Inequality Basics

Learning goals

  • Compare two quantities with <<, >>, ≤\le or ≥\ge
  • Describe a solution set as a range of values
  • Show a solution in words, on a line, and as an interval
  • Bracket an included endpoint and parenthesize an excluded one
  • Reverse the direction when scaling by a negative number
  • Read a compound a<x<ba < x < b as a band between bounds

The four inequality symbols

An inequality compares two quantities that need not be equal. In place of the equals sign it uses one of four symbols.

SymbolRead asMeaning
<<”less than”strictly smaller, not equal
>>”greater than”strictly larger, not equal
≤\le”less than or equal to”, “at most”smaller, or exactly equal
≥\ge”greater than or equal to”, “at least”larger, or exactly equal

So 5<85 < 8 says five is less than eight, and 8>58 > 5 says eight is greater than five. Those are the same fact read from opposite ends. The symbol always opens toward the larger amount and narrows to a point at the smaller one. So the pointed end aims at the smaller number: in 5<85 < 8 the point sits on the 55.

That two-ended reading is worth stating as a rule, because it lets you turn any inequality around. Take 7>x7 > x: swap the two sides and reverse the symbol, and you get x<7x < 7, the same fact read from the other end. You will use this often to move a variable to the left. In general, the statements a≤ba \le b and b≥ab \ge a say exactly the same thing: ”aa is at most bb” is the same claim as ”bb is at least aa.”

The two symbols with a bar underneath, ≤\le and ≥\ge, allow equality as well as the strict comparison. The statement x≤6x \le 6 is true when xx is 66 and also true when xx is anything below 66. We call << and >> strict inequalities, because they exclude the boundary, and ≤\le and ≥\ge inclusive (or non-strict), because they let it in. That one bar is the whole difference between admitting the boundary number and shutting it out.

Every number that makes an inequality true is part of its solution set. Ask which numbers satisfy x>3x > 3: the value 44 works, and so do 55, 3.13.1, 3.00013.0001, and a million. Every number to the right of 33 qualifies, so this solution set holds infinitely many numbers. That is why we describe the answer as a range and draw it, rather than listing it.

Worked example 1 Turn each phrase into an inequality

Let nn be the number in question. Match the wording to the symbol, watching for whether the boundary is allowed.

“A rider must be at least 4848 inches tall” allows exactly 4848 and anything taller, so it includes the boundary:

n≥48.n \ge 48.

“You may spend at most 2020 dollars” allows exactly 2020 and anything less, again including the boundary:

n≤20.n \le 20.

“There are more than 22 players” excludes 22 itself, since “more than” is strict:

n>2.n > 2.

Finally, rewrite 15≥n15 \ge n with the variable on the left. Swap the sides and reverse the symbol:

15≥n⟺n≤15.15 \ge n \quad\Longleftrightarrow\quad n \le 15.

Both forms say ”nn is at most 1515.”

Picturing the solution on a number line

A number line turns a solution set into a picture you can take in at a glance. Locate the boundary number from the inequality, decide whether the boundary itself belongs, and shade the direction that holds the rest of the solutions. Three decisions capture every case.

The circle at the boundary records whether the boundary is a solution. Draw an open (hollow) circle for a strict inequality, << or >>, because the boundary is excluded. Draw a filled (closed) circle for an inclusive inequality, ≤\le or ≥\ge, because the boundary is included. The shaded ray then runs in the direction of the solutions: to the right for a “greater” relation (>> or ≥\ge), toward the larger numbers. The ray runs to the left for a “less” relation (<< or ≤\le). The arrowhead means the solutions run on forever that way.

Here is the graph of x>2x > 2. The circle on 22 is open, because 22 is not a solution, and the shading runs right across every number larger than 22.

Number-line graph of x greater than 2A number line with an open circle at 2 and shading extending right, representing all numbers greater than 2.-10123456
The graph of x > 2. The open circle on 2 marks it as excluded, and the shaded ray runs right to cover every number greater than 2.

The only thing that changes for x≥2x \ge 2 is the circle: it becomes filled, because now 22 belongs to the solution. Switching between a strict and an inclusive relation never moves the shading, it only opens or closes the boundary circle.

That last sentence is the one to test rather than take on trust, and the figure below is built for it. Its three controls are exactly the three decisions described above. “Endpoint” sets where the boundary sits, “Side” picks which way to shade, and “Endpoint is” switches the boundary in or out.

Set “Endpoint” to 22 and “Side” to right, and you have the graph of x>2x > 2 from the picture above. Now switch “Endpoint is” between out and in, and watch the whole figure. The ray does not move a pixel, the sign in the readout changes from >> to ≥\ge, and the circle fills in. One decision, one thing on the screen, and it is the decision students most often lose. Then set “Endpoint” to −1-1, “Side” to left, and “Endpoint is” to in: that is the graph the next worked example asks for before you read it.

The three decisions behind a solution graph

x > 2. Every number to the right of 2 is a solution. 2 itself is not, so its circle is hollow: the sign is strict. A number line from -6 to 6. A thick ray covers every number on one side of a marked endpoint, and the circle at that endpoint is filled when it is part of the solution and hollow when it is not. Use the controls below the figure to move the endpoint, switch which side is covered, or switch between a strict sign and one that allows equality. -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 2
Endpoint Side Endpoint is

x > 2. Every number to the right of 2 is a solution. 2 itself is not, so its circle is hollow: the sign is strict.

A number line carrying the solution set of an inequality. The circle at the boundary is hollow when the boundary is excluded and filled when it is included, and the shaded ray runs toward the numbers that solve the inequality.

Worked example 2 Graph x≤−1x \le -1 and describe its parts

The symbol is ≤\le, which is inclusive, so the boundary −1-1 is a solution and gets a filled circle. It is a “less” relation, so the solutions are the numbers below −1-1, which lie to the left. Shade left from the filled circle.

Number-line graph of x less than or equal to negative 1A number line with a filled circle at negative 1 and shading extending left, representing negative 1 and all numbers below it.-5-4-3-2-1012
The graph of x is less than or equal to negative 1. The filled circle on negative 1 includes it, and the shading runs left over every number below negative 1.

So the picture is a filled dot on −1-1 with a ray heading left. In words it reads ”xx is at most −1-1,” and the boundary −1-1 is part of the answer.

Check your understanding

Which number line shows the solution of x<3x < 3?

Answer choices

Interval notation

Drawing a picture every time is slow, so there is a compact written shorthand for the same solution set, called interval notation. An interval is written as its left and right bounds inside a pair of fences, with the kind of fence recording whether each bound is included.

A square bracket includes the endpoint, matching a filled circle and an inclusive symbol. A parenthesis excludes the endpoint, matching an open circle and a strict symbol. So 2≤x≤52 \le x \le 5 becomes [2,5][2, 5], while 2<x<52 < x < 5 becomes (2,5)(2, 5). The two can be mixed: 2≤x<52 \le x < 5 is the half-open interval [2,5)[2, 5), closed on the left and open on the right.

When the solution runs on forever in one direction, we mark that end with the infinity symbol, ∞\infty for the right and −∞-\infty for the left. Here is the one firm rule: infinity always takes a parenthesis, never a bracket. Infinity is not a number, so there is no endpoint to reach or to include; the parenthesis records that the interval is open and unbounded. That is why x>2x > 2 is written (2,∞)(2, \infty) and x≤−1x \le -1 is written (−∞,−1](-\infty, -1], with a bracket on the real endpoint and a parenthesis at infinity. The table lines up all three ways of saying the same thing.

InequalityIn wordsNumber lineInterval
x>2x > 2greater than 22open circle at 22, shade right(2,∞)(2, \infty)
x≥2x \ge 2at least 22filled circle at 22, shade right[2,∞)[2, \infty)
x<2x < 2less than 22open circle at 22, shade left(−∞,2)(-\infty, 2)
x≤2x \le 2at most 22filled circle at 22, shade left(−∞,2](-\infty, 2]
−1<x≤3-1 < x \le 3above −1-1 and at most 33open at −1-1, filled at 33, shade between(−1,3](-1, 3]

One caution about reading. Written on its own, a symbol pair like (2,5)(2, 5) could be an interval on a line or a point (x,y)(x, y) in the plane from the last chapter. The surrounding words tell you which is meant: an interval names a set of numbers on one axis, while a coordinate point names one location in the plane.

Worked example 3 Move between symbols, number line, and interval

Take the inequality x≥−2x \ge -2. It is inclusive, so the boundary −2-2 belongs, and it is a “greater” relation, so the solutions run right. On a number line that is a filled circle on −2-2 with a ray to the right. In interval notation the real endpoint −2-2 is included, so it takes a bracket, and the right end runs to infinity, which always takes a parenthesis:

x≥−2⟺[−2,∞).x \ge -2 \quad\Longleftrightarrow\quad [-2, \infty).

Now run the translation the other way, starting from the interval (−∞,4)(-\infty, 4). The left end is −∞-\infty, so the set is unbounded below, and the right endpoint 44 carries a parenthesis, so 44 is excluded. That is exactly the strict “less” relation

(−∞,4)⟺x<4,(-\infty, 4) \quad\Longleftrightarrow\quad x < 4,

an open circle on 44 with the shading running left. Reading the fences carefully is the whole skill: a bracket is a filled circle, a parenthesis is an open circle, and an infinity end is always open.

Check your understanding

Which interval represents x≤5x \le 5?

Answer choices

The rules for changing an inequality

To rearrange an inequality without breaking it, you need to know which moves keep the statement true. Three of the rules match your equation habits exactly, and one carries a twist. In each case the reason is visible on the number line, where a<ba < b simply means aa sits to the left of bb.

Adding or subtracting the same quantity from both sides keeps the direction. Sliding both numbers the same distance in the same direction does not change which one is on the left. Starting from the true statement 3<73 < 7 and adding 44 to each side gives 7<117 < 11, still true, and subtracting 1010 from each side of 3<73 < 7 gives −7<−3-7 < -3, also true. The order survives every shift.

Multiplying or dividing both sides by a positive number keeps the direction. Scaling by a positive factor stretches or shrinks every distance from zero but never carries a number across zero, so the left-right order holds. From 3<73 < 7, multiplying both sides by 22 gives 6<146 < 14, and dividing 6<146 < 14 by 22 returns 3<73 < 7. Both stay true.

Multiplying or dividing both sides by a negative number reverses the direction. This is the twist, and it is the single most error-prone fact about inequalities, so it deserves a careful look. Take the true statement 3<53 < 5 and multiply both sides by −1-1. The left becomes −3-3 and the right becomes −5-5. If the symbol stayed put we would be claiming −3<−5-3 < -5, but that is false. On the number line −5-5 lies to the left of −3-3, so −3-3 is the larger number. The true statement is −3>−5-3 > -5, with the symbol flipped.

Multiplying an inequality by a negative reverses the orderA number line showing 3 and 5 and their reflections across zero at negative 3 and negative 5, where the left-right order is swapped.-5-4-3-2-10123453 < 5-3 > -5reflecting across 0 reverses the order
Multiplying by -1 reflects each number across zero. 3 sits left of 5, but their opposites -3 and -5 land in swapped order, with -3 now to the right of -5, so the direction reverses.

Why multiplying by a negative reverses the inequality#

Try it with numbers first: 4<104 < 10, and multiply both sides by the positive number 22. The difference 10−4=610 - 4 = 6 is positive. Multiplying that difference by 22 gives 2×6=122 \times 6 = 12, still positive, and 2×62 \times 6 is the same amount as 2×10−2×42 \times 10 - 2 \times 4, so

20−8=12.20 - 8 = 12.

So 2020 is larger than 88: multiplying 4<104 < 10 by the positive number 22 keeps the order, giving 8<208 < 20. Now negate each side: 88 becomes −8-8 and 2020 becomes −20-20, and −8>−20-8 > -20. Multiplying 4<104 < 10 by the negative number −2-2 gave −8>−20-8 > -20, with the symbol reversed.

Nothing above depended on the numbers 44, 1010, and 22. The same three moves work for any a<ba < b and any positive kk. The difference b−ab - a is positive, so multiplying it by kk gives k(b−a)>0k(b - a) > 0, which distributes to

kb−ka>0.kb - ka > 0.

So kbkb is larger than kaka: multiplying a<ba < b by the positive number kk keeps the order, exactly as the earlier rule said. Now take the opposite of each side. Taking opposites sends kaka to −ka-ka and kbkb to −kb-kb, and the positive difference kb−kakb - ka equals (−ka)−(−kb)(-ka) - (-kb). So (−ka)−(−kb)(-ka) - (-kb) is positive too, which means

−ka>−kb.-ka > -kb.

Multiplying a<ba < b by the negative number −k-k gives −ka>−kb-ka > -kb, with the symbol reversed. The same argument works with ≤\le in place of <<, giving ≥\ge: the boundary case a=ba = b turns into −ka=−kb-ka = -kb either way.

Dividing by a negative number reverses the direction for the same reason. Dividing by −k-k is the same as multiplying by the positive number 1k\tfrac{1}{k} and then by −1-1. Only that last step flips anything.

Transitivity chains comparisons together. If a<ba < b and b<cb < c, then a<ca < c. On the line, aa is left of bb and bb is left of cc, so aa must be left of cc. A quick check: since 2<52 < 5 and 5<95 < 9, transitivity gives 2<92 < 9, which is plainly true. You will use this soon to check whether a chain like a<x<ba < x < b can describe any numbers at all.

These rules are enough to take a short step on an inequality, the same way you take one step on an equation. For x+3<7x + 3 < 7, subtract 33 from both sides, and since subtraction never disturbs the symbol, x<4x < 4. For −2x<6-2x < 6, divide both sides by −2-2, and because −2-2 is negative you reverse the symbol, giving x>−3x > -3. Carrying such a problem through several steps in a reliable order is the job of the next lesson. Here the point is only that each individual move is governed by the rules above.

Worked example 4 Apply one rule to an inequality

Start with −13x≤2-\tfrac{1}{3}x \le 2. The variable is multiplied by the negative number −13-\tfrac{1}{3}, so to isolate xx you multiply both sides by −3-3. Multiplying by a negative reverses the direction, so ≤\le turns into ≥\ge:

−3⋅(−13x)≥−3⋅2⟹x≥−6.-3 \cdot \left(-\tfrac{1}{3}x\right) \ge -3 \cdot 2 \quad\Longrightarrow\quad x \ge -6.

The flip is the entire lesson of this step. To confirm the direction, test one value against the original inequality. Try x=0x = 0, which should satisfy x≥−6x \ge -6: the original gives −13(0)=0-\tfrac{1}{3}(0) = 0, and 0≤20 \le 2 is true. Try x=−9x = -9, which should fail x≥−6x \ge -6: the original gives −13(−9)=3-\tfrac{1}{3}(-9) = 3, and 3≤23 \le 2 is false, exactly as predicted. The solution set is [−6,∞)[-6, \infty).

Check your understanding

You multiply both sides of the true statement −4<1-4 < 1 by −2-2. Which statement is correct?

Answer choices

Compound inequalities

Sometimes a value is squeezed from both sides at once, larger than one number and smaller than another. A compound inequality writes both bounds in a single line. The notation a<x<ba < x < b means x>ax > a and x<bx < b at the same time, so xx lies strictly between aa and bb. Read it as two claims joined by “and”: xx is above aa, and xx is below bb.

Both symbols in the chain point the same way: never one “less than” symbol and one “greater than” symbol. Use << or ≤\le throughout, or use >> or ≥\ge throughout; that is the syntax. But writing it correctly is not enough to guarantee it describes any numbers. If x>ax > a and x<bx < b are both true, transitivity forces a<ba < b: the left bound must be smaller than the right bound. We write −2<x≤3-2 < x \le 3, which reads left to right as ”−2-2 is less than xx, and xx is at most 33,” and it makes sense because −2<3-2 < 3. Now look at 3<x<−23 < x < -2: both symbols point the same way, so the syntax looks fine, but the numbers do not cooperate. It asks for a number above 33 and below −2-2, and since 33 is not less than −2-2, nothing satisfies both parts at once. Check the numbers, not just the arrows, before trusting a compound inequality.

Each end keeps its own circle: an open circle where the bound is strict and a filled circle where it is inclusive. On the number line the solution is the band between the two boundaries rather than a ray running off to one side. In interval notation the same band is just the pair of bounds in one set of fences, each fence chosen by its own symbol. So −2<x≤3-2 < x \le 3, open on the left and closed on the right, is the half-open interval (−2,3](-2, 3].

Number-line graph of a compound inequality between negative 2 and 3A number line with an open circle at negative 2, a filled circle at 3, and shading covering the numbers between them.-3-2-101234
The graph of negative 2 < x is less than or equal to 3. An open circle on negative 2 and a filled circle on 3 bound a shaded band covering the numbers in between, matching the interval (-2, 3].

Worked example 5 A compound inequality in three forms

Read the statement ”xx is at least −1-1 and less than 44.” “At least −1-1” is inclusive, so the left bound −1-1 is included; “less than 44” is strict, so the right bound 44 is excluded. Written as one line with the smaller bound on the left, that is

−1≤x<4.-1 \le x < 4.

On the number line this is a filled circle on −1-1, an open circle on 44, and a shaded band between them. In interval notation the left endpoint is included, so it takes a bracket, and the right endpoint is excluded, so it takes a parenthesis:

−1≤x<4⟺[−1,4).-1 \le x < 4 \quad\Longleftrightarrow\quad [-1, 4).

Test a value to make sure both parts hold. Is x=−1x = -1 a solution? The left part −1≤−1-1 \le -1 is true and the right part −1<4-1 < 4 is true, so −1-1 is in, matching the filled circle. Is x=4x = 4 a solution? The right part 4<44 < 4 is false, so 44 is out, matching the open circle. The value x=0x = 0 sits comfortably inside, since −1≤0-1 \le 0 and 0<40 < 4 both hold.

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Core practice

Practice problems at the level of the course, to be worked out on paper. Hints one at a time, then the answer or the full worked solution, with your progress kept in this browser.

Core practice Work it out on paper 10 problems Start →
More practice (optional)

Extra sets, as hard as the Challenge set. Each one opens on its own page.

More resources (optional)

Other explanations of this lesson, if you want a second take.

A bit of history (optional)

Look again at the four signs in this lesson. Two of them are the other two with a line drawn underneath.

That line has a history of its own. The plain signs << and >> reached print in 16311631. They appear in an algebra book of Thomas Harriot’s, published ten years after the English mathematician died. For most of the century that followed, nobody had a settled mark for “smaller, or else exactly equal”. A writer set down the comparison, then explained the exception in ordinary words.

The mathematician John Wallis tried a version in 16701670, drawing the extra line above the sign instead of below. It did not catch on. The form close to the one you have been using came from Pierre Bouguer, a French scientist who taught navigation and measured the shape of the planet. Around 17341734 he printed the signs with a bar underneath to mark “or equal to,” and a line under the sign is the idea that stuck.

It is a tiny addition, and a surprising amount rides on it. A bridge posted at twelve tons either admits a twelve ton truck or turns it away. Nothing else in the notice settles that question.

You have been drawing that bar all lesson without calling it that. The filled circle on the number line is the bar, and so is the square bracket in an interval. Three notations, one repeated decision: does the boundary belong to the answer?