EXERCISE 1.2 Relations and Functions Questions
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EXERCISE 1.2 Relations and Functions Questions solutions

1. Show that the function f : ℝ* → ℝ* defined by f(x) = 1/x is one-to-one and onto:

To prove that the function f(x) = 1/x from the set of all non-zero real numbers to itself is one-to-one and onto:

  1. One-to-one: Let x1 and x2 be two distinct non-zero real numbers such that f(x1) = f(x2). Then, 1/x1 = 1/x2. Solving for x1 and x2, we get x1 = x2. Therefore, f is one-to-one.
  2. Onto: For any non-zero real number y, let x = 1/y. Then, f(x) = 1/(1/y) = y. Therefore, for every y in ℝ*, there exists an x in ℝ* such that f(x) = y. Hence, f is onto.

Therefore, the function f : ℝ* → ℝ* defined by f(x) = 1/x is both one-to-one and onto.

Regarding the replacement of the domain ℝ* with ℕ:

The result may not hold true if the domain ℝ* is replaced by ℕ with the same co-domain ℝ*. This is because the function f(x) = 1/x is not defined for x = 0, which is an element of ℕ but not of ℝ*. Therefore, the function would not be onto when the domain is ℕ.

2. Check the injectivity and surjectivity of the following functions:

  1. f : ℕ → ℕ, f(x) = x2
    • Injectivity: The function f(x) = x2 is not injective because, for example, f(1) = 12 = 1 and f(-1) = (-1)2 = 1, but 1 ≠ -1.
    • Surjectivity: The function f(x) = x2 is not surjective because, for example, there is no natural number whose square is 2.
  2. f : ℤ → ℤ, f(x) = x2
    • Injectivity: The function f(x) = x2 is not injective because, for example, f(1) = 12 = 1 and f(-1) = (-1)2 = 1, but 1 ≠ -1.
    • Surjectivity: The function f(x) = x2 is not surjective because, for example, there is no integer whose square is -1.
  3. f : ℝ → ℝ, f(x) = x2
    • Injectivity: The function f(x) = x2 is not injective because, for example, f(1) = 12 = 1 and f(-1) = (-1)2 = 1, but 1 ≠ -1.
    • Surjectivity: The function f(x) = x2 is not surjective because, for example, there is no real number whose square is -1.
  4. f : ℕ → ℕ, f(x) = x3
    • Injectivity: The function f(x) = x3 is injective because, for every x1 ≠ x2, f(x1) ≠ f(x2).
    • Surjectivity: The function f(x) = x3 is surjective because, for every y in ℕ, there exists an x in ℕ such that f(x) = y (every natural number is a cube of some other natural number).
  5. f : ℤ → ℤ, f(x) = x3
    • Injectivity: The function f(x) = x3 is injective because, for every x1 ≠ x2, f(x1) ≠ f(x2).
    • Surjectivity: The function f(x) = x3 is surjective because, for every y in ℤ, there exists an x in ℤ such that f(x) = y (every integer is a cube of some other integer).

3. Proof that the Greatest Integer Function f : ℝ → ℝ, given by f(x) = [x], is neither one-to-one nor onto:

To prove that the Greatest Integer Function f(x) = [x] from the set of real numbers to itself is neither one-to-one nor onto:

  1. One-to-one: Consider the cases:
    • f(1.5) = [1.5] = 1 and f(1.7) = [1.7] = 1, but 1.5 ≠ 1.7. Therefore, f is not one-to-one.
  2. Onto: Consider the case:
    • There is no real number x such that f(x) = 1. For example, for any x in the interval (1, 2), f(x) = [x] = 1. Therefore, f is not onto.

Hence, the Greatest Integer Function f is neither one-to-one nor onto.

4. Proof that the Modulus Function f : ℝ → ℝ, given by f(x) = |x|, is neither one-to-one nor onto:

To prove that the Modulus Function f(x) = |x| from the set of real numbers to itself is neither one-to-one nor onto:

  1. One-to-one: Consider the cases:
    • f(1) = |1| = 1 and f(-1) = |-1| = 1, but 1 ≠ -1. Therefore, f is not one-to-one.
  2. Onto: Consider the case:
    • There is no real number x such that f(x) = -1. Since the modulus of any real number is non-negative, there is no x such that f(x) = -1. Therefore, f is not onto.

Hence, the Modulus Function f is neither one-to-one nor onto.

5. Show that the Signum Function f : ℝ → ℝ is neither one-to-one nor onto:

To prove that the Signum Function f(x) = { 1 if x > 0, 0 if x = 0, -1 if x < 0 } is neither one-to-one nor onto:

  1. One-to-one: Consider the cases:
    • f(1) = f(-1) = 1, so f is not one-to-one.
  2. Onto: Since f(x) can only take values -1, 0, or 1, it cannot cover all real numbers. For example, there is no x such that f(x) = 2. Therefore, f is not onto.

Hence, the Signum Function f is neither one-to-one nor onto.

6. Proof that the function f from A to B is one-to-one:

To prove that the function f = {(1, 4), (2, 5), (3, 6)} from set A = {1, 2, 3} to set B = {4, 5, 6, 7} is one-to-one:

Let's examine each element of A:

  • f(1) = 4
  • f(2) = 5
  • f(3) = 6

Since each element of A is mapped to a unique element of B, f is one-to-one.