $a > 0$ by the positive number $a$ preserves the direction and gives $a^{2} > 0$ . If $a < 0$ , multiplying $a < 0$ by the negative number $a$ flips the direction and again gives $a^{2} > 0$ . Both cases land in the same place, so a square is never negative, and it is zero only when $a = 0$ . This fact is the engine behind many later inequalities, including the proof that $x + \frac{1}{x} \geq 2$ for positive $x$ .

What is the most common mistake when solving an inequality?

Forgetting to flip the sign after multiplying or dividing by a negative quantity. A close second is multiplying both sides by a variable expression such as $x$ or $x - 3$ , whose sign is unknown. Doing so silently assumes the expression is positive, and the answer is wrong whenever it is not. The safe approach is to move everything to one side and analyze the sign of the resulting expression, rather than clearing denominators the way one would in an equation.

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