Greatest Integer Function (GIF): Definition, Graph, Domain, Range, Properties & Examples

 The Greatest Integer Function (GIF), also known as the Floor Function, is one of the most important concepts in the study of functions for competitive examinations such as JEE Main, JEE Advanced, WBJEE, and Olympiads. Despite its simple definition, it frequently appears in questions involving graphs, domain and range, continuity, inequalities, transformations, inverse functions, and composite functions.

1. The Core Definition

The Greatest Integer Function is denoted by square brackets: f(x) = [x].

The Golden Rule: [x] outputs the greatest integer that is less than or equal to x.

Think of it as a rounding-down machine that maps any real number to the nearest integer to its left on the number line.

Case A: When x is already an integer

If the input is an integer, it stays exactly the same.

• [5] = 5
• [0] = 0
• [-3] = -3

Case B: When x is a positive decimal

Look for the integer immediately to its left.

• [2.3] = 2 (Since 2 is the greatest integer ≤ 2.3)
• [0.99] = 0 (Don't round up! It must be less than or equal to 0.99)

Case C: When x is a negative decimal (The Danger Zone!)

Students often make mistakes here. Remember to go left on the number line.

• [-1.2] = -2 (Because -2 is less than -1.2. Note that -1 is greater than -1.2)
• [-0.4] = -1

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2. The Step-Graph (Visualizing [x])

If we plot y = [x] for various intervals, we see a fascinating pattern emerge. Let's break it down interval by interval:

• For x ∈ [0, 1), y = 0
• For x ∈ [1, 2), y = 1
• For x ∈ [2, 3), y = 2
• For x ∈ [-1, 0), y = -1

Because of this piece-wise nature, the graph looks like a flight of stairs. It is universally called the Step Function.

Crucial Graph Features for JEE:

1. Solid Dot (•): Indicates that the endpoint is included (e.g., at x = 1, y jumps up to 1).
2. Open Circle (◦): Indicates that the endpoint is excluded (e.g., as x approaches 1 from the left, y stays at 0 until the very last micro-moment).
3. Domain: ℝ (You can plug in any real number).
4. Range: ℤ (The output is strictly an integer. The graph only exists at integer heights like y = ..., -2, -1, 0, 1, 2, ...).

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3. The First Level of Application (Algebraic Equations)

Now let's see how WBCHSE and JEE Main frame basic questions using this definition.

Example 1: Solving a Quadratic Equation involving [x]

Question: Solve for x: [x]² - 5[x] + 6 = 0.

Step 1: Let [x] = t. Since [x] behaves like a variable here, we rewrite the equation:

t² - 5t + 6 = 0

Step 2: Factorize the quadratic equation:

(t - 2)(t - 3) = 0 ⇒ t = 2 or t = 3

Step 3: Substitute back t = [x]:

• Case 1: [x] = 2 ⇒ x ∈ [2, 3)
• Case 2: [x] = 3 ⇒ x ∈ [3, 4)

Step 4: Combine the intervals. If x can be anything from 2 up to (but excluding) 3, AND anything from 3 up to (but excluding) 4, the combined continuous solution is:

x ∈ [2, 4)

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Your Turn to Practice!

To make sure you've locked down this foundation, try solving this problem:

Find the domain of the function:

f(x) = 1 / √( [x]² - [x] - 6 )

Hint: For a square root in the denominator, the expression inside must be strictly greater than 0 (> 0). Treat [x] as t first.

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4. The Core Properties of [x]

Think of these properties as the algebraic rules that allow you to simplify expressions containing the Greatest Integer Function.

Property 1: The Integer Jump

[x + n] = [x] + n    (where n is an integer)

Explanation: If an integer is added or subtracted inside the function, you can simply pull it outside. Decimals stay locked inside.

• Example: [x + 5] = [x] + 5
• Example: [x - 3] = [x] - 3

Property 2: The Negative Variable Identity

[x] + [-x] = 0     if x ∈ ℤ (x is an integer)
[x] + [-x] = -1    if x ∉ ℤ (x is not an integer)

Explanation: Let's test this with numbers:

• If x = 4 (integer): [4] + [-4] = 4 + (-4) = 0.
• If x = 2.3 (not an integer): [2.3] + [-2.3] = 2 + (-3) = -1.

Property 3: The Basic Inequality Bounds

x - 1 < [x] ≤ x

Explanation: The value of [x] is always strictly greater than x - 1, and less than or equal to x itself. This property is heavily used in solving calculus limits using the Sandwich Theorem.

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5. Solving GIF Inequalities

Inequalities involving [x] require careful attention to the boundary points. Let's look at the rules for converting [x] inequalities into standard x intervals.

Rule A: The Upper Bound Case

If [x] ≤ n (where n is an integer), then x < n + 1.

Why? If [x] can be n, then x can be any decimal up to, but not including, n + 1. For example, if [x] ≤ 4, then x can be 4.999... which means x < 5.

Rule B: The Lower Bound Case

If [x] ≥ n (where n is an integer), then x ≥ n.

Why? If [x] must be at least n, x must start exactly at n or be greater. For example, if [x] ≥ 3, x can be 3.0, 3.5, etc., so x ≥ 3.

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6. Step-by-Step JEE Main Example

Question: Solve the inequality for x: 2 ≤ [x] < 5

Step 1: Break the compound inequality into two separate parts:

1. [x] ≥ 2
2. [x] < 5  ⇒  [x] ≤ 4 (since [x] must be an integer)

Step 2: Apply the lower bound rule to [x] ≥ 2:

x ≥ 2

Step 3: Apply the upper bound rule to [x] ≤ 4:

x < 5

Step 4: Combine the two intervals to find the final domain solution:

x ∈ [2, 5)

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Practice Quiz: Greatest Integer Function (GIF)

Test your fundamentals and property applications! Select an option for each question to instantly see the correct answer and step-by-step explanation.

Question 1: Basic Calculations

Find the value of the expression: [-2.4] + [1.8] - [-0.2]

Explanation:

Let's evaluate each term carefully using the "round-down to nearest left integer" rule:

• [-2.4] = -3 (Remember, for negatives, we go left on the number line!)
• [1.8] = 1
• [-0.2] = -1

Now, substitute these back into the expression:

(-3) + 1 - (-1) = -3 + 1 + 1 = -1

Correct Answer: C

Question 2: Solving Inequalities

If 3 ≤ [2x - 1] < 5, then find the correct interval for x:

Explanation:

Since the greatest integer value [2x - 1] must be an integer, the inequality 3 ≤ [2x - 1] < 5 implies:

[2x - 1] ∈ {3, 4}

Using the fundamental definition, if [y] is either 3 or 4, then y must lie in the range:

3 ≤ 2x - 1 < 5

Now, solve for x step-by-step:

• Add 1 to all sides: 4 ≤ 2x < 6
• Divide all sides by 2: 2 ≤ x < 3

Therefore, x ∈ [2, 3).

Correct Answer: A

Question 3: Applying Properties

Find the number of integral solutions to the equation: [x] + [-x] = x² - 9

Explanation:

Recall Property 2 from Class 2: [x] + [-x] has two possible values:

1. It equals 0 if x is an integer.
2. It equals -1 if x is not an integer.

Since the question specifically asks for integral solutions (meaning x must be an integer), we set [x] + [-x] = 0:

0 = x² - 9

x² = 9 ⇒ x = 3 or x = -3

Since both 3 and -3 are integers, both are valid solutions. Thus, there are exactly 2 solutions.

Correct Answer: C

Question 4: Domain of a Function

Find the domain of the function: f(x) = 1 / √( [x]² - [x] - 6 )

Explanation:

This is the challenge problem from Class 1! For the square root in the denominator to yield real values, the expression inside must be strictly positive:

[x]² - [x] - 6 > 0

Let [x] = t. The inequality becomes: t² - t - 6 > 0

Factorizing: (t - 3)(t + 2) > 0

This gives: t < -2 or t > 3

Substituting back t = [x]:

• Case 1: [x] < -2 ⇒ [x] ≤ -3 (since [x] must be an integer) ⇒ x < -2
• Case 2: [x] > 3 ⇒ [x] ≥ 4 ⇒ x ≥ 4

Combining the intervals: x ∈ (-∞, -2) ∪ [4, ∞)

Correct Answer: A

Question 5: Conceptual Connection

If [x] + {x} = 2.5 (where {x} denotes the fractional part function), find the value of x:

Explanation:

Any real number x can be split into its integer and fractional parts: x = [x] + {x}.

The equation given is directly the identity formulation:

[x] + {x} = 2.5 ⇒ x = 2.5

Let's verify this output: for x = 2.5, [2.5] = 2, and {2.5} = 0.5. Since 2 + 0.5 = 2.5, the solution is perfectly valid.

Correct Answer: B

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Part 1: JEE Main & Advanced Actual Questions

JEE Main 2024

The Continuity Disruption

Question: Let f: [-1, 2] → ℝ be given by:

f(x) = 2x² + x + [x²] - [x]

where [t] denotes the greatest integer less than or equal to t. Find the number of points in the interval [-1, 2] where f is not continuous.

Step-by-Step Solution:

The potential points of discontinuity for f(x) occur where the expressions inside the GIF, i.e., x and x², become integers.

In the domain [-1, 2]:

• The integers for [x] are: x = -1, 0, 1, 2.
• The integers for [x²] are: x² = 0, 1, 2, 3, 4 ⇒ x = -1, 0, 1, √2, √3, 2.

Combining these, our critical transition points to check for continuity are: x = 0, 1, √2, √3 (endpoints -1 and 2 are checked via one-sided continuity).

Checking the critical points:

1. At x = 0: LHL = -1, RHL = 0. (Discontinuous)
2. At x = 1: LHL = 2, RHL = 2. (Continuous!)
3. At x = √2: LHL = 2 + √2, RHL = 3 + √2. (Discontinuous)
4. At x = √3: LHL = 4 + √3, RHL = 5 + √3. (Discontinuous)

Thus, the function is discontinuous at exactly 4 points: {0, √2, √3, 2}.

JEE Main 2022

Simplifying with Trigonometry

Question: If f: ℝ → ℝ is a function defined by:

f(x) = [x - 1] * cos( ((2x - 1)/2) * π )

where [.] denotes the greatest integer function, then analyze the continuity of f(x) at any integer point.

Step-by-Step Solution:

Usually, [x - 1] makes the function discontinuous at all integers. However, let's look at the cosine term:

cos( ((2x - 1)/2) * π ) = cos( xπ - π/2 ) = sin(xπ)

Now rewrite the function:

f(x) = [x - 1] * sin(xπ)

For any integer n, we know that sin(nπ) = 0. Since the limit of sin(xπ) is 0 at all integers, it successfully "neutralizes" the sudden step-jump discontinuity of [x - 1].

Therefore, the function f(x) is actually continuous for every real x.

JEE Advanced 2018

Limits Meeting GIF

Question: For each positive integer n, let:

y_n = (1/n) * [ (n + 1)(n + 2)...(n + n) ]^(1/n)

If lim (n → ∞) y_n = L, then find the value of [L] (where [.] is the Greatest Integer Function).

Step-by-Step Solution:

First, we solve for the limit L using the integration limit definition. Let's rewrite y_n:

ln(y_n) = (1/n) * Σ [ ln(1 + r/n) ] from r=1 to n

Converting this Riemann sum to a definite integral as n → ∞:

ln(L) = ∫₀¹ ln(1 + x) dx

Using integration by parts:

ln(L) = [ x*ln(1+x) - x + ln(1+x) ] evaluated from 0 to 1
ln(L) = 2*ln(2) - 1 = ln(4/e)

Therefore, L = 4/e.

Since e ≈ 2.718, we have L = 4 / 2.718 ≈ 1.47.

Applying the Greatest Integer Function: [1.47] = 1.

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Part 2: Board Exam Level Actual Questions (WBCHSE)

WBCHSE / HS Board Level

Finding Domain and Range

Question: Find the domain of the real-valued function:

f(x) = 1 / [x]

where [x] represents the greatest integer function. Also, find its range.

Step-by-Step Solution:

1. Finding the Domain:

The function is defined for all real values of x except where the denominator becomes zero:

[x] ≠ 0

By definition, [x] = 0 when x lies in the interval [0, 1).

Therefore, we exclude this interval from the set of real numbers ℝ.

Domain = ℝ - [0, 1)

2. Finding the Range:

Since [x] only outputs integer values (excluding 0), the output values of f(x) are of the form 1/n, where n is a non-zero integer.

Range = { 1/n : n ∈ ℤ, n ≠ 0 } (e.g., ..., -1, -1/2, 1, 1/2, 1/3, ...)