SOLUTION: Alternative (B) is right. A = {a, b} and A x A = {(a,a), (a,b),(b,b),(b,a)} Number of components = 4 Along these lines, number of subsets = 2^4 = 16.
Number of binary operations on the set {a, b} are : (A) 10 (B) 16 (C) 20 (D ) 8
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Miscellaneous
Let f : R → R be the Signum Function defined as and g : R → R be the Greatest Integer Function given by g (x) = [x], where [x] is greatest integer less than or equal to x. Then, does fog and gof coincide in (0, 1]?
SOLUTION: Given: f : R → R be the Signum Function characterized as also, g : R → R be the Greatest Integer Function given by g (x) = [x], where [x] is most prominent number not exactly or equivalent...
Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is : (A) 1 (B) 2 (C) 3 (D) 4
solution: The correct answer is option(B)
Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
(A) 1 (B) 2 (C) 3 (D) 4 solution: Alternative (A) is right. As 1 is reflexive and symmetric yet not...
Let A = {– 1, 0, 1, 2}, B = {– 4, – 2, 0, 2} and f, g : A → B be functions defined by f(x) = x2 – x, x ∈ A and g(x) = 2|x – ½| – 1, x ∈ A. Are f and g equal? Justify your answer. (Hint: One may note that two functions f : A → B and g : A → B such that f(a) = g (a) ∀ a ∈ A, are called equal functions).
solution: Given capacities are: f(x) = x2 – x and g(x) = 2|x – ½| - 1 At x = - 1 f(- 1) = 12 + 1 = 2 and g(- 1) = 2|-1 – ½| - 1 = 2 At x = 0 F(0) = 0 and g(0) = 0 At x = 1 F(1) = 0 and g(1) = 0 At x...
Define a binary operation * on the set {0, 1, 2, 3, 4, 5} as a * b = { ???? + ???? ???????? ???? + ???? < ???? ???? + ???? − ???? ???????? ???? + ???? ≥ ???? . Show that zero is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 – a being the inverse of a.
Arrangement: Let x = {0, 1, 2, 3, 4, 5} and activity * is characterized as a * b = { ???? + ???? ???????? ???? + ???? < 6 ???? + ???? − 6 ???????? ???? + ???? ≥ 0 Let us say, is the personality...
Given a non-empty set X, let * : P(X) × P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), ∀ A, B ∈ P(X). Show that the empty set ϕ is the identity for the operation * and all the elements A of P(X) are invertible with A–1 = A. (Hint : (A – ϕ) ∪ (ϕ– A) = A and (A – A) ∪ (A – A) = A * A = ϕ).
solution: x ∈ P(x) Furthermore, ϕ is the character component for the activity * on P(x). Additionally A*A= = Each component An of P(X) is invertible with A-1 = A.
Consider the binary operations * : R × R → R and o : R × R → R defined as a * b = |a– b| and a o b =a, ∀ a, b ∈ R. Show that ∗ is commutative but not associative, o is associative but not commutative. Further, show that ∀ a, b, c ∈ R, a * (b o c) = (a * b) o (a* c). [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over *? Justify your answer.
Solution: Stage 1: Check for commutative and cooperative for activity *. a * b = |a – b| and b * a = |b – a| = (a, b) Activity * is commutative. a(bc) = a|b-c| = |a-(b-c)| = |a-b+c| and (ab)c =...
let S = {a, b, c} and T = {1, 2, 3}. Find F–1 of the following functions F from S to T, if it exists.
(i) F = {(a, 3), (b, 2), (c, 1)} (ii) F = {(a, 2), (b, 1), (c, 1)} solution: (I) F = {(a, 3), (b, 2), (c, 1)} F(a) = 3, F(b) = 2 and F(c) = 1 F-1 (3) = a, F-1 (2) = b and...
Find the number of all onto functions from the set {1, 2, 3, , n} to itself.
solution: The quantity of onto capacities that can be characterized from a limited
Given a non-empty set X, consider the binary operation * : P(X) × P(X) → P(X) given by A * B = A ∩ B ∀ A, B in P(X), where 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 * .
Solution: Leave T alone a non-void set and P(T) be its force set. Let any two subsets An and B of T. A ∪ B ⊂ T Along these lines, A ∪ B ∈ P(T) Thusly, ∪ is a twofold procedure on P(T). Additionally,...
Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows:For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X)? Justify your answer.
Give examples of two functions f : N → Z and g : Z → Z such that g o f is injective but g is not injective.
(Hint : Consider f(x) = x +1 and g (x) = ???? − ???? ???????? ???? > ???? { ???? ???????? ???? = ???? } solution: Given: Two capacities f : N → Z and g : Z → Z Say f(x) = x+ 1 Also, g (x) = ????...
Give examples of two functions f : N → Z and g : Z → Z such that g o f is injective but g is not injective.
(Hint : Consider f(x) = x and g (x) = | x |) solution: Given: two capacities are f : N → Z and g : Z → Z Allow us to say, f(x) = x and g(x) = x gof = (gof)(x) = f(f(x)) = g(x) Here gof is injective...
Show that the function f : R → R given by f(x) = x3 is injective.
The capacity f : R → R given by f(x) = x3 Let x , y This suggests , x3 = y3 x = y f is one-one. So f is injective.
Show that the function f : R → {x ∈ R : – 1 < x < 1} defined by f(x) = ????/1+|????|,x ∈ R is one and onto function.
solution: The capacity f : R → {x ∈ R : – 1 < x < 1} characterized by f(x) = ???? / 1+|????| ,x ∈ R For one-one: Say x, y ∈ R ...
If f : R → R is defined by f(x) = x2 – 3x + 2, find f (f(x)).
solution: Given: f(x) = \[x2\text{ }\text{ }3x\text{ }+\text{ }2\text{ }f\text{ }\left( f\left( x \right) \right)\text{ }=\text{ }f\left( x2\text{ }\text{ }3x\text{ }+\text{ }2 \right)\]...
Let f : W → W be defined as f(n) = n – 1, if n is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.
Solution: f : W → W be characterized as f(n) = n – 1, in case n is odd and f(n) = n + 1, in case n is even. Capacity can be characterized as: f is invertible, in case f is one-one and onto. For...
Let f : R → R be defined as f(x) = 10x + 7. Find the function g : R → R such that g o f = f o g = IR.
Solution: Right off the bat, Find the opposite of f. Let say, g is reverse of f and y = f(x) = 10x + 7 y = 10x + 7 or on the other hand x = (y-7)/10 or on the other hand g(y) = (y-7)/10; where g : Y...