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Relations and Functions - One-to-one and onto functions

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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One-to-one (Injective) Functions: A function f:A→Bf: A \rightarrow B is called one-to-one if distinct elements in AA have distinct images in BB. Formally, f(x1)=f(x2)  ⟹  x1=x2f(x_1) = f(x_2) \implies x_1 = x_2. Visually, no two arrows from the domain point to the same element in the codomain.

A mapping diagram showing a one-to-one relationship where each element of the domain maps to a unique element in the codomain.
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Onto (Surjective) Functions: A function f:A→Bf: A \rightarrow B is onto if every element of the codomain BB is the image of at least one element of the domain AA. This means the range of ff is exactly equal to the codomain BB.

A mapping diagram showing an onto function where every element in the codomain has an incoming arrow.
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Bijective Functions: A function is bijective if it is both one-to-one (injective) and onto (surjective). Such functions establish a perfect pairing between the elements of the domain and the codomain.

A simple bijection diagram with two elements in each set paired exactly one-to-one.
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Horizontal Line Test: To check if a function is one-to-one using its graph, draw horizontal lines. If any horizontal line intersects the graph at more than one point, the function is not one-to-one.

📐Formulae

Condition for Injectivity: f(x1)=f(x2)impliesx1=x2forallx1,x2inDomainf(x_1) = f(x_2) \\implies x_1 = x_2 \\forall x_1, x_2 \\in Domain

Condition for Surjectivity: f(A)=Bf(A) = B or Range=CodomainRange = Codomain

Number of injective functions from AA to BB (where n(A)=m,n(B)=nn(A)=m, n(B)=n): nPm=fracn!(n−m)!^nP_m = \\frac{n!}{(n-m)!} if ngeqmn \\geq m, else 00

Number of bijective functions from AA to BB (where n(A)=n(B)=nn(A) = n(B) = n): n!n!

Number of onto functions from AA to BB (where n(A)=m,n(B)=nn(A)=m, n(B)=n): sumr=0n(−1)r,nCr,(n−r)m\\sum_{r=0}^{n} (-1)^r \\, ^nC_r \\, (n-r)^m

💡Examples

Problem 1:

Show that the function f:mathbbRrightarrowmathbbRf: \\mathbb{R} \\rightarrow \\mathbb{R} defined by f(x)=2x+3f(x) = 2x + 3 is a bijective function.

Solution:

Step 1: Check for Injectivity (One-to-one). Let f(x1)=f(x2)f(x_1) = f(x_2) for some x1,x2inmathbbRx_1, x_2 \\in \\mathbb{R}.
2x1+3=2x2+32x_1 + 3 = 2x_2 + 3
2x1=2x22x_1 = 2x_2
x1=x2x_1 = x_2. Since f(x1)=f(x2)impliesx1=x2f(x_1) = f(x_2) \\implies x_1 = x_2, the function is injective. Step 2: Check for Surjectivity (Onto). Let yy be an arbitrary element in the codomain mathbbR\\mathbb{R}. Set y=f(x)=2x+3y = f(x) = 2x + 3. Solve for xx: x=fracy−32x = \\frac{y-3}{2}. Since yy is a real number, x=fracy−32x = \\frac{y-3}{2} is also a real number (belongs to the domain).
f(x)=f(fracy−32)=2(fracy−32)+3=y−3+3=yf(x) = f(\\frac{y-3}{2}) = 2(\\frac{y-3}{2}) + 3 = y - 3 + 3 = y. Thus, for every yy in the codomain, there exists an xx in the domain such that f(x)=yf(x) = y. The function is surjective. Conclusion: Since ff is both injective and surjective, it is bijective.

Explanation:

To prove bijectivity, we must algebraically demonstrate that the function satisfies both the one-to-one condition and the onto condition.

Problem 2:

Check the injectivity and surjectivity of the function f:mathbbZrightarrowmathbbZf: \\mathbb{Z} \\rightarrow \\mathbb{Z} given by f(x)=x2f(x) = x^2.

Solution:

Step 1: Check Injectivity. Let x1=1x_1 = 1 and x2=−1x_2 = -1.
f(1)=(1)2=1f(1) = (1)^2 = 1 and f(−1)=(−1)2=1f(-1) = (-1)^2 = 1. Here f(1)=f(−1)f(1) = f(-1) but 1neq−11 \\neq -1. Therefore, ff is not injective (it is many-to-one). Step 2: Check Surjectivity. The codomain is mathbbZ\\mathbb{Z} (all integers). Consider y=−2y = -2 in the codomain. Is there an xinmathbbZx \\in \\mathbb{Z} such that x2=−2x^2 = -2? No integer squared results in a negative number. Since there is no xinmathbbZx \\in \\mathbb{Z} such that f(x)=−2f(x) = -2, the element −2-2 has no pre-image. Therefore, ff is not surjective (it is an 'into' function).

Explanation:

Using counter-examples is the most efficient way to show that a function fails to be injective or surjective. In this case, the square function on integers fails both.

Problem 3:

Check if the function f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} defined by f(x)=x3f(x) = x^3 is one-to-one and onto.

Graph of y = x^3 showing it passes the horizontal line test.

Solution:

  1. For Injectivity: Let f(x1)=f(x2)f(x_1) = f(x_2). Then x13=x23x_1^3 = x_2^3. Taking the cube root of both sides, we get x1=x2x_1 = x_2. Thus, ff is one-to-one.
  2. For Surjectivity: Let y∈Ry \in \mathbb{R}. We set y=x3y = x^3. This implies x=y1/3x = y^{1/3}. Since the cube root of any real number is also a real number, every yy has a pre-image xx in the domain. Thus, ff is onto. Since ff is both one-to-one and onto, it is a bijective function.

Explanation:

The cubic function is strictly increasing, which ensures it passes the horizontal line test (one-to-one). Because it extends from −∞-\infty to +∞+\infty on the y-axis, its range is all real numbers (onto).

Problem 4:

Let A={1,2,3}A = \{1, 2, 3\} and B={a,b,c,d}B = \{a, b, c, d\}. Is the function f:A→Bf: A \rightarrow B defined by f={(1,a),(2,b),(3,c)}f = \{(1, a), (2, b), (3, c)\} onto?

Mapping diagram showing element 'd' in the codomain with no pre-image.

Solution:

  1. Range of f={a,b,c}f = \{a, b, c\}.
  2. Codomain B={a,b,c,d}B = \{a, b, c, d\}.
  3. Since Range ≠\neq Codomain (the element dd has no pre-image in AA), the function is not onto.

Explanation:

For a function to be onto, every element in the destination set (codomain) must be 'hit' by at least one arrow from the source set (domain). Here, dd is left out.