In short

An interval is a connected piece of the real number line — every point between its endpoints is included. Intervals come in four flavours depending on whether the endpoints are included (closed brackets [\ ]) or excluded (open brackets (\ )). Inequalities are the algebraic tool that produce intervals: solving 2x - 3 < 7 gives x < 5, which is the interval (-\infty, 5). The rules are the same as for equations — do the same thing to both sides — except for one critical trap: multiplying or dividing by a negative number flips the inequality sign. Absolute-value inequalities like |x - 4| < 3 reduce to a double inequality 1 < x < 7, and the answer is always an interval centred on the number inside the absolute value.

A mobile plan charges Rs 299 per month for 2 GB of data, plus Rs 12 for every extra GB. You have a budget of Rs 500. How many extra gigabytes can you use? The constraint is 299 + 12g \le 500, and the answer is not a single number — it is every value of g from 0 up to 16.75. The answer to an inequality is a set of numbers, and that set is almost always an interval.

This article is the notation and the toolkit: the four types of interval, bracket notation, how to solve linear inequalities (including the critical trap when you multiply by a negative), and how to handle absolute-value inequalities — the type that produces a bounded interval instead of a ray.

What is an interval?

An interval is a subset of \mathbb{R} with no gaps. Formally: a set I of real numbers is an interval if, whenever a and b are in I with a < b, every number between a and b is also in I.

The set \{1, 2, 3\} is not an interval — it contains 1 and 3 but not 2.5. The set of all x with 1 \le x \le 3 is an interval — it contains every real number from 1 to 3 with no missing points.

The four bounded intervals

Given two real numbers a < b, there are four ways to build a bounded interval depending on which endpoints you include.

Notation Name Set Endpoints
[a, b] Closed \{x : a \le x \le b\} Both included
(a, b) Open \{x : a < x < b\} Both excluded
[a, b) Half-open (closed-open) \{x : a \le x < b\} Left included, right excluded
(a, b] Half-open (open-closed) \{x : a < x \le b\} Left excluded, right included

The square bracket [ means "include this endpoint" and the round bracket ( means "exclude it." The same convention holds everywhere in mathematics, and it maps directly to the filled-dot / hollow-dot picture on the number line: a filled dot at the endpoint means it is in the set, a hollow dot means it is not.

Four bounded intervals between a and b on the number lineFour horizontal number lines stacked vertically, each showing the interval from a to b. The first shows the closed interval with filled circles at both endpoints. The second shows the open interval with hollow circles at both endpoints. The third shows the half-open interval closed at a and open at b, with a filled circle at a and hollow at b. The fourth shows the half-open interval open at a and closed at b, with a hollow circle at a and filled at b. Each interval's name and notation appear to the right. [a, b] — closed a b (a, b) — open a b [a, b) — half-open a b (a, b] — half-open a b
The four types of bounded interval. Filled dots (included) vs. hollow dots (excluded) at the endpoints are the only difference. The thick segment in the middle — the body of the interval — is the same in all four cases.

Unbounded intervals and the infinity symbol

Not every interval has two finite endpoints. The solution to x < 5 is a ray extending to the left forever. The notation uses -\infty and +\infty (or just \infty) as stand-ins for "keeps going."

Notation Set Description
(-\infty, b) \{x : x < b\} Everything left of b
(-\infty, b] \{x : x \le b\} Everything left of and including b
(a, \infty) \{x : x > a\} Everything right of a
[a, \infty) \{x : x \ge a\} Everything right of and including a
(-\infty, \infty) \mathbb{R} The entire real line

An important convention: \infty always gets a round bracket, never a square one. You write (-\infty, 5], not [-\infty, 5]. The reason is that \infty is not a real number — it is a notational device meaning "no boundary here." Since \infty is not an element of \mathbb{R}, it cannot be "included" in the interval, so a square bracket would be meaningless.

Two unbounded intervals shown as rays on the number lineTwo number lines stacked vertically. The top shows the interval from negative infinity to 5, exclusive, drawn as a thick ray pointing left from a hollow circle at 5 with an arrow at the far left. The bottom shows the interval from negative 2 inclusive to positive infinity, drawn as a thick ray pointing right from a filled circle at negative 2 with an arrow at the far right. (−∞, 5) 5 [−2, ∞) −2
Top: $(-\infty, 5)$ is every real number less than $5$, shown as a leftward ray with a hollow circle at $5$. Bottom: $[-2, \infty)$ is every real number greater than or equal to $-2$, shown as a rightward ray with a filled circle at $-2$. Both arrows mean "extends forever."

Solving linear inequalities

The rules for manipulating inequalities are almost the same as for equations. You can add or subtract the same number on both sides. You can multiply or divide both sides by the same positive number. The one critical difference — drilled into you in Operations and Properties — is that multiplying or dividing by a negative number reverses the inequality sign.

Here is the complete set of legal moves:

The same rules hold for \le, >, and \ge.

A simple example without the flip

Solve 3x + 7 \le 22.

Subtract 7: 3x \le 15. Divide by 3 (positive, no flip): x \le 5. The solution is (-\infty, 5].

An example with the flip

Solve -4x + 1 > 9.

Subtract 1: -4x > 8. Divide by -4 (negative — flip): x < -2. The solution is (-\infty, -2).

If you had forgotten to flip, you would have written x > -2, which is the interval (-2, \infty) — the entire wrong half of the number line. This is the most common error in inequality problems, and exam graders in board exams and JEE see it every year.

Absolute value inequalities

The absolute value |x| is the distance of x from 0 on the number line. More generally, |x - c| is the distance of x from c. This geometric reading is the key to solving absolute-value inequalities.

The "less than" case: |x - c| < r

The inequality |x - c| < r says "the distance from x to c is less than r." Geometrically, that means x lies within r units of c — which is the open interval (c - r,\ c + r).

Absolute Value Inequality (Less Than)

For r > 0:

|x - c| < r \iff c - r < x < c + r

The solution is the open interval (c - r,\ c + r), centred at c with radius r.

The "greater than" case: |x - c| > r

The inequality |x - c| > r says "the distance from x to c is more than r." The point x must be outside the interval of radius r around c — so the solution is two rays:

x < c - r \quad\text{or}\quad x > c + r

In interval notation: (-\infty,\ c - r) \cup (c + r,\ \infty).

Geometric meaning of absolute value inequalities on the number lineTwo number lines stacked vertically. The top line shows the solution of absolute value of x minus c less than r as a shaded interval from c minus r to c plus r, with the centre c marked. The bottom line shows the solution of absolute value of x minus c greater than r as two shaded rays extending left from c minus r and right from c plus r, leaving a gap of width 2r around c. |x − c| < r c − r c c + r radius r |x − c| > r c − r c c + r Inside r: one interval. Outside r: two rays.
The "less than" case gives a single interval centred at $c$ (top). The "greater than" case gives two rays pointing away from $c$ (bottom). The geometric picture — distance from $c$ — makes both cases immediate.

Combining intervals: intersection and union

Sometimes solving an inequality produces two conditions that must both hold (intersection, \cap) or at least one must hold (union, \cup).

For example, [1, 5] \cap [3, 8] = [3, 5] — the overlap of the two intervals. And (-\infty, -2) \cup (2, \infty) is the solution to |x| > 2: two rays with a gap in the middle.

The intersection of two intervals is always an interval (or empty). The union of two intervals is an interval only when they overlap or share an endpoint; otherwise it is two disjoint pieces and you write it with \cup.

An interactive exploration

Drag the red point along the number line below. The readout shows x, |x - 3|, and whether the point satisfies the inequality |x - 3| \le 4 — that is, whether x lies in the interval [-1, 7].

Interactive absolute value inequality explorerA number line from negative 5 to 11 with tick marks at every integer. The interval from negative 1 to 7 is highlighted. A draggable red point moves along the line, and readouts above show x, the absolute value of x minus 3, and whether the inequality absolute value of x minus 3 is less than or equal to 4 is satisfied. −5 −3 −1 1 3 5 7 9 11 ↔ drag the red point
The shaded band marks the interval $[-1, 7]$. Drag the red point and watch: when $|x - 3| \le 4$, the point is inside the shaded region. Move it outside, and the distance from $3$ exceeds $4$. The inequality is a "stay within radius $4$ of the centre $3$" condition.

Two worked examples

Example 1: Solve $\dfrac{2x + 1}{3} - 1 \ge \dfrac{x}{2}$ and express the answer in interval notation

This is a linear inequality with fractions. The first job is to clear the denominators.

Step 1. Multiply every term by 6 (the LCM of 3 and 2). Since 6 > 0, the inequality direction stays.

6 \cdot \frac{2x + 1}{3} - 6 \cdot 1 \ge 6 \cdot \frac{x}{2}
2(2x + 1) - 6 \ge 3x

Why: multiplying through by the LCM eliminates all fractions at once, turning the inequality into one with integer coefficients. Since the multiplier is positive, no flip.

Step 2. Expand and simplify the left side.

4x + 2 - 6 \ge 3x
4x - 4 \ge 3x

Why: the distributive law expands 2(2x + 1) to 4x + 2, then you combine the constants 2 - 6 = -4.

Step 3. Subtract 3x from both sides.

4x - 3x - 4 \ge 0
x - 4 \ge 0

Why: collecting x-terms on one side isolates the variable. Subtracting the same value preserves the inequality.

Step 4. Add 4 to both sides.

x \ge 4

Result. x \ge 4, which is the interval [4, \infty).

Solution x greater than or equal to 4 on the number lineA number line from negative 2 to 10 with integer tick marks. A thick red ray starts at a filled circle at 4 and extends rightward with an arrow, representing the solution set x is greater than or equal to 4, or the interval from 4 to positive infinity inclusive at 4. −2 −1 0 1 2 3 4 5 6 7 8 9 10 x ≥ 4 → [4, ∞)
The solution ray starts at $4$ (included, filled dot) and extends forever to the right. Every $x \ge 4$ satisfies the original inequality. You can check: at $x = 4$, the left side is $\frac{9}{3} - 1 = 2$ and the right side is $\frac{4}{2} = 2$, so $2 \ge 2$ holds.

The boundary check confirms the filled dot: at x = 4 the two sides are equal, so the \ge is satisfied as equality.

Example 2: Solve $|2x - 5| < 3$ and express the answer in interval notation

This is an absolute-value inequality. The geometric reading: the distance from 2x - 5 to 0 is less than 3.

Step 1. Convert the absolute-value inequality to a double inequality.

|2x - 5| < 3 \iff -3 < 2x - 5 < 3

Why: |E| < r means E is within r units of 0, which means -r < E < r. This splits the single absolute-value inequality into two simultaneous linear inequalities.

Step 2. Add 5 to all three parts of the double inequality.

-3 + 5 < 2x - 5 + 5 < 3 + 5
2 < 2x < 8

Why: adding the same number to all three parts preserves both inequality signs. The -5 and +5 cancel in the middle.

Step 3. Divide all three parts by 2 (positive, no flip).

1 < x < 4

Why: dividing by a positive number does not change the direction of the inequalities.

Step 4. Write the answer in interval notation.

x \in (1, 4)

Result. The solution is the open interval (1, 4).

Solution of absolute value of 2x minus 5 less than 3 as an open interval on the number lineA number line from negative 2 to 7 with integer tick marks. A thick accent-coloured segment runs between hollow circles at 1 and 4, representing the open interval from 1 to 4. The centre of the interval is at 2.5, marked with a small dot, corresponding to the point where 2x minus 5 equals 0, that is x equals 2.5. −2 −1 0 1 2 3 4 5 6 7 centre 2.5 1 < x < 4 → (1, 4)
The solution is the open interval $(1, 4)$, centred at $x = 2.5$ (where $2x - 5 = 0$) with "radius" $1.5$ on each side. Both endpoints are hollow because the inequality is strict ($<$, not $\le$). Checking: at $x = 1$, $|2(1) - 5| = 3$, which is *not* less than $3$, confirming $x = 1$ is excluded.

The check at the boundary matters: at x = 1 the distance is exactly 3, not less than 3, so the endpoint is correctly excluded (hollow dot). The same holds at x = 4. The picture matches the algebra.

Common confusions

Going deeper

If you came here for interval notation and the technique of solving linear and absolute-value inequalities, you have it. The rest is for readers who want to see the connection between intervals and the structure of \mathbb{R}.

Intervals and completeness

The fact that intervals have no internal gaps is a direct consequence of the completeness of \mathbb{R}. In \mathbb{Q}, the set \{q \in \mathbb{Q} : 1 < q < 2\} looks like an interval but has holes — \sqrt{2} is missing, for instance. When you move to \mathbb{R}, those holes fill in, and every interval is genuinely gap-free. The definition of an interval — "if a and b are in I, so is everything between them" — only gives you a nice connected stretch because \mathbb{R} is complete. On the rationals, the "everything between them" part fails for irrational points.

Compound inequalities and systems

A single inequality gives a single interval or ray. A system of inequalities — two or more inequalities that must hold simultaneously — gives an intersection of intervals. For example, the system

x > 1 \qquad\text{and}\qquad x \le 5

gives x \in (1, 5]. The solution is the overlap of (1, \infty) and (-\infty, 5].

When the two conditions point in opposite directions — like x > 7 and x < 3 — the intersection is empty, and the system has no solution. Geometrically, the two rays don't overlap.

When the conditions are joined by "or" instead of "and," you take the union. x < -1 or x > 4 gives (-\infty, -1) \cup (4, \infty).

Quadratic and polynomial inequalities

Once you move beyond linear expressions, the solution sets can be unions of intervals. The inequality x^2 - 4 > 0 factors as (x - 2)(x + 2) > 0, which holds when both factors are positive or both are negative. The solution is (-\infty, -2) \cup (2, \infty). The technique — factor, find the roots, test the sign in each interval — is the subject of Quadratic Inequalities, and it builds directly on the interval language from this article.

The triangle inequality revisited

The absolute-value inequality |a + b| \le |a| + |b| from Operations and Properties is the reason the "distance" interpretation of |x - c| works. That inequality says that distances on the number line obey the triangle inequality, which is the minimum requirement for something to be a genuine distance function (a metric). Intervals, as subsets of \mathbb{R} defined by distance conditions like |x - c| < r, inherit their clean geometric behaviour from this single inequality.

Where this leads next

Intervals and inequalities are the language in which most of the rest of algebra states its answers.