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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Relations and Functions
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...
Consider a binary operation ∗ on N defined as a ∗ b = a3 + b3. Choose the correct answer.
Is ∗ both associative and commutative?Is ∗ commutative but not associative?Is ∗ associative but not commutative?Is ∗ neither commutative nor associative? solution: A two fold activity ∗ on N...
State whether the following statements are true or false. Justify.
For an arbitrary binary operation * on a set N, a * a = a ∀ a ∈ N.If * is a commutative binary operation on N, then a * (b * c) = (c * b) * a solution: (i) Given: * being a paired procedure on N, is...
Let A = N × N and ∗ be the binary operation on A defined by (a, b) ∗ (c, d) = (a + c, b + d) Show that ∗ is commutative and associative. Find the identity element for ∗ on A, if any.
solution: A = N x N and * is a paired activity characterized on A. (a, b) * (c, d) = (a + c, b + d) (c, d) * (a, b) = (c + a, d + b) = (a + c, b + d) The activity * is commutative Once more, ((a, b)...
Find which of the operations given above has identity.
Solution: Let I be the identity. a * I = a – I ≠ aa * I = a2 – I2 ≠ aa * I = a + a I ≠ aa * I = (a – I) 2 ≠ aa * I = aI/4 ≠ a Which is only possible at I = 4 i.e. a * I = aI/4 = a(4)/4 = a 6. a * I...
Let ∗ be a binary operation on the set Q of rational numbers as follows and Find which of the binary operations are commutative and which are associative.
a ∗ b = ab/4a ∗ b = ab2 (v) a ∗ b = abdominal muscle/4 b * a = ba/2 = abdominal muscle/2 a ∗ b = b * a The activity * is commutative. Check for affiliated: (a * b) * c = abdominal muscle/4 * c =...
Let ∗ be a binary operation on the set Q of rational numbers as follows and Find which of the binary operations are commutative and which are associative.
a ∗ b = a + aba ∗ b = (a – b)2 (iii) a ∗ b = a + stomach muscle a ∗ b = a + stomach muscle = a(1 + b) b * a = b + ba = b (1+a) a ∗ b ≠ b * a The activity * isn't commutative Check for cooperative:...
Let ∗ be a binary operation on the set Q of rational numbers as follows and Find which of the binary operations are commutative and which are associative.
a ∗ b = a – ba ∗ b = a2 + b2 Arrangement: (i) a ∗ b = a – b a ∗ b = a – b = - (b – a) = - b * c ≠ b * a (Not commutative) (a * b) * c = (a – b) * c = (a – (b – c) = a – b + c ≠ a * (b *c) (Not...
Let ∗ be the binary operation on N defined by a ∗ b = H.C.F. of a and b. Is ∗ commutative? Is ∗ associative? Does there exist identity for this binary operation on N?
solution: The activity ∗ be the double procedure on N characterized by a ∗ b = H.C.F. of an and b a * b = H.C.F. of an and b = H.C.F. of b and a = b * a Therefore, activity * is commutative. Once...
Is ∗ defined on the set {1, 2, 3, 4, 5} by a ∗ b = L.C.M. of a and b a binary operation? Justify your answer.
solution: The activity ∗ characterized on the set {1, 2, 3, 4, 5} by a ∗ b = L.C.M. of an and b Assume, a = 2 and b = 3 2 * 3 = L.C.M. of 2 and 3 = 6 However, 6 doesn't has a place with the set A....
Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b. Find
Which elements of N are invertible for the operation ∗? (v) Which components of N are invertible for the activity ∗? Just the component 1 in N is invertible for the activity * in light of the fact...
Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b. Find
Is ∗ associative?Find the identity of ∗ in N (iii) Is ∗ affiliated? For a,b, c ∈ N (a ∗ b) * c = (L.C.M. of an and b) * c = L.C.M. of a, b and c a ∗ (b * c) = a * (L.C.M. of b and c) = L.C.M. of a,...
Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b. Find
5 ∗ 7, 20 ∗ 16Is ∗ commutative? solution: (i) 5 ∗ 7 = LCM of 5 and 7 = 35 20 ∗ 16 = LCM of 20 and 16 = 80 (ii) Is ∗ commutative? a ∗ b = L.C.M. of an and b ∗ a = L.C.M. of b and a ∗ b = b ∗ a Along...
Let ∗′ be the binary operation on the set {1, 2, 3, 4, 5} defined by a ∗′ b = H.C.F. of a and b. Is the operation ∗′ same as the operation ∗ defined in Exercise 4 above? Justify your answer.
solution: Let A = {1, 2, 3, 4, 5} and a ∗′ b H.C.F. of an and b. Plot a table qualities, we have Activity ∗′ same as the activity *.
Consider a binary operation ∗ on the set {1, 2, 3, 4, 5} given by the following multiplication table (Table 1.2).
(i) Compute (2 ∗ 3) ∗ 4 and 2 ∗ (3 ∗ 4)Is ∗ commutative? (ii)Compute (2 ∗ 3) ∗ (4 ∗ 5). (Hint: use the following table) solution: (I) Compute (2 ∗ 3) ∗ 4 and 2 ∗ (3 ∗ 4) From table: (2 ∗ 3) = 1 and...
Consider the binary operation ∧ on the set {1, 2, 3, 4, 5} defined by a ∧ b = min {a, b}. Write the operation table of the operation ∧ .
SOLUTION: The paired activity ∧ on the set, say A = {1, 2, 3, 4, 5} characterized by a ∧ b = min {a, b}. the activity table of the activity ∧ as follow:
For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.
On Z+, define a ∗ b = abOn R – {– 1}, define a ∗ b = a/(b+1) (v) On Z+, characterize a ∗ b = abdominal muscle Stage 1: Check for commutative Consider ∗ is commutative, then, at that point a ∗ b = b...
For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.
On Q, define a ∗ b = ab/2On Z+, define a ∗ b = 2ab (iii) On Q, characterize a ∗ b = stomach muscle/2 Stage 1: Check for commutative Consider ∗ is commutative, then, at that point a ∗ b = b * a Which...
For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.
On Z, define a ∗ b = a – bOn Q, define a ∗ b = ab + 1 (i) On Z, characterize a ∗ b = a – b Stage 1: Check for commutative Consider ∗ is commutative, then, at that point a ∗ b = b * a Which implies,...
Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.(v)
On Z+, define ∗ by a ∗ b = a On Z+ = {1, 2, 3, 4, 5,… … .} Let a = 2 and b = 1 Therefore, a ∗ b = 2 ∈ Z+ Activity * is a paired procedure on Z+ .
Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.
On R, define ∗ by a ∗ b = ab2On Z+, define ∗ by a ∗ b = | a – b | (iii) On R, characterize ∗ by a ∗ b = ab2 R = { - ∞, … , - 1, 0, 1, 2,… … , ∞} Let a = 1.2 and b = 2 Accordingly, a ∗ b = ab2 =...
Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.
On Z+, define ∗ by a ∗ b = a – bOn Z+, define ∗ by a ∗ b = ab (i) On Z+ = {1, 2,3 , 4, 5,… … .} Let a = 1 and b = 2 Subsequently, a ∗ b = a – b = 1 – 2 = - 1 ∉ Z+ activity * is certifiably not a...
Let f : R – { -4/3 } → R be a function defined as f(x) = ????????/3????+????. The inverse of f is the map g : Range f → R – { -4/3 } given by
(A) g(y) = 3y/(3-4y) (B) g(y) = 4y/(4-3y) (C) g(y) =...
Consider f : {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c. Find f–1 and show that (f–1)–1 = f.
solution: Think about f : {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c So f = {(a, 1), (b, 2), (c, 3)} Subsequently f-1 (a) = 1, f-1 (b) = 2 and f-1 (c) = 3 Now, f-1 = {(a, 1), (b,...
Let f : X → Y be an invertible function. Show that f has unique inverse.
(Hint: suppose g1 and g2 are two inverses of f. Then for all y ∈ Y,fog1(y) = 1Y(y) = fog2(y). Use one-one ness of f) solution: Given, f : X → Y be an invertible capacity. What's more, g1 and g2 are...
Consider f : R+ → [– 5, ∞) given by f (x) = 9×2 + 6x – 5. Show that f is invertible with
solution: Think about f : R+ → [–5, ∞) given by f (x) = 9x2 + 6x – 5 Consider f : R+ → [4, ∞) given by f(x) = x2 + 4 Let x, y ∈ R → [-5, ∞) then, at that point f(x) = 9x2 + 6x – 5 and f(y) = 9y2 +...
Consider f : R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with the inverse f–1 of f given by f–1(y) = √???? − ???? , where R+ is the set of all non-negative real numbers.
solution: Think about f : R+ → [4, ∞) given by f(x) = x2 + 4 Let x, y ∈ R → [4, ∞) then, at that point f(x) = x2 + 4 and f(y) = y2 + 4 on the off chance that f(x) = f(y) x2 + 4 = y2 + 4 or x = y f...
Consider f : R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
solution: Think about f : R → R given by f(x) = 4x + 3 Say, x, y ∈ R Let f(x) = f(y) then, at that point 4x + 3 = 4y + 3 x = y f is one-one capacity. Let y ∈ Range of f y = 4x + 3 or on the other...
Show that f : [–1, 1] → R, given by f (x) = x/(x+2) is one-one. Find the inverse of the function f : [–1, 1] → Range f. some x in [–1, 1], i.e., x = 2y/(1-y).
solution: Given capacity: (x) = x/(x+2) Let x, y ∈ [–1, 1] Let f(x) = f(y) x/(x+2) = y/(y+2) xy + 2x = xy + 2y x = y f is one-one. Once more, Since f : [–1, 1] → Range f is onto say, y = x/(x+2) yx...
State with reason whether following functions have inverse
(i) f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)} (ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)} (iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with...
If f(x) = (????????+????)/(60-4) , x ≠ 2/3, Show that fof(x) = x, for all x ≠ 2/3. What is the inverse of f.
solution: f(x) = (4????+3)/(60-4), x ≠ 2/3, (6????−4) In this way, fof(x) = x for all x ≠ 2/3. Once more, fof = I The backwards of the given capacity, f will be f.
Find gof and fog, if
(i) f(x) = | x | and g(x) = | 5x – 2 | (ii) f(x) = 8x3 and g(x) = x1/3 . solution: (I) f(x) = | x | and g(x) = | 5x – 2 | gof =(gof)(x) = g(f(x) = g(|x|) = |5 |x| - 2| mist = (fog)(x) = f(g(x)) =...
Let f, g and h be functions from R to R. Show that (f + g) oh = foh + goh (f . g) oh = (foh) . (goh)
solution: LHS = (f + g) gracious = (f+g)(h(x)) = f(h(x)) + g(h(x)) = foh + goh = RHS Once more, LHS = (f . g) gracious = f.g(h(x)) = f(h(x)) . g(h(x)) = (foh) . (goh) =...
Let f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3} be given by and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.
solution: Given capacity, f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)} Find gof. At f(1) = 2 and g(2) = 3, gof is...
Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6}. Choose the correct answer. (A) (2, 4) ∈ R (B) (3, 8) ∈ R (C) (6, 8) ∈ R (D) (8, 7) ∈ R
solution: R = {(a, b) : a = b – 2, b > 6} (A) Incorrect : Value of b = 4, false. (B) Incorrect : a = 3 and b = 8 > 6 a = b – 2 => 3 = 8 – 2 and 3 = 6, which is bogus. (C) Correct: a = 6 and b...
Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1, L2) : L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.
solution: Once more, The arrangement of all lines identified with the line y = 2x + 4, is the arrangement of all its equal lines. Incline of given line is m = 2. As we probably are aware slant of...
Show that the relation R defined in the set A of all polygons as R = {(P1, P2) :P1 and P2 have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?
solution: Case I: R = {(P1, P2) :P1 and P2 have same number of sides} Check for reflexive: P1 and P1 have same number of sides, So R is reflexive. Check for symmetric: P1 and P2 have same number of...
Show that the relation R defined in the set A of all triangles as R = {(T1, T2) : T1 is similar to T2}, is equivalence relation. Consider three right angle triangles T1 with sides 3, 4, 5, T2 with sides 5, 12, 13 and T3 with sides 6, 8, 10. Which triangles among T1, T2 and T3 are related?
solution: Case I: T1, T2 are triangle. R = {(T1, T2): T1 is like T2} Check for reflexive: As We realize that every triangle is like itself, so (T1, T1) ∈ R is reflexive. Check for symmetric:...
Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.
solution: R = {(P, Q): distance of the point P from the beginning is equivalent to the distance of the point Q from the origin} Say "O" is beginning Point. Since the distance of the point P from the...
Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by R = {(a, b) : |a – b| is a multiple of 4} ;R = {(a, b) : a = b} is an equivalence relation. Find the set of all elements related to 1 in each case.
solution: (I) A = {x ∈ Z : 0 ≤ x ≤ 12} In this way, A = {0, 1, 2, 3, … … , 12} Presently R = {(a, b) : |a – b| is a different of 4} R = {(4, 0), (0, 4), (5, 1), (1, 5), (6, 2), (2, 6), ….., (12, 9),...
Show that the relation R in the set A = {1, 2, 3, 4, 5} given by = {(a, b) : |a – b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}
Solution:A = {1, 2, 3, 4, 5} and R = {(a, b) : |a – b| is even} We get, R = {(1, 3), (1, 5), (3, 5), (2, 4)} For (a, a), |a – b| = |a – a| = 0 is even. Therfore, R is reflexive. If |a – b| is even,...
Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y) : x and y have same number of pages} is an equivalence relation.
solution: Books x and x have same number of pages. (x, x) ∈ R. R is reflexive. In the event that (x, y) ∈ R and (y, x) ∈ R, so R is symmetric. Since, Books x and y have same number of pages and...
Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.
solution: R = {(1, 2), (2, 1)} (x, x) ∉ R. R isn't reflexive. (1, 2) ∈ R and (2,1) ∈ R. R is symmetric. Once more, (x, y) ∈ R and (y, z) ∈ R then, at that point (x, z) doesn't suggest to R. R isn't...
Check whether the relation R in R defined by R = {(a, b) : a ≤ b3} is reflexive, symmetric or transitive.
solution: R = {(a, b) : a ≤ b3} a ≤ a3: which is valid, (a, a) ∉ R, So R isn't reflexive. a ≤ b3 however b ≤ a3 (bogus): (a, b) ∈ R yet (b, a) ∉ R, So R isn't symmetric. Once more, a ≤ b3 and b ≤ c3...
Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric.
Solution: a ≤ a: which is valid, (a, a) ∈ R, So R is reflexive. a ≤ b yet b ≤ a (bogus): (a, b) ∈ R however (b, a) ∉ R, So R isn't symmetric. Once more, a ≤ b and b ≤ c then a ≤ c : (a, b) ∈ R and...
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.
solution: R = {(a, b) : b = a + 1} R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} At the point when b = a, a = a + 1: which is bogus, So R isn't reflexive. In the event that (a, b) = (b,a), b = a+1...
Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b2} is neither reflexive nor symmetric nor transitive.
Solution: R = {(a, b) : a ≤ b2} , Relation R is characterized as the arrangement of genuine numbers. (a, a) ∈ R then a ≤ a2 , which is bogus. R isn't reflexive. (a, b)=(b, a) ∈ R then a ≤ b2 and b ≤...
Determine whether each of the following relations are reflexive, symmetric and transitive:
Relation R in the set A of human beings in a town at a particular time given by R = {(x, y) : x and y work at the same place}R = {(x, y) : x and y live in the same locality}R = {(x, y) : x is...
Determine whether each of the following relations are reflexive, symmetric and transitive:
Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is divisible by x}Relation R in the set Z of all integers defined as R = {(x, y) : x – y is an integer} (iii) R = {(x, y) : y is...
Determine whether each of the following relations are reflexive, symmetric and transitive:
Relation R in the set A = {1, 2, 3,….... , 13, 14} defined as R = {(x, y) : 3x – y = 0Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4} Solution: (i)R =...
Let f : R → R be defined as f(x) = 3x. Choose the correct answer.
(A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto. solution: f : R → R be...
Let f : R → R be defined as f(x) = x4. Choose the correct answer.
(A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto. Solution: f : R...
Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by f(x) = (x-2)/(x-3). Is f one-one and onto? Justify your answer. Solution: A = R – {3} and B = R – {1}
f : A → B characterized by f(x) = (x-2)/(x-3) Let (x, y) ∈ A then, at that point ????(????) = ???? − 2 ???? − 3 ???????????? ????(????) = ???? − 2 ???? − 3 For f(x) = f(y) Once more, f(x) =...
Let f : N → N be defined by and State whether the function f is bijective. Justify your answer
Arrangement: For n = 1, 2 f(1) = (n+1)/2 = (1+1)/2 = 1 and f(2) = (n)/2 = (2)/2 = 1 f(1) = f(2), yet 1 ≠ 2 f isn't one-one. For a characteristic number, "a" in co-space N In case n is odd n = 2k + 1...
Let A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function.
Solution: Stage 1: Check for Injectivity: Let (a1, b1) and (a2, b2) ∈ A x B with the end goal that f (a1, b1) = (a2, b2) This suggests, (b1 , a1) and (b2, a2 ) b1 = b2 and a1 = a2 (a1, b1) = (a2,...
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
f : R → R defined by f(x) = 3 – 4xf : R → R defined by f(x) = 1 + x2 Solution: (i) f : R → R characterized by f(x) = 3 – 4x On the off chance that x1, x2 ∈ R f(x1) = 3 – 4x1 and f(x2) = 3 – 4x2 On...
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is on
Solution: A = {1, 2, 3} B = {4, 5, 6, 7} and f = {(1, 4), (2, 5), (3, 6)} f(1) = 4, f(2) = 5 and f(3) = 6 Here, additionally particular components of A have unmistakable pictures in B. In this...
Show that the Modulus Function f : R → R, given by f(x) = | x |, is neither one-one nor onto, where | x | is x, if x is positive or 0 and |x | is – x, if x is negative.
solution: f : R → R, given by f(x) = | x |, characterized as f contains values like (- 1, 1),(1, 1),(- 2, 2)(2,2) f(- 1) = f(1), yet - 1 f isn't one-one. R contains some adverse numbers which are...
Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one- one nor onto, where [x] denotes the greatest integer less than or equal to x.
Solution: Capacity f : R → R, given by f(x) = [x] f(x) = 1, since 1 ≤ x ≤ 2 f(1.2) = [1.2] = 1 f(1.9) = [1.9] = 1 Yet, 1.2 ≠ 1.9 f isn't one-one. There is no division appropriate or inappropriate...
Check the injectivity and surjectivity of the following functions:
f : N → N given by f(x) = x2f : Z → Z given by f(x) = x2f : R → R given by f(x) = x2f : N → N given by f(x) = x3f : Z → Z given by f(x) = x3 Solution: (i) f : N → N given by f(x) = x2 For x, y ∈ N...
Show that the function f : R∗ → R∗ defined by f(x) = 1/x is one-one and onto, where R∗ is the set of all non-zero real numbers. Is the result true, if the domain R∗ is replaced by N with co-domain being same as R∗?
SOLUTON: Given: f : R∗ → R∗ characterized by f(x) = 1/x Check for One-One ????(????1) = ???? ???????????? ????(????2) = ????1 ???????? ????(????1) = ????(????2) ????ℎ???????? ????= x2 This infers...