## NCERT Solutions Class 12th Maths Chapter 1: Relations and Functions## Exercise 1.11. 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 = 0}**Relation R in the set N of natural numbers defined as** R = {(x, y) : y = x + 5 and x < 4}**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}**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 exactly 7 cm taller than y}****R = {(x, y) : x is wife of y}****R = {(x, y) : x is father of y}**
R = {(x, y): 3x - y = 0} Therefore, R = {(1, 3), (2, 6), (3, 9), (4, 12)}. R is not reflexive as (1, 1), (2, 2), (3, 3), …, (13, 13), (14, 14) ∉ R. R is not symmetric as (1, 3) ∈ R but (3, 1) ∉ R. R is not transitive as (1, 3), (3, 9) ∈ R but (1, 9) ∉ R.
Therefore, R = {(1, 6), (2, 7), (3, 8)} R is not reflexive as (1, 1), (2, 2), … ∉ R. R is not symmetric as (1, 6) ∈ R but (6, 1) ∉ R. R is not transitive as there is no pair as (x, y) and (y, z) ∈ R.
R = {(x, y): y is divisible by x} Therefore, R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (5, 5), (6, 6)} R is reflexive as (1, 1), (2, 2), (3, 3), …, (6, 6) ∈ R. R is not symmetric as (1, 2) ∈ R but (2, 1) ∉ R. R is transitive as (x, y), (y, z) ∈ R, which implies that z is divisible by x, so (x, z) ∈ R.
R is reflexive as for any integer x ∈ Z, (x, x) ∈ R because x - x = 0 which is an integer. R is symmetric as for any (x, y) ∈ R, (y - x) = - (x - y) which is an integer. So, (y, x) ∈ R. R is transitive as for (x, y), (y, z) ∈ R, (x - z) = (x - y) + (y - z) which are both integers. So, (x, z) ∈ R.
R is reflexive as (x, x) ∈ R. R is symmetric as for (x, y) ∈ R, it implies that y and x also work at the same place. So, (y, x) ∈ R. R is transitive as for (x, y), (y, z) ∈ R, it implies that x and z also work at the same place. So, (x, z) ∈ R.
R is reflexive as (x, x) in R R is symmetric as for (x, y) ∈ R, it implies that y and x also live in the same locality. So, (y, x) ∈ R. R is transitive as for (x, y), (y, z) ∈ R, it implies that x and z also live in the same locality. So, (x, z) ∈ R.
R is not reflexive as (x, x) ∉ R because same person cannot have different heights. R is not symmetric as (x, y) ∈ R implies that y is taller than x. So, (y, x) ∉ R. R is not transitive as (x, y), (y, z) ∈ R implies that y is 7 cm taller than x and z is 7 cm taller than y which means that z is 14 cm taller than x. So, (x, z) ∉ R.
R is not reflexive as (x, x) ∉ R because same person cannot have themselves as their wife. R is not symmetric as (x, y) ∈ R implies that x is the wife of y which means that y is the husband of x. So, (y, x) ∉ R. R is not transitive as (x, y), (y, z) ∈ R implies that x is the wife of y and y is the wife of z which is not possible. So, (x, z) ∉ R as well.
R is not reflexive as (x, x) ∉ R because the same person cannot be the father of himself. R is not symmetric as (x, y) ∈ R implies that x is the father of y which means that y is the offspring of x. So, (y, x) ∉ R. R is not transitive as (x, y), (y, z) ∈ R implies that x is the father of y and y is the father of z which means that x is the grandfather of z. So, (x, z) ∉ R. 2. Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b For a real number 1/x, (1/x) So, (1/x, 1/x) ∉ R. Therefore, R is not reflexive. R is not symmetric as (1, 4) ∈ R implies that 1 < 4 R is not transitive as (3, 2), (2, 1.5) ∈ R implies that 3 < 2 Hence, R is neither reflexive, nor symmetric, nor transitive. 3. 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.
R = {(a, b): b = a + 1} Therefore, R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} R is not reflexive as (1, 1), (2, 2), …, (6, 6) ∉ R. R is not symmetric as (1, 2) ∈ R but (2, 1) ∉ R. R is not transitive as (1, 2), (2, 3) ∈ R but (1, 3) ∉ R. 4. Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric.
R = {(a, b): a ≤ b} R is reflexive as a = a so (a, a) ∈ R. R is not symmetric as (a, a + 1) ∈ R implies that a ≤ a + 1 but (a + 1, a) ∈ R implies that a + 1 ≤ a which is false. So, (a + 1, a) ∉ R. R is transitive as (a, b), (b, c) ∈ R implies that a ≤ b ≤ c which means that a ≤ c. So, (a, c) ∈ R. Hence, R is reflexive and transitive but not symmetric. 5. Check whether the relation R in R defined by R = {(a, b) : a ≤ b
R = {(a, b): a ≤ b For a real number 1/x, (1/x) So, (1/x, 1/x) ∉ R. Therefore, R is not reflexive. R is not symmetric as (1, 4) ∈ R implies that 1 < 4 R is not transitive as (3, 1.5), (1.5, 1.2) ∈ R implies that 3 < (1.5) Hence, R is neither reflexive, nor symmetric, nor transitive. 6. 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.
Let A = {1, 2, 3} Relation R on set A = R = {(1, 2), (2, 1)} R is not reflexive as (1, 1), (2, 2), (3, 3) ∉ R. R is symmetric as (1, 2) ∈ R and (2, 1) ∈ R. R is not transitive as (1, 2), (2, 1) ∈ R but (1, 1) ∉ R. Hence, R = {(1, 2), (2, 1)} is symmetric but neither reflexive not transitive. 7. 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.
R = {(x, y): x and y have same number of pages} R is reflexive as one book cannot have two different numbers of pages, so (x, x) ∈ R. R is symmetric as (x, y) ∈ R implies that x and y have the same number of pages which means y and x also have same number of pages. So, (y, x) ∈ R. R is transitive as (x, y), (y, z) ∈ R implies that x and y have the same number of pages and y & z have the same number of pages which means that x and x also have the same number of pages. So, (x, z) ∈ R. Hence, R s an equivalence relation. 8. Show that the relation R in the set A = {1, 2, 3, 4, 5} given by
A = {1, 2, 3, 4, 5} R = {(a, b): |a - b| is even} R is reflexive as for any element a ∈ A, |a - a| = 0 which is even, so (a, a) ∈ R. R is symmetric as (a, b) ∈ R implies that |a - b| is even |-(a - b)| = |b - a| is also even So (b, a) ∈ R. R is transitive as (a, b), (b, c) ∈ R implies that |a - b| and |b - c| are even |a - b + b - c| is also even = |a - c| is even So (a, c) ∈ R. Therefore, R is an equivalence relation. Elements of {1, 3, 5} are related to each other because each element is odd. Modulus of difference between any two elements will be even. Elements of {2, 4} are related to each other because each element is even. Modulus of difference between any two elements will be even. Since, elements of {1, 3, 5} are odd and elements of {2, 4} are even. Therefore, there is no relation between any element of {1, 3, 5} to any element of {2, 4}. Modulus of difference between two elements (one from each set) will not be odd. 9. 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}**
A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
R is reflexive as for any element a ∈ A, |a - a| = 0 which is a multiple of 4, so (a, a) ∈ R. R is symmetric as (a, b) ∈ R implies that |a - b| is a multiple of 4 |-(a - b)| = |b - a| is also a multiple of 4 So (b, a) ∈ R. R is transitive as (a, b), (b, c) ∈ R implies that |a - b| and |b - c| are multiples of 4 |a - b + b - c| is also a multiple of 4 = |a - c| is a multiple of 4 So (a, c) ∈ R. Therefore, R is an equivalence relation. The set of elements related to 1 is {1, 5, 9} since |1 - 1| = 0 which is a multiple of 4 |5 - 1| = 4 which is a multiple of 4 |9 - 1| = 8 which is a multiple of 4
R is reflexive as for any element a ∈ A, a = a, so (a, a) ∈ R. R is symmetric as (a, b) ∈ R implies that a = b which means that b = a So (b, a) ∈ R. R is transitive as (a, b), (b, c) ∈ R implies that a = b and b = c which means that a = c So (a, c) ∈ R. Therefore, R is an equivalence relation. The set of elements related to 1 is {1} since only the elements that are equal to 1 are related to 1. 10. Give an example of a relation. Which is **Symmetric but neither reflexive nor transitive.****Transitive but neither reflexive nor symmetric.****Reflexive and symmetric but not transitive.****Reflexive and transitive but not symmetric.****Symmetric and transitive but not reflexive.**
R = {(1, 2), (2, 1)} R is not reflexive as (1, 1), (2, 2), (3, 3), (4, 4) ∉ R. R is symmetric as (1, 2) ∈ R as well as (2, 1) ∈ R. R is not transitive as (1, 2), (2, 1) ∈ R but (1, 1) ∉ R.
R = {(a, b): a > b} R is not reflexive as for any element a ∈ A, a > a is false because a = a. So, (a, a) ∉ R. R is not symmetric as (a, b) ∈ R implies that a > b which makes b > a impossible. So, (b, a) ∉ R. R is transitive as (a, b), (b, c) ∈ R implies that a > b and b > c which means that a > c. So, (a, c) ∈ R.
R = {(1, 1), (2, 2), (3, 3), (2, 1), (1, 2), (3, 2), (2, 3)} R is reflexive as (1, 1), (2, 2), (3, 3) ∈ R. R is symmetric as (a, b) ∈ R as well as (b, a) ∈ R ∀ a, b ∈ R. R is not transitive as (1, 2), (2, 3) ∈ R but (1, 3) ∉ R.
R = {(a, b): a R is reflexive as a R is not symmetric as (a, b) ∈ R implies that a R is transitive as (a, b), (b, a) ∈ R implies that a
R = {(1, 2), (2, 1), (1, 1)} R is not reflexive as (2, 2) ∉ R. R is symmetric as (1, 2) in R as well as (2, 1) ∈ R. R is transitive as (1, 2), (2, 1) ∈ R and also (1, 1) ∈ R. 11. 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.
R = {(P, Q): distance of the point P from the origin is same as the distance of the point Q from the origin} R is reflexive as the distance of a point P from the origin is fixed and remains the same. So, (P, P) ∈ R. R is symmetric as (P, Q) ∈ R implies that the distance of the point P from the origin is same as the distance of the point Q from the origin Which means that distance of point Q from the origin is same as distance of the point P from the origin. So, (Q, P) ∈ R. R is transitive as (P, Q), (Q, R) ∈ R implies that distance of P from the origin is equal to the distance of Q from the origin and the distance of Q from the origin is equal o the distance of R from the origin Which means that distance of P from the origin is same as distance of R from the origin. So, (P, R) ∈ R. Therefore, R is an equivalence relation. The set of points related to P ≠ (0, 0) will be those points which are at the same distance from the origin as the point P from the origin, i.e., for the origin O (0, 0), the distance of each point of the set will be equal to OP. Hence, the set of points forms a circle passing through point P with origin as the centre. 12. Show that the relation R defined in the set A of all triangles as R = {(T
R = {(T R is reflexive as any triangle T R is symmetric as (T R is transitive as (T Therefore, R is an equivalence relation. In triangle T 3/5 ≠ 4/12 T In triangle T 5/6 ≠ 12/8 T In triangle T 3/6 = 4/8 = 5/10 = T Hence, T 13. Show that the relation R defined in the set A of all polygons as R = {(P
R = {(P R is reflexive as the number of sides a polygon P R is symmetric as (P R is transitive as (P Therefore, R is an equivalence relation. T is a right-angled triangle with three sides 3, 4 and 5 so, the set of elements in A that are related to T are the polygons that have three sides, i.e., the set of all triangles. 14. Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L
R = {(L R is reflexive as any line L R is symmetric as (L R is transitive as (L Therefore, R is an equivalence relation. The set of lines related to the line y = 2x + 4 are the ones that are parallel to it, i.e., the set of lines related to y = 2x + 4 is given by y = 2x + c, c ∈ 15. Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer. **R is reflexive and symmetric but not transitive.****R is reflexive and transitive but not symmetric.****R is symmetric and transitive but not reflexive.****R is an equivalence relation.**
A = {1, 2, 3, 4} R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)} R is reflexive as (1, 1), (2, 2), (3, 3), (4, 4) ∈ R. R is not symmetric as (1, 2) ∈ R but (2, 1) ∉ R. R is transitive as for each element a, b, c ∈ A, (a, b), (b, c) ∈ R then (a, c) ∈ R as well. Therefore, R is reflexive and transitive but not symmetric. Hence, (B) is the correct answer. 16. Let R be the relation in the set N given by R = {(a, b) : a = b - 2, b > 6}. Choose the correct answer. **(2, 4) ∈ R****(3, 8) ∈ R****(6, 8) ∈ R****(8, 7) ∈ R**
R = {(a, b): a = b - 2, b > 6} (2, 4) ∉ R as b > 6 (3, 8) ∉ R as 3 ≠ 8 - 2 = 6 (6, 8) ∈ R as 6 = 8 - 2 = 6 (8, 7) ∉ R as 8 ≠ 7 - 2 = 5 Hence, (C) is the correct answer. ## Exercise 1.21. 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∗?
f: R∗ → R∗ is defined by f (x) = 1/x Check for One-one: f (x) = f (y) 1/x = 1/y x = y Therefore, f is one-one Check for Onto: For any y ∈ R∗, x = 1/y ∈ R∗ also exists such that f (x) = 1/(1/y) f (x) = y Therefore, f is onto. Thus, the function f is one-one and onto. g: N → R∗ is defined by g (x) = 1/x Check for One-one: g (x) = g (y) 1/x = 1/y x = y Therefore, g is one-one. Check for Onto: g is not onto because for 1.2, 1.3, 1.4, … ∈ R∗ no x exists in N such that g (x) = 1/1.2, g (x) = 1/1.3, … etc Hence, the function is one-one only and not onto anymore. 2. Check the injectivity and surjectivity of the following functions: **f : N → N given by f (x) = x**^{2}**f : Z → Z given by f (x) = x**^{2}**f : R → R given by f (x) = x**^{2}**f : N → N given by f (x) = x**^{3}**f : Z → Z given by f (x) = x**^{3}
For x, y ∈ N, f (x) = f (y) x x = y Therefore, f is injective. For 2 ∈ N, there exists no x in N such that f (x) = x Therefore, f is not surjective.
f (-1) = f (1) (-1) But -1 ≠ 1 Therefore, f is not injective. For -2 ∈ Z, there exists no x in Z such that f (x) = x Therefore, f is not surjective.
f (-1) = f (1) (-1) But -1 ≠ 1 Therefore, f is not injective. For -2 ∈ R, there exists no x in R such that f (x) = x Therefore, f is not surjective.
For x, y ∈ N, f (x) = f (y) x x = y Therefore, f is injective. For 2 ∈ N, there exists no x in N such that f (x) = x Therefore, f is not surjective.
For x, y ∈ Z, f (x) = f (y) x x = y Therefore, f is injective. For 2 ∈ Z, there exists no x in Z such that f (x) = x Therefore, f is not surjective. 3. 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.
f: R → R is given by f (x) = [x] Check for One-one: f (1.5) = [1.5] = 1 f (1.9) = [1.1] = 1 f (1.5) = f (1.9) But 1.5 ≠ 1.9 Therefore, f is not one-one. Check for Onto: For 0.5 ∈ R, there exists no x ∈ R such that f (x) = 0.7 because [x] is an integer. Therefore, f is not onto. Hence, the Greatest Integer Function is neither one-one nor onto. 4. 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.
f: R → R is given by f (x) = |x| = Check for One-one: f (-1) = |-1| = 1 f (1) = |1| = 1 f (-1) = f (1) But -1 ≠ 1 Therefore, f is not one-one. Check for Onto: For -1 ∈ R, there exists no element x ∈ R such that f (x) = |x| = -1 because |x| is always positive. Therefore, f is not onto. Hence, the modulus function is neither one-one nor onto. 5. Show that the Signum Function f : R → R, given by
Check for One-one f (1) = 1 f (2) = 1 f (1) = f (2) But 1 ≠ 2 Therefore, f is not one-one. Check for Onto: f only takes 3 values (1, 0, -1). So, for -2 ∈ R, there exists no element x ∈ R such that f (x) = -2 Therefore, f is not onto. Hence, the signum function is neither one-one nor onto. 6. 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 one-one.
A = {1, 2, 3} and B = {4, 5, 6, 7} f: A → B is given by f = {(1, 4), (2, 5), (3, 6)} f (1) = 4 f (2) = 5 f (3) = 6 Each element of A has a distinct image in the function f. Therefore, f is one-one. 7. 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 - 4x****f : R → R defined by f (x) = 1 + x**^{2}
Check for One-one: Take any two elements x f (x 3 - 4x -4x x Therefore, f is one-one. Check for Onto: For any real y ∈ R, there exists (3 - y)/4 ∈ R such that f ((3 - y)/4) = 3 - 4(3 - y)/4 = 3 - 3 + y = y Therefore, f is onto. Hence, f is bijective.
Check for One-one: Take any two elements x f (x 1 + x x x x x Therefore, f is not one-one. Check for Onto: For -2 ∈ R, there exists no element x ∈ R such that f (x) = -2 because f (x) = 1 + x Therefore f is not onto. Hence, f is neither one-one nor onto. 8. Let A and B be sets. Show that f : A × B → B × A such that f (a, b) = (b, a) is bijective function.
f: A × B → B × A is defined as f (a, b) = (b, a) Check for One-one: Take (a f (a (b b (a Therefore, f is one-one. Check for Onto: For (b, a) ∈ B × A, there exists (a, b) ∈ A × B such that f (a, b) = (b, a) Therefore, f is onto. Hence, f is bijective. 9. Let f : N → N be defined by f (n) = for all n ∈ N.
f: N → N is defined by f (n) = Check for One-one: f (1) = (1 + 1)/2 = 2/2 = 1 f (2) = 2/2 = 1 f (1) = f (2) But 1 ≠ 2 Therefore, f is not one-one. Check for Onto: For n ∈ N, If n is odd: n = 2r + 1 for some r ∈ N, there exists 4r + 1 ∈ N such that f (4r + 1) = (4r + 1 + 1)/2 = 2(2r + 1)/2 = 2r + 1 If n is even: n = 2r for some r ∈ N, there exists 4r ∈ N such that f (4r) = 4r/2 = 2r Therefore, f is onto. Hence, f is not bijective. 10. 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.
A = R - {3} and B = R - {1} f: A → B is defined by f (x) = (x - 2)/(x - 3) Check for One-one: Take any two elements x, y ∈ A such that f (x) = f (y) (x - 2)/(x - 3) = (y - 2)/(y - 3) (x - 2)(y - 3) = (y - 2)(x - 3) xy - 2y - 3x + 6 = xy - 2x - 3y + 6 -3x - 2y = -2x - 3y 3x - 2x = 3y - 2y x = y Therefore, f is one-one. Check for Onto: For y ∈ B, y ≠ 1. If there exists x ∈ A such that f (x) = y, then the function f is onto. f (x) = y (x - 2)/(x - 3) = y x - 2 = y(x - 3) x - 2 = xy - 3y x - xy = 2 - 3y x(1 - y) = 2 - 3y x = (2 - 3y)/(1 - y) ∈ A Therefore, for y ∈ B, there exists (2 - 3y)/(1 - y) ∈ A such that f ((2 - 3y)/(1 - y)) = [(2 - 3y)/(1 - y) - 2]/[(2 - 3y)/(1 - y) - 3] = [(2 - 3y - 2 - 2y)/(1 - y)]/[(2 - 3y - 3 - 3y)/(1 - y)] = -y/-1 = y Therefore, f is onto. Hence, the function f is both one-one and onto. 11. Let f : R → R be defined as f (x) = x **f is one-one onto****f is many-one onto****f is one-one but not onto****f is neither one-one nor onto**
f: R → R is defined as f (x) = x Check for One-one: f (-1) = (-1) f (1) = (1) f (-1) = f (1) (-1) But -1 ≠ 1 Therefore, f is not one-one. Check for Onto: For -2 ∈ R, there exists no x in R such that f (x) = x Therefore, f is not onto. Hence, the correct answer is (D). 12. Let f : R → R be defined as f (x) = 3x. Choose the correct answer. **f is one-one onto****f is many-one onto****f is one-one but not onto****f is neither one-one nor onto**
f: R → R is defined as f (x) = 3x Check for One-one: Take x, y ∈ R such that f (x) = f (y) 3x = 3y x = y Therefore, f is one-one. Check for Onto: For y ∈ R, there exists y/3 ∈ R such that f (y/3) = 3 × y/3 = y Therefore, f is onto. Hence, the correct answer is (A). ## Exercise 1.31. Let 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)}. Write down gof.
f: {1, 3, 5} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} are given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3) (2, 3), (5, 1)} f (1) = 2 and g(2) = 3 gof = g(f (1)) = g(2) = 3 f (3) = 5 and g(5) = 1 gof (3) = g(f (3)) = g(5) = 1 f (4) = 1 and g(1) = 1 gof (4) = g(f (4)) = g(1) = 3 Hence, gof = {(1, 3), (3, 1), (4, 3)} 2. Let f, g and h be functions from R to R. Show that
(f + g) oh = (f + g)(h (x)) = f (h (x)) + g(h (x)) = foh + goh LHS = RHS Therefore, proved that (f + g) oh = foh + goh (f . g) oh = f.g(h (x)) = f(h (x)). g(h (x)) = (foh) . (goh) = RHS LHS = RHS Therefore, proved that (f . g) oh = (foh) . (goh) 3. Find gof and fog, if **f (x) = |x| and g (x) = |5x - 2|****f (x) = 8x**^{3}and g (x) = x^{1/3}
gof (x) = g(f (x)) = g(|x|) = |5|x| - 2| fog (x) = f(g (x)) = f (|5x - 2|) = ||5x - 2|| = |5x - 2|
gof (x) = g(f (x)) = g(8x = (8x fog (x) = f(g (x)) = f (x = 8(x 4. If f (x) = (4x + 3)/(6x - 4), x ≠ 2/3, show that f o f (x) = x for all x ≠ 2/3. What is the inverse of f?
f (x) = (4x + 3)/(6x - 4), x ≠ 2/3 fof (x) = f(f (x)) = f ((4x + 3)/(6x - 4)) = [4(4x + 3)/(6x - 4) + 3]/[6(4x + 3)/(6x - 4) - 4] = [(16x + 12 + 18x - 12)/(6x - 4)]/[(24x + 18 - 24x + 16)/(6x - 4)] = 34x/34 = x Therefore, fof (x) = x for all x ≠ 2/3. Hence, fof = 1. The given function f is invertible and its inverse is f itself. 5. State with reason whether following functions have inverse **f: {1, 2, 3, 4} → {10} with** f = {(1, 10), (2, 10), (3, 10), (4, 10)}**g: {5, 6, 7, 8} → {1, 2, 3, 4} with** g = {(5, 4), (6, 3), (7, 4), (8, 2)}**h: {2, 3, 4, 5} → {7, 9, 11, 13} with** h = {(2, 7), (3, 9), (4, 11), (5, 13)}
f = {(1, 10), (2, 10), (3, 10), (4, 10)} f (1) = f (2) = f (3) = f (4) = 10 So, f is a many one function. Therefore, f is not one-one which implies that it does not have an inverse.
g = {(5, 4), (6, 3), (7, 4), (8, 2)} g(5) = g(7) = 4 So, g is a many one function. Therefore, g is not one-one which implies that it does not have an inverse.
h = {(2, 7), (3, 9), (4, 11), (5, 13)} Each element of the set {2, 3, 4, 5} has a distinct image under the function h. So, h is one-one. For every element y in the set {7, 9, 11, 13}, there exists an element x ∈ {2, 3, 4, 5} such that h(x) = y. So, h is onto. Therefore, h is one-one and onto which implies that it has an inverse. 6. 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.
f: [-1, 1] → R is given by f (x) = x/(x + 2) Let f (x) = f (y) x/(x + 2) = y/(y + 2) x(y + 2) = y(x + 2) xy + 2x = xy + 2y 2x = 2y x = y Therefore, f is one-one. The function f: [-1, 1] → Range f is clearly an onto function. Thus, the function f: [-1, 1] → Range f is both one-one and onto which implies that the inverse of the function f: [-1, 1] → Range f exists. Let function g: Range f → [-1, 1] be the inverse of f: [-1, 1] → Range f. Let y be an arbitrary element of the range f. So, y = f (x) for x ∈ [-1, 1] y = x/(x + 2) xy + 2y = x xy - x = -2y x(y - 1) = -2y x(1 - y) = 2y x = 2y/(1 - y), y ≠ 1 g: Range f → [-1, 1] is defined as g (y) = 2y/(1 - y), y ≠ 1 Therefore, gof (x) = g(f (x)) = g (x/(x + 2)) = 2(x/(x + 2))/(1 - x/(x + 2)) = [2x/(x + 2)]/[(x + 2 - x)/(x + 2)] = 2x/(x + 2 - x) = 2x/2 = x fog (x) = f(g (x)) = f (2y/(1 - y)) = [2y/(1 - y)]/[2y/(1 - y) + 2] = [2y(1 - y)]/[(2y + 2 - 2y)/(1 - y)] = 2y/(2y - 2y + 2) = 2y/2 = y Therefore, gof = I Thus, f f 7. Consider f: R → R is given by f (x) = 4x + 3. Show that f is invertible. Find the inverse of f.
f: R → R is given by f (x) = 4x + 3 Let f (x) = f (y) 4x + 3 = 4y + 3 4x = 4y x = y Therefore, f is one-one. For some y ∈ R, let y = 4x + 3 4x = y - 3 x = (y - 3)/4 ∈ R For y ∈ R, there exists x = (y - 3)/4 ∈ R such that f (x) = f ((y - 3)/4) = 4 (y - 3)/4 + 3 = y - 3 + 3 = y Therefore, f is onto. Thus, f is both one-one and onto which implies that the inverse of f exists. Let a function g: R → R be defined by g (y) = (y - 3)/4 gof (x) = g(f (x)) = g(4x + 3) = [(4x + 3) - 3]/4 = 4x/4 = x fog (y) = f(g (y)) = f ((y - 3)/4) = 4 (y - 3)/4 + 3 = y - 3 + 3 = y Therefore, gof = I Hence, the inverse of f is given by f 8. Consider f: R
f: R Let f (x) = f (y) x x x = y as x, y ∈ R Therefore, f is one-one. For y ∈ [4, ∞), let y = x x x = √(y - 4) For y ∈ [4, ∞), there exists x = √(y - 4) ∈ R such that f (x) = f (√(y - 4)) = (√(y - 4)) Therefore, f is onto. Thus, f is both one-one and onto which implies that the inverse of f exists. Let a function g: [4, ∞) → R gof (x) = g(f (x)) = g(x fog (y) = f(g (y)) = f (√(y - 4)) = √(y - 4) Therefore, gof = I Hence, the inverse of f is given by f 9. Consider f: R
f: R Let y be an arbitrary element of [-5, ∞) Let y = 9x y = 9x y = (3x) y = (3x + 1) (3x + 1) 3x + 1 = √(y + 6) as y ≥ -5 which implies that y + 6 > 0 3x = √(y + 6) - 1 x = (√(y + 6) - 1)/3 Therefore, f is onto. Let a function g: [-5, ∞) → R gof (x) = g(f (x)) = g (9x = (√((3x + 1) = (√(3x + 1) = (3x + 1 - 1)/3 = 3x/3 = x fog (y) = f(g (y)) = f ([√(y + 6) - 1]/3) = [3(√(y + 6) - 1)/3 + 1] = (√(y + 6)) = y Therefore, gof = I Hence, f is invertible and the inverse of f is given by f 10. Lef f: X → Y be an invertible function. Show that f has unique inverse.
Let f: X → Y be an invertible function. Let us suppose that f has two inverses g fog f(g g g Therefore, g is one-one. Hence, f has a unique inverse. 11. Consider f: {1, 2, 3} → {a, b, c} given by f (1) = a, f (2) = b and f (3) = c. Find f
f: {1, 2, 3} → {a, b, c} is given by f (1) = a, f (2) = b and f (3) = c Let us define g: {a, b, c} → {1, 2, 3} as g (a) = 1, g (b) = 2 and g (c) = 3 then fog (a) = f(g (a)) = f (1) = a fog (b) = f(g (b)) = f (2) = b fog (c) = f(g (c)) = f (3) = c gof (1) = g(f (1)) = g (a) = 1 gof (2) = g(f (2)) = g (b) = 2 gof (3) = g(f (3)) = g (c) = 3 Therefore, goh = I The inverse of g exists g f = h Hence, f = (f 12. Let f: X → Y be an invertible function. Show that the inverse of f
f: X → Y is an invertible function, then there exists a function g: Y → X such that gof = I f gof = I f Hence, f 13. If f: R → R be given by f (x) = (3 - x **1/x**^{3}**x**^{3}**x****(3 - x**^{3})
f: R → R is given by f (x) = (3 - x fof (x) = f(f (x)) = f ((3 - x = [3 - ((3 - x = [3 - 3 + x = (x = x Hence, (C) is the correct answer. 14. Let f: R - {-4/3} → R be a function defined as f (x) = 4x/(3x + 4). The inverse of f is map g: Range f → R - {-4/3} given by **g (y) = 3y/(3 - 4y)****g (y) = 4y/(4 - 3y)****(C) g (y) = 4y/(3 - 4y)****g (y) = 3y/(4 - 3y)**
f: R - {-4/3} → R is defined as f (x) = 4x/(3x + 4) Let y b an arbitrary element of Range f, then there exists x ∈ R - {-4/3} such that f (x) = y y = 4x/(3x + 4) y(3x + 4) = 4x 3xy + 4y = 4x 4x - 3xy = 4y x(4 - 3y) = 4y x = 4y/(4 - 3y) Let there be a function g: Range f → R - {-4/3} defined as g (y) = 4y/(4 - 3y) gof (x) = g(f (x)) = g (4x/(3x + 4)) = 4(4x/(3x + 4))/(4 - 3(4x/(3x + 4))) = [16x/(3x + 4)]/[(4(3x + 4) - 12x)/(3x + 4)] = 16x/(12x + 16 - 12x) = 16x/16 = x fog (y) = f(g (y)) = f (4y/(4 - 3y)) = 4(4y/(4 - 3y))/(3(4y/(4 - 3y)) + 4) = [16y/(4 - 3y)]/[(12y + 4(4 - 3y))/(4 - 3y)] = 16y/(12y + 16 - 12y) = 16y/16 = y Therefore, gof = I Therefore, the inverse of f is map g: Range f → R - {-4/3} given by g (y) = 4y/(4 - 3y). Hence, (B) is the correct answer. ## Exercise 1.41. 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 - b**On Z**^{+}, define * by a * b = ab**On R, define * by a * b = ab**^{2}**On Z**^{+}, define * by a * b = |a - b|**On Z**^{+}, define * by a * b = a
Consider (1, 2) under * 1 * 2 = 1 - 2 = -1 The image of (1, 2) under * is -1 ∉ Z Therefore, * is not a binary operation.
For each a, b ∈ Z Therefore, * is a binary operation.
For each a, b ∈ Z Therefore, * is a binary operation.
For each a, b ∈ Z Therefore, * is a binary operation.
For each a, b ∈ Z Therefore, * is a binary operation. 2. For each binary operation * defined below, determine whether * is commutative or associative. **On Z, define a * b = a - b****On Q, define a * b = ab + 1****On Q, define a * b = ab/2****On Z**^{+}, define a * b = 2^{ab}**On Z**^{+}, define a * b = a^{b}**On R - {-1}, define a * b = a/(b + 1)**
Consider (1, 2) ∈ Z under * 1 * 2 = 1 - 2 = -1 Consider (2, 1) ∈ Z under * 2 * 1 = 2 - 1 = 1 1 * 2 ≠ 2 * 1 Therefore, the operation * is not commutative. Now, (1 * 2) * 3 = (1 - 2) - 3 = -1 - 3 = -4 1 * (2 * 3) = 1 - (2 - 3) = 1 - (-1) = 1 + 1 = 2 (1 * 2) * 3 ≠ 1 * (2 * 3) Therefore, the operation * is not associative.
We know that ab = ba for all a, b ∈ Q ab + 1 = ba + 1 for all a, b ∈ Q a * b = ab + 1 b * a = ba + 1 So, a * b = b * a for all a, b ∈ Q. Therefore, the operation * is commutative. Now, (1 * 2) * 3 = (1 × 2 + 1) × 3 + 1 = (2 + 1) (3) + 1 = 3(3) + 1 = 9 + 1 = 10 1 * (2 * 3) = 1 × (2 × 3 + 1) + 1 = (6 + 1) + 1 = 8 (1 * 2) * 3 ≠ 1 * (2 * 3) Therefore, the operation * is not associative.
We know that ab = ba for all a, b ∈ Q ab/2 = ba/2 for all a, b ∈ Q a * b = ab/2 b * a = ba/2 So, a * b = b * a for all a, b ∈ Q. Therefore, the operation * is commutative. Now, (a * b) * c = (ab/2) * c = (ab/2)c/2 = abc/4 a * (b * c) = a * (bc/2) = a(bc/2)/2 = abc/4 (a * b) * c = a * (b * c) Therefore, the operation * is associative.
We know that ab = ba for all a, b ∈ Z 2 a * b = 2 b * a = 2 So, a * b = b * a for all a, b ∈ Z Therefore, the operation * is commutative. Now, (1 * 2) * 3 = (2 1 * (2 * 3) = 1 * (2 (1 * 2) * 3 ≠ 1 * (2 * 3) Therefore, the operation * is not associative.
Consider (1, 2) ∈ Z 1 * 2 = 1 Consider (2, 1) ∈ Z 2 * 1 = 2 1 * 2 ≠ 2 * 1 Therefore, the operation * is not commutative. Now, (2 * 3) * 4 = 2 2 * (3 * 4) = 2 * 3 (2 * 3) * 4 ≠ 2 * (3 * 4) Therefore, the operation * is not associative.
Consider (1, 2) ∈ R - {-1} under * 1 * 2 = 1/(2 + 1) = 1/3 Consider (2, 1) ∈ r - {-1} under * 2 * 1 = 2/(1 + 1) = 2/2 = 1 1 * 2 ≠ 2 * 1 Therefore, the operation * is not commutative. Now, (1 * 2) * 3 = 1/(2 + 1) * 3 = 1/3 * 3 = (1/3)/(3 + 1) = (1/3)/4 = 1/12 1 * (2 * 3) = 1 * 2/(3 + 1) = 1 * 2/4 = 1 * 1/2 = 1/(1/2 + 1) = 1/(3/2) = 2/3 (1 * 2) * 3 ≠ 1 * (2 * 3) Therefore, the operation * is not associative. 3. Consider the binary operation v on the set {1, 2, 3, 4, 5} defined by a v b = min {a, b}. Write the operation table of the operator v.
Binary operator v on the set {1, 2, 3, 4, 5} is defined by a v b = min {a, b} for all a, b ∈ {1, 2, 3, 4, 5} Therefore, the operation table will be:
4. Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table. **Compute (2 * 3) * 4 and 2 * (3 * 4)****Is * commutative?****Compute (2 * 3) * (4 * 5).**
2 * (3 * 4) = 2 * 1 = 1
(2 * 3) * (4 * 5) = 1 * 1 = 1 5. 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.
The binary operation *' on the set {1, 2, 3, 4, 5} is defined by a *' b = H.C.F of a and b. Therefore, the operation table for the operation *' will be:
The operation table for the operation *' is same as that of operation *. Hence, the operation *' is the same as the operation *. 6. Let * be the binary operation on N given by a * b = L.C.M. of a and b. Find **5 * 7, 20 * 16****Is * commutative?****Is * associative?****Find the identity of * in N.****Which elements of N are invertible for the operation *?**
The binary operation * on N is defined as a * b = L.C.M. of a and b.
20 * 16 = L.C.M. of 20 and 16 = 80
Therefore, a * b = b * a
(a * b) * c = (L.C.M. of a 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, b and c (a * b) * c = a * (b * c) Therefore, the operation * is associative.
a * 1 = 1 * a Therefore, the identity of * in N is 1.
Now, e = 1. So, L.C.M. of a and b = 1 = L.C.M. of b and a This is only possible if a and b are both equal to 1. Therefore, 1 is the only invertible element of N under the operation *. 7. 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.
The operation * on the set {1, 2, 3, 4, 5} is defined as a * b = L.C.M. of a and b. The operation table for the operation * will be:
3 * 2 = 2 * 3 = 6 ∉ {1, 2, 3, 4, 5} 5 * 2 = 2 * 5 = 10 ∉ {1, 2, 3, 4, 5} 3 * 5 = 5 * 3 = 15 ∉ {1, 2, 3, 4, 5} 3 * 4 = 4 * 3 = 12 ∉ {1, 2, 3, 4, 5} 5 * 4 = 4 * 5 = 20 ∉ {1, 2, 3, 4, 5} Hence, the operation * is not a binary operation. 8. Let * be the binary operator 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?
The binary operation * on N is defined as a * b = H.C.F. of a and b H.C.F. of a and b = H.C.F. of b and a for all a, b ∈ N. So, a * b = b * a Therefore, the operation * is commutative. For a, b, c ∈ N: (a * b) * c = (H.C.F. of a and b) * c = H.C.F. of a, b and c a * (b * c) = a * (H.C.F. of b and c) = H.C.F. of a, b and c So, (a * b) * c = a * (b * c) Therefore, the operation * is associative. Now, an element i ∈ N will be the identity for the operation * if a * i = a = i * a for all a ∈ N. But a * i = a = i * a is not true for any a ∈ n. Therefore, he operation * does not have any identity in N. 9. Let * be a binary operation on the set Q of rational numbers as follows: **a * b = a - b****a * b = a**^{2}+ b^{2}**a * b = a + ab****a * b = (a - b)**^{2}**a * b = ab/4****a * b = ab**^{2}
1/2 * 1/3 = 1/2 - 1/3 = (3 - 2)/6 = 1/6 1/3 * 1/2 = 1/3 - 1/2 = (2 - 3)/6 = -1/6 1/2 * 1/3 ≠ 1.3 * 1/2 Therefore, operation * is not commutative. For 1/2, 1/3, 1/4 ∈ Q: (1/2 * 1/3) * 1/4 = (1/2 - 1/3) * 1/4 = 1/6 * 1/4 = 1/6 - 1/4 = (2 - 3)/12 = -1/12 1/2 * (1/3 * 1/4) = 1/2 * (1/3 - 1/4) = 1/2 * (4 - 3)/12 = 1/2 * 1/12 = 1/2 - 1/12 = (6 - 1)/12 = 5/12 (1/2 * 1/3) * 1/4 ≠ 1/2 * (1/3 * 1/4) Therefore, the operation * is not associative.
For a, b ∈ Q: a * b = a b * a = b Clearly, a a * b = b * a Therefore, the operation * is commutative. For 1, 2, 3 ∈ Q: (1 * 2) * 3 = (1 = 5 1 * (2 * 3) = 1 * (2 = 1 (1 * 2) * 3 ≠ 1 * (2 * 3) Therefore, the operation * is not associative.
1 * 2 = 1 + 1 × 2 = 1 + 2 = 3 2 * 1 = 2 + 2 × 1 = 2 + 2 = 4 1 * 2 ≠ 2 * 1 Therefore, the operation * is not commutative. For 1, 2, 3 ∈ Q: (1 * 2) * 3 = (1 + 1 × 2) * 3 = (1 + 2) * 3 = 3 * 3 = 3 + 3 × 3 = 3 + 9 = 12 1 * (2 * 3) = 1 * (2 + 2 × 3) = 1 * (2 + 6) = 1 * 8 = 1 + 1 × 8 = 1 + 8 = 9 (1 * 2) * 3 ≠ 1 * (2 * 3) Therefore, the operation * is not associative.
For a, b ∈ Q: a * b = (a - b) b * a = (b - a) (b - a) a * b = b * a Therefore, the operation * is commutative. For 1, 2, 3, ∈ Q: (1 * 2) * 3 = (1 - 2) = (1 - 3) 1 * (2 * 3) = 1 * (2 - 3) = (1 - 1) (1 * 2) * 3 ≠ 1 * (2 * 3) Therefore, the operation * is not associative.
For a, b ∈ Q: a * b = ab/4 b * a = ba/4 ab = ba ab/4 = ba/4 a * b = b * a Therefore, the operation * is commutative. For a, b, c ∈ Q: (a * b) * c = ab/4 * c = abc/4(4) = abc/16 a * (b * c) = a * bc/4 = abc/4(4) = abc/16 (a * b) * c = a * (b * c) Therefore, the operation * is associative.
For 1/2, 1/3 ∈ Q: 1/2 * 1/3 = 1/2 × (1/3) 1/3 * 1/2 = 1/3 × (1/2) 1/2 * 1/3 ≠ 1/3 * 1/2 Therefore, the operation * is not commutative. For 1/2, 1/3, 1/4 ∈ Q: (1/2 * 1/3) * 1/4 = 1/2 × (1/3) = 1/18 × (1/4) 1/2 * (1/3 * 1/4) = 1/2 * (1/3 × (1/4) = 1/2 × (1/48) (1/2 * 1/3) * 1/4 ≠ 1/2 * (1/3 * 1/4) Therefore, the operation * is not associative. 10. Find which of the operations given above has identity.
An element e ∈ Q will be the identity element for an operation * on Q if a * e = e = e * a for all a ∈ Q. There is no such element in any of the six operations given above. Hence, none of the six operations have identity. 11. Let A = N × N and * be the binary operation on A defined by (a, b) * (c, d) = (a + c, b + d)
A = N × N The operation * on A is defined by (a, b) * (c, d) = (a + c, b + d) For (a, b), (c, d) ∈ A: (a, b) * (c, d) = (a + c, b + d) (c, d) * (a, b) = (c + a, d + b) c + a = a + c and b + d = d + b (a, b) * (c, d) = (c, d) * (a, b) Therefore, the operation * is commutative. For (a, b), (c, d), (e, f) ∈ A: ((a, b) * (c, d)) * (e, f) = (a + c, b + d) * (e, f) = (a + c + e, b + d + f) (a, b) * ((c, d) * (e, f)) = (a, b) * (c + e, d + f) = (a + c + e, b + d + f) ((a, b) * (c, d)) * (e, f) = (a, b) * ((c, d) * (e, f)) Therefore, the operation * is associative. An element e = (e Hence, the operation * does not have any identity element. 12. State whether the following statements are true or false. Justify. **For an arbitrary binary operation * on a set N, a * a = a for all a****∈ N****If * is commutative binary operation on N, then a * (b * c) = (c * b) * a**
For a = b = 3 ∈ N: 3 * 3 = 3 + 3 = 6 ≠ 3 Therefore, the given statement is false.
a * (c * b) = (c* b) * a = RHS as the operation * is commutative Therefore, the given statement is true. 13. Consider a binary operation * on N defined as a * b = a *** is both associative and commutative***** is commutative but not associative***** is associative but not commutative***** is neither commutative nor associative**
The operation * on N is defined as a * b = a For a, b ∈ N: a * b = a b * a = b Clearly, a a * b = b * a Therefore, the operation * is commutative. For 1, 2, 3 ∈ N: (1 * 2) * 3 = (1 = 9 1 * (2 * 3) = 1 * (2 = 1 (1 * 2) * 3 ≠ 1 * (2 * 3) Therefore, the operation * is not associative. Hence, the correct answer is (B). ## Miscellaneous Exercise1. Show that the function f : R → {x ∈ R : - 1 < x < 1} defined by f (x) = x/(1 + |x|), x ∈ R is one one and onto function.
f: R → {x ∈ R: -1 < 1} is defined as f (x) = x/(1 + |x|), x ∈ R Check for One-one: Take x, y ∈ R such that f (x) = f (y) x/(1 + |x|) = y/(1 + |y|) If x is positive and y is negative: x/(1 + x) = y/(1 - y) x(1 - y) = y(1 + x) x - xy = y + xy 2xy = x - y 2xy is negative as y is negative, i.e., 2xy < 0 But, x > 0 and y < 0, so x > y which implies that x - y > 0 Thus, 2xy ≠ x - y If x is negative and y is positive: x/(1 - x) = y/(1 + y) x + xy = y - xy 2xy = y - x 2xy is negative as x is negative, i.e., 2xy < 0 But, y > 0 and x < 0, so y > x which implies that y - x > 0 Thus, 2xy ≠ y - x If x and y are both positive: f (x) = f (y) x/(1 + x) = y/(1 + y) x + xy = y + x x = y If x and y are both negative: f (x) = f (y) x/(1 - x) = y/(1 - y) x - xy = y - xy x = y Therefore, f is one-one. Check for Onto: For y ∈ R such that -1 < y < 1 If y is negative, there exists x = y/(1 + y) ∈ R such that f (x) = f (y/(1 + y)) = (y/(1 + y))/(1 + |y/(1 + y)|) = (y/(1 + y))/(1 - y/(1 + y)) = y/(1 + y - y) = y If y is positive, there exists x = y/(1 - y) ∈ R such that f (x) = f (y/(1 - y)) = (y/(1 - y))/(1 + |y/(1 - y)|) = (y/(1 - y))/(1 + y/(1 - y)) = y/(1 - y + y) = y Therefore, f is onto. Hence, the given function f is both one-one and onto. 2. Show that the function f : R → R given by f (x) = x
f: R → R is given by f (x) = x Take x, y ∈ R such that f (x) = f (y) x Take cubic root on both sides x = y Hence, f is injective. 3. 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.
We know that every set is a subset of itself, ARA ∀ A ∈ P(X). Therefore, the relation R is reflexive. Let ARB imply that A ⊂ B. This implication will not be possible for B ⊂ A. If A = B + {x} then A ⊂ B but B ⊄ A. Therefore, the relation R is not symmetric. If ARB and BRC, then it can be implied that A ⊂ B and B ⊂ C, which means that A ⊂ C. So, ARC. Therefore, the relation R is transitive. Hence, R is not an equivalence relation. 4. Find the number of all onto functions from the set {1, 2, 3,......, n} to itself.
The number of onto functions from the set {1, 2, 3, …, n} to itself will be the equal to the number of permutations on n elements 1, 2, 3, …, n. Therefore, the number of all onto functions from the set {1, 2, 3, …, n} to itself is n. 5. Let A = {- 1, 0, 1, 2}, B = {- 4, - 2, 0, 2} and f, g : A → B be functions defined by f (x) = x
A = {-1, 0, 1, 2} and B = {-4, -2, 0, 2} f, g: A → B are defined as f (x) = x f (-1) = (-1) g (-1) = 2 |-1 - 1/2| - 1 = 2 |-3/2| - 1 = 2 (3/2) - 1 = 3 - 1 = 2 f (-1) = g (-1) f (0) = 0 g (0) = 2|0 - 1/2| - 1 = 2(1/2) - 1 = 1 - 1 = 0 f (0) = g (0) f (1) = 1 g (1) = 2|1 - 1/2| - 1 = 2(1/2) - 1 = 1 - 1 = 0 f (1) = g (1) Therefore, f (a) = g (a) ∀ a ∈ A. Hence, the functions f and g are equal. 6. Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is **1****2****3****4**
A = {1, 2, 3} Smallest relation containing (1, 2) and (1, 3) which is reflexive and symmetric but not transitive = R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)} R is reflexive as (1, 1), (2, 2), (3, 3) ∈ R. R is symmetric as for (1, 2), (1, 3) ∈ R, (2, 1), (3, 1) ∈ R is also true. R is not transitive as (3, 1), (1, 2) ∈ R but (3, 2) ∉ R. Therefore, number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is 1. Hence, the correct answer is A. 7. Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is **1****2****3****4**
A = {1, 2, 3} Smallest equivalence relation containing (1, 2) = R = {(1, 1), (2, 2), (1, 2), (2, 1), (3, 3)} R is reflexive as (1, 1), (2, 2), (3, 3) ∈ R. R is symmetric as (1, 2) ∈ R, then (2, 1) ∈ R. R is transitive as (2, 1), (1, 2) ∈ R then (2, 2) ∈ R. R is an equivalence relation. The only other possible equivalence relation containing (1, 2) will be the universal relation. Therefore, number of equivalence relations containing (1, 2) are 2. Hence, the correct answer is B. |