Q: 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.

R={(a,b):a≤b2} It can be observed that (21,21)∈/R, since 21>(21)2=41. ∴R is not reflexive.Now, (1,4)∈R as 1 (1.5)2=2.25 ∴(3,1.5)∈/R∴R is not transitive.Hence, R is neither reflexive, nor symmetric, nor transitive.

Q: 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.

Let A={1,2,3,4,5,6}A relation R is defined on set A as:R={(a,b):b=a+1}∴R={(1,2),(2,3),(3,4),(4,5),(5,6)}We can find (a,a)∈/R, where a∈A.For instance,(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)∈/R∴R is not reflexive.It can be observed that (1,2)∈R, but (2,1)∈/R.∴R is not symmetric.Now, (1,2),(2,3)∈RBut,(1,3)∈/R∴R is not transitiveHence, R is neither reflexive, nor symmetric, nor transitive.

Q: Show that the relation R in the set A={1,2,3,4,5} given by R={(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}.

A={1,2,3,4,5}R={(a,b):∣a−b∣ is even }It is clear that for any element a∈A, we have ∣a−a∣=0 (which is even).∴R is reflexive.Let (a,b)∈R. ⇒∣a−b∣ is even. ⇒∣−(a−b)∣=∣b−a∣ is also even. ⇒(b,a)∈R∴R is symmetric.Now, let (a,b)∈R and (b,c)∈R⇒∣a−b∣ is even and ∣b−c∣ is even.⇒(a−b) is even and (b−c) is even.⇒(a−c)=(a−b)+(b−c) is even.[Sum of two even integers is even]⇒∣a−c∣ is even.⇒(a,c)∈R∴R is transitive.Hence, R is an equivalence relation.Now, all elements of the set {1,2,3} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even.Similarly, all elements of the set {2,4} are related to each other as all the elements of this subset are even.Also, no element of the subset {1,3,5} can be related to any element of {2,4} as all elements of {1,3,5} are odd and all elements of {2,4} are even. Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.

Q: 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 point related to a point P=0, 0 ) is the circle passing through P with origin as centre.

R={(P,Q): distance of point P from the origin is the same as the distance of point Q from the origin }Clearly, (P,P)∈R since the distance of point P from the origin is always the same as the distance of the same point P from the origin.R is reflexive.Now,⇒ The distance of point P from the origin is the same as the distance of point Q from the origin.⇒ The distance of point Q from the origin is the same as the distance of point P from the origin.⇒(Q,P)∈R: R is symmetric.Now,Let (P,Q),(Q,S)∈R⇒ The distance of points P and Q from the origin is the same and also, the distance of points Q and S from the origin is the same.⇒ The distance of points P and S from the origin is the same.⇒(P,S)∈R:R is transitive.Therefore, R is an equivalence relation.The set of all points related to P=(0,0) will be those points whose distance from the origin is the same as the distance of point P from the origin.In other words, if O(0,0) is the origin and OP=k, then the set of all points related to P is at a distance of k from the origin.Hence, this set of points forms a circle with the centre as the origin and this circle passes through point P.

Question 1

Determine whether each of the following relations are reflexive, symmetric and transitive:(i)Relation R in the set A={1,2,3…13,14} defined asR={(x,y):3x−y=0} (ii) Relation R in the set N of natural numbers defined asR={(x,y):y=x+5 and x<4} (iii) Relation R in the set A={1,2,3,4,5,6} as R={(x,y):y is divisible by x} (iv) Relation R in the set Z of all integers defined as R={(x,y):x−y is as integer } (v) Relation R  in the set A of human beings in a town at a particular time given by(a) R={(x,y):x  and y work at the same place}(b) R={(x,y):x  and y live in the same locality} (c) R={(x,y):x  is exactly 7 cm taller than y} (d) R={(x,y):x  is wife of y} (e) R={(x,y):x  is father of y}

Accepted Answer

​ (i) A={1,2,3…13,14}R={(x,y):3x−y=0}∴R={(1,3),(2,6),(3,9),(4,12)}​ R is not reflexive since (1,1),(2,2)…(14,14)∈R.Also, R is not symmetric as (1,3)∈R, but (3,1)∈/R⋅[3(3)−1=0]Also, R  is not transitive as (1,3),(3,9)∈R, but (1,9)∈/R.[3(1)−9=0]Hence, R is neither reflexive, nor symmetric, nor transitive.(ii) R={(x,y):y=x+5 and x<4}={(1,6),(2,7),(3,8)}It is seen that (1,1)∈/R.∴R is not reflexive.(1,6)∈RBut,(1,6)∈/R ∴R is not symmetric.Now, since there is no pair in R such that (x,y) and (y,z)∈R, then (x,z) cannot belong to R.∴R is not transitive.Hence, R is neither reflexive, nor symmetric, nor transitive.(iii) A={1,2,3,4,5,6}R={(x,y):y is divisible by x}We know that any number (x) is divisible by itself.⇒(x,x)∈R∴R is reflexive.Now,(2,4)∈R[ as 4 is divisible by 2 ]But,(4,2)∈/ R. [ as 2 is not divisible by 4] ∴R is not symmetric.∴z is divisible by x.⇒(x,z)∈R∴R is transitive.Hence, R is reflexive and transitive but not symmetric.(iv) R={(x,y):x−y is an integer }Now, for every x∈Z,(x,x)∈R as x−x=0 is an integer.∴R is reflexive.Now, for every x,y∈Z if (x,y)∈R, then x−y is an integer.⇒−(x−y) is also an integer.⇒(y−x) is an integer.∴(y,x)∈R∴R is symmetric.Now,Let (x,y) and (y,z)∈R, where x,y,z∈Z⇒(x−y) and (y−z) are integers.⇒x−z=(x−y)+(y−z) is an integer.∴(x,z)∈R :R  is transitive.Hence, R is reflexive, symmetric, and transitive.(v) (a) R={(x,y):x and y work at the same place }⇒(x,x)∈R∴R is reflexive.If (x,y)∈R, then x and y work at the same place.⇒y and x work at the same place.⇒(y,x)∈R ∴R is symmetric.Now, let (x,y),(y,z)∈R⇒x and y work at the same place and y and z work at the same place.⇒x and z work at the same place.⇒(x,z)∈R∴R is transitive.Hence, R is reflexive, symmetric, and transitive.(b) R={(x,y):x and y live in the same locality }Clearly (x,x)∈R as x and x is the same human being.∴R is reflexive.If (x,y)∈R, then x and y live in the same locality.⇒y and x live in the same locality.⇒(y,x)∈R∴R is symmetric.Now, let (x,y)∈R and (y,z)∈R.⇒x and y live in the same locality and y and z live in the same locality.⇒X and z live in the same locality.⇒(x,z)∈R∴R  is transitive.Hence, R is reflexive, symmetric, and transitive.(c) R={(x,y):x is exactly 7 cm taller than y}Now,(x,x)∈/RSince human being x cannot be taller than himself.∴R is not reflexive.Now, let (x,y)∈R.⇒x is exactly 7 cm taller than y.Then, y is not taller than x.∴(y,x)∈/RIndeed if x is exactly 7 cm taller than y, then y is exactly 7 cm shorter than x. ∴R is not symmetric.Now,⇒x is exactly 7 cm taller than y and y is exactly 7 cm taller than z.⇒x is exactly 14 cm taller than z.∴(x,z)∈/R∴R is not transitive.Hence, R is neither reflexive, nor symmetric, nor transitive.(d) R={(x,y):x is the wife of y}Now,(x,x)∈/RSince x cannot be the wife of herself.∴R is not reflexive.Now, let (x,y)∈R⇒x is the wife of y.Clearly y is not the wife of x ∴(y,x)∈/R Indeed if x is the wife of y, then y is the husband of x.∴R is not transitive.Let (x,y),(y,z)∈R⇒x is the wife of y and y is the wife of z.This case is not possible. Also, this does not imply that x is the wife of z.∴(x,z)∈/R ∴R is not transitive.Hence, R is neither reflexive, nor symmetric, nor transitive.(e) R={(x,y):x is the father of y}(x,x)∈/R As x cannot be the father of himself.∴R is not reflexive.Now, let (x,y)∈R.⇒x is the father of y.⇒y cannot be the father of y.Indeed, y is the son or the daughter of y.∴(y,x)∈/R ∴R  is not symmetric.Now, let (x,y)∈R and (y,z)∈R⇒x is the father of y and y is the father of z.⇒x is not the father of zIndeed x is the grandfather of z∴(x,z)∈/R∴R is not transitive.Hence, R is neither reflexive, nor symmetric, nor transitive.

Question 2

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.

Accepted Answer

R={(a,b):a≤b2} It can be observed that (21,21)∈/R, since 21>(21)2=41. ∴R is not reflexive.Now, (1,4)∈R as 1<42But, 4 is not less than 12.∴(4,1)∈/R∴R is not symmetric.Now,(3,2),(2,1.5)∈R (as 3<22=4 and 2<(1.5)2=2.25)But, 3>(1.5)2=2.25 ∴(3,1.5)∈/R∴R is not transitive.Hence, R is neither reflexive, nor symmetric, nor transitive.

Question 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.

Accepted Answer

Let A={1,2,3,4,5,6}A relation R is defined on set A as:​R={(a,b):b=a+1}∴R={(1,2),(2,3),(3,4),(4,5),(5,6)}​We can find (a,a)∈/R, where a∈A.For instance,(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)∈/R∴R is not reflexive.It can be observed that (1,2)∈R, but (2,1)∈/R.∴R is not symmetric.Now, (1,2),(2,3)∈RBut,(1,3)∈/R∴R is not transitiveHence, R is neither reflexive, nor symmetric, nor transitive.

Question 4

Show that the relation R in R defined as R={(a,b):a≤b}, is reflexive and transitive but not symmetric.

Accepted Answer

R={(a,b);a≤b} Clearly (a,a)∈R as a=a ∴R is reflexive.Now,But, (4,2)∈/R as 4 is greater than 2 .∴R is not symmetric.Now, let (a,b),(b,c)∈R.Then,Hence, R is reflexive and transitive but not symmetric.

Question 5

Check whether the relation R in R defined as R={(a,b):a≤b3} is reflexive, symmetric or transitive.

Accepted Answer

R={(a,b):a≤b3}It is observed that (21,21)∈/R as 21>(21)3=81∴R is not reflexive.Now,(1,2)∈R( as 1<23=8)But,(2,1)∈/R( as 23>1)∴R is not symmetric. We have (3,23),(23,56)∈R as 3<(23)3 and 23<(56)3. But (3,56)∈/R as 3>(56)3∴R is not transitive.Hence, R is neither reflexive, nor symmetric, nor transitive.

Question 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.

Accepted Answer

Let A={1,2,3}A relation R on A is defined as R={(1,2),(2,1)}It is seen that (1,1),(2,2),(3,3)∈/R.∴R is not reflexive.Now, as (1,2)∈R and (2,1)∈R, then R is symmetric.Now, (1,2) and (2,1)∈RHowever,(1,1)∈/R ∴R is not transitive.Hence, R is symmetric but neither reflexive nor transitive.

Question 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.

Accepted Answer

Set A is the set of all books in the library of a college.R={x,y):x and y have the same number of pages }Now, R is reflexive since (x,x)∈R as x and x has the same number of pages.Let (x,y)∈R⇒x and y have the same number of pages.⇒y and x have the same number of pages.⇒(y,x)∈R∴R is symmetric.Now, let (x,y)∈R and (y,z)∈R⇒x and y and have the same number of pages and y and z have the same number of pages.⇒x and z have the same number of pages.⇒(x,z)∈R∴R is transitive.Hence, R is an equivalence relation.

Question 8

Show that the relation R in the set A={1,2,3,4,5} given by R={(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}.

Accepted Answer

​A={1,2,3,4,5}R={(a,b):∣a−b∣ is even }​It is clear that for any element a∈A, we have ∣a−a∣=0 (which is even).∴R is reflexive.Let (a,b)∈R. ⇒∣a−b∣ is even. ⇒∣−(a−b)∣=∣b−a∣  is also even. ⇒(b,a)∈R∴R is symmetric.Now, let (a,b)∈R and (b,c)∈R⇒∣a−b∣ is even and ∣b−c∣ is even.⇒(a−b) is even and (b−c) is even.⇒(a−c)=(a−b)+(b−c) is even.[Sum of two even integers is even]⇒∣a−c∣ is even.⇒(a,c)∈R∴R is transitive.Hence, R is an equivalence relation.Now, all elements of the set {1,2,3} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even.Similarly, all elements of the set {2,4} are related to each other as all the elements of this subset are even.Also, no element of the subset {1,3,5} can be related to any element of {2,4} as all elements of {1,3,5} are odd and all elements of {2,4} are even. Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.

Question 9

Show that each of the relation R in the set A={x∈Z:0≤x≤12}, given by(i) R={(a,b):∣a−b∣ is a multiple of 4}(ii) R={(a,b):a=b}is an equivalence relation. Find the set of all elements related to 1 in each case.

Accepted Answer

A={x∈Z:0≤x≤12}={0,1,2,3,4,5,6,7,8,9,10,11,12}(i) R={(a,b):∣a−b∣ is a multiple of 4}For any element a∈A, we have (a,a)∈R as  ∴R is reflexive.Now, let (a,b)∈R⇒∣a−b∣is a multiple of ​4⇒∣−(a−b)∣=∣b−a∣ is a multiple of 4⇒(b,a)∈R∴R is symmetric.Now, let (a,b),(b,c)∈R.⇒∣a−b∣ is a multiple of 4 and ∣b−c∣ is a multiple of 4⇒(a−b) is a multiple of 4 and (b−c) is a multiple of 4 is a multiple of 4⇒∣a−c∣ is a multiple of 4 .⇒(a,c)∈R∴R is transitive.Hence, R is an equivalence relation.The set of elements related to 1 is {1,5,9} since∣1−1∣=0 is a multiple of 4∣5−1∣=4 is a multiple of 4, and∣9−1∣=8 is a multiple of 4(ii) R={(a,b):a=b}For any element a∈A, we have (a,a)∈R, since a=a∴R is reflexive.Now, let (a,b)∈R.⇒a=b ⇒b=a⇒(b,a)∈R∴R is symmetric.Now, let (a,b)∈R and (b,c)∈R.⇒a=b and b=c⇒a=c⇒(a,c)∈R∴R is transitive.Hence, R is an equivalence relation.The elements in R that are related to 1 will be those elements from set A which are equal to 1 .Hence, the set of elements related to 1 is {1}.

Question 10

Given an example of a relation. Which is(i) Symmetric but neither reflexive nor transitive.(ii) Transitive but neither reflexive nor symmetric.(iii) Reflexive and symmetric but not transitive.(iv) Reflexive and transitive but not symmetric.(v) Symmetric and transitive but not reflexive.

Accepted Answer

Hence, relation R is symmetric but not reflexive or transitive.(ii)Consider a relation R in R defined as:(^​(a,b):a<b} For any a∈R, we have (a,a)∈/R since a cannot be strictly less than a itself. In fact, a= a Hence, relation R is transitive but not reflexive and symmetric.(iii) Let A={4,6,8}.Define a relation R on A as:Relation R is reflexive since for every a∈A,(a,a)∈R i.e., (4,4),(6,6),(8,8)}∈RRelation R is symmetric since (a,b)∈R⇒(b,a)∈R for all a,b∈R.Relation R is not transitive since (4,6),(6,8)∈R, but (4,8)∈/R.Hence, relation R is reflexive and symmetric but not transitive.(iv) Define a relation R in R as:R={a,b):a3≥b3}Clearly (a, a) \in R as a3=a3R is reflexive.Now,(2,1)∈R( as 23≥13)But, (1,2)∈/R(as 13<23)R is not symmetric.Now,​ Let (a,b),(b,c)∈R.⇒a3≥b3 and b3≥c3⇒a3≥c3⇒(a,c)∈R​∴R is transitive.Hence, relation R is reflexive and transitive but not symmetric.(v) Let A={−5,−6}.Define a relation R on A as:R={(−5,−6),(−6,−5),(−5,−5)}Relation R is not reflexive as (−6,−6)∈/R.Relation R is symmetric as (−5,−6)∈R and (−6,−5}∈R. It is seen that  (−5,−6),(−6,−5)∈R. Also, (−5,−5)∈R. ∴  The relation R is transitive.Hence, relation R is symmetric and transitive but not reflexive.

Question 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 point related to a point P=0, 0 ) is the circle passing through P with origin as centre.

Accepted Answer

R={(P,Q): distance of point P from the origin is the same as the distance of point Q from the origin }Clearly, (P,P)∈R since the distance of point P from the origin is always the same as the distance of the same point P from the origin.R is reflexive.Now,⇒ The distance of point P from the origin is the same as the distance of point Q from the origin.⇒ The distance of point Q from the origin is the same as the distance of point P from the origin.⇒(Q,P)∈R: R is symmetric.Now,Let (P,Q),(Q,S)∈R⇒ The distance of points P and Q from the origin is the same and also, the distance of points Q and S from the origin is the same.⇒ The distance of points P and S from the origin is the same.⇒(P,S)∈R:R is transitive.Therefore, R is an equivalence relation.The set of all points related to P=(0,0) will be those points whose distance from the origin is the same as the distance of point P from the origin.In other words, if O(0,0) is the origin and OP=k, then the set of all points related to P is at a distance of k from the origin.Hence, this set of points forms a circle with the centre as the origin and this circle passes through point P.

Question 12

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?

Accepted Answer

R={(T1​,T2​):T1​ is similar to T2​}R is reflexive since every triangle is similar to itself.Further, if (T1​,T2​)∈R, then T1​ is similar to T2​.⇒T2​ is similar to T1​.⇒(T2​,T1​)∈R∴R is symmetric.Now,Let (T1​,T2​),(T2​,T3​)∈R.⇒T1​ is similar to T2​ and T2​ is similar to T3​.⇒T1​ is similar to T3​⇒(T1​,T3​)∈R∴R is transitive.Thus, R is an equivalence relation.Now, we can observe that:63​=84​=105​(=21​)∴ The corresponding sides of triangles T1​ and T3​ are in the same ratio. Then, triangle T1​ is similar to triangle T3​Hence, T1​ is related to T3​.

Question 13

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?

Accepted Answer

R={(P1​,P2​):P1​ and P2​ have same the number of sides }R is reflexive since (P1​,P1​)∈R as the same polygon has the same number of sides with itself.Let (P1​,P2​)∈R⇒P1​ and P2​ have the same number of sides.⇒P2​ and P1​ have the same number of sides.⇒(P2​,P1​)∈R∴R is symmetric.Now,Let (P1​,P2​),(P2​,P3​)∈R⇒P1​ and P2​ have the same number of sides. Also, P2​ and P3​ have the same number of sides.⇒P1​ and P3​ have the same number of sides.⇒(P1​,P3​)∈R∴R is transitive.Hence, R is an equivalence relation.The elements in A related to the right-angled triangle (T) with sides 3,4, and 5 are those polygons which have 3 sides (since T is a polygon with 3 sides).Hence, the set of all elements in A related to triangle T is the set of all triangles.

Question 14

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.

Accepted Answer

R={(L1​,L2​):L1​ is parallel to L2​}R is reflexive as any line L1​ is parallel to itself i.e., (L1​,L1​)∈R.Now,Let (L1​,L2​)∈R.⇒L1​ is parallel to L2​⇒L2​ is parallel to L1​⇒(L2​,L1​)∈R∴R is symmetric.Now,Let (L1​,L2​),(L2​,L3​)∈R.⇒L1​ is parallel to L2​. Also, L2​ is parallel to L3​⇒L1​ is parallel to L3​∴R is transitive.Hence, R is an equivalence relation.The set of all lines related to the line y=2x+4 is the set of all lines that are parallel to the line y=2x+4Slope of line y=2x+4 is m=2It is known that parallel lines have the same slopes.The line parallel to the given line is of the form y=2x+c, where c∈R.Hence, the set of all lines related to the given line is given by y=2x+c, where c∈R.

Question 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 option.

Accepted Answer

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)} Step 1:(1,1),(2,2),(3,3),(4,4)∈R . R is reflexive.Step 2:(1,2)∈R but (2,1) ∈/  R. R is not symmetric.Step 3: Consider any set of points, (1,3)∈R and (3,2)∈R then (1,2)∈R. So R is transitive.

Question 16

Let R be the relation in the set N given by R={(a,b):a=b−2,b>6}. Choose the correct answer.

Accepted Answer

R={(a,b):a=b−2,b>6}Now, since b>6,(2,4)∈/RAlso, as 3=8−2,(3,8)∈/RAnd, as 8=7−2∴(8,7)∈/R Now, consider (6,8).We have 8>6 and also, 6=8−2∴(6,8)∈R The correct answer is C.

Question 17

Show that the function f:R∗​→R∗​ defined by f(x)=x1​ 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∗?

Accepted Answer

It is given that f:R.⟶R. is defined by f(x)=x1​.One-one:​f(x)=f(y)⇒x1​=y1​⇒x=y​af is one-one.Onto:x=y1​∈R.( Exists as y=0)such thatf(x)=(y1​)1​=yaf is onto.Thus, the given function (t) is one-one and onto.Now, consider function g:N→ R.defined byg(x)=x1​We have,g(x1​)=g(x2​)⇒x1​1​=x2​1​⇒x1​=x2​sg is one-one.Further, it is clear that g is not onto as for 1.2∈R. there does not exit any x in N such that g(x)=1.21​Hence, function g is one-one but not onto.

Question 18

Check the injectivity and surjectivity of the following functions:(i) f:N→N given by f(x)=x2(ii) f:Z→Z given by f(x)=x2(iii) f:R→R given by f(x)=x2(iv)f:N→N given by f(x)=x3(v)f:z→Z given by f(x)=x3

Accepted Answer

(i) f:N→N is given by,f(x)=x2 It is seen that for x,y∈N,f(x)=f(y)⇒x2=y2⇒x=y∴f is injective.Now, 2∈N. But, there does not exist any x in N such that f(x)=x2=2∴f is not surjective.Hence, function f is injective but not surjective.(ii) f:Z→Z is given by,f(x)=x2It is seen that f(−1)=f(1)=1, but −1=1∴f is not injective.Now, −2∈Z. But, there does not exist any element x∈Z such that f(x)=x2=−2.∴f is not surjective.Hence, function f is neither injective nor surjective.(iii) f:R→R is given by,f(x)=x2It is seen that f(−1)=f(1)=1, but −1=1∴f is not injective.Now, −2∈R. But, there does not exist any element x∈R such that f(x)=x2=−2. ∴f is not surjective.Hence, function f is neither injective nor surjective.(iv) f:N→N given by,f(x)=x3It is seen that for x,y∈N,f(x)=f(y)⇒x3=y3⇒x=y∴f is injective.Now, 2∈N. But, there does not exist any element x in domain N such that f(x)=x3= 2. ∴f  is not surjectiveHence, function f is injective but not surjective.(v) f:Z→Z is given by,f(x)=x3It is seen that for x,y∈Z,f(x)=f(y)⇒x3=y3⇒x=y∴f is injective.Now, 2∈Z. But, there does not exist any element x in domain Z such that f(x)=x3=2.∴f is not surjective.Hence, function f is injective but not surjective.

Question 19

Prove that the Greatest Integer Function f:R→R given by f(x)=[x], is neither one once nor onto, where [x] denotes the greatest integer less than or equal to x.

Accepted Answer

f:R→R is given byf(x)=[x] It is seen that f(1.2)=[1.2]=1,f(1.9)=[1.9]=1∴f(1.2)=f(1.9), but 1.2=1.9∴f is not one-one.Now, consider 0.7∈R.It is known that f(x)=[x] is always an integer. Thus, there does not exist any element x ∈R such that f(x)=0.7∴f is not onto.Hence, the greatest integer function is neither one-one nor onto.

Question 20

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.

Accepted Answer

f:R→R is given by,f(x)=∣x∣={x, if x≥0−x, if x<0​It is seen that f(−1)=∣−1∣=1,f(1)=∣1∣=1∴f(−1)=f(1), but −1=1∴f is not one-one.Now, consider −1∈R.It is known that f(x)=∣x∣ is always non-negative. Thus, there does not exist any element x in domain R such that f(x)=∣x∣=−1∴f is not onto.Hence, the modulus function is neither one-one nor onto.

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