64. Given a non-empty set X, consider the binary operation *: P(X) * P(X) → P(X) given by A * B = A ∩ B &mn For E; A, B in P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation*.
64. Given a non-empty set X, consider the binary operation *: P(X) * P(X) → P(X) given by A * B = A ∩ B &mn For E; A, B in P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation*.
Let S be a non-empty set and P (S) be its power set. Let any two subsets A and B of S.
It is given that: is defined as
We know that
Thus, X is the identity element for the given binary operation*.
Now, an element is invertible if there exists such that
(As X is the identity element)
i.e.
This case
Similar Questions for you
R1 = { (1, 1) (1, 2), (1, 3)., (1, 20), (2, 2), (2, 4). (2, 20), (3, 3), (3, 6), . (3, 18),
(4, 4), (4, 8), . (4, 20), (5, 5), (5, 10), (5, 15), (5, 20), (6, 6), (6, 12), (6, 18), (7. 7),
(7, 14), (8, 8), (8, 16), (9, 9), (9, 18), (10, 10), (10, 20), (11, 11), (12, 12)


⇒ (y, x) ∈ R V (x, y) ∈ R
(x, y) ∈ R ⇒ 2x = 3y and (y, x) ∈ R ⇒ 3x = 2y
Which holds only for (0, 0)
Which does not belongs to R.
∴ Value of n = 0
f is increasing function
x < 5x < 7x

f (x) < f (5x) < f (7x)
->
Given f (k) =
Case I : If x is even then g (x) = x . (i)
Case II : If x is odd then g (x + 1) = x + 1 . (ii)
From (i) & (ii), g (x) = x, when x is even
So total no. of functions = 105 × 1 = 105
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Maths Relations and Functions 2025
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