Relations and Functions Exercise 01
Question:
Check the injectivity and surjectivity of the following functions: (i) f:N→N given by f(x)=x^{2} (ii) f:Z→Z given by f(x)=x^{2} (iii) f:R→R given by f(x)=x^{2} (iv) f:N→N given by f(x)=x^{3} (v) f:Z→Z given by f(x)=x^{3}
Answer:
(i) f:N→N given by f(x)=x^2 Injectivity: Not injective, since f(2)=f(−2)=4. Surjectivity: Not surjective, since for any y∈N, there is no x∈N such that f(x)=y.
(ii) f:Z→Z given by f(x)=x^2 Injectivity: Not injective, since f(2)=f(−2)=4. Surjectivity: Not surjective, since for any y∈Z, there is no x∈Z such that f(x)=y.
(iii) f:R→R given by f(x)=x^2 Injectivity: Not injective, since f(2)=f(−2)=4. Surjectivity: Surjective, since for any y∈R, there exists an x∈R such that f(x)=y.
(iv) f:N→N given by f(x)=x^3 Injectivity: Not injective, since f(2)=f(−2)=−8. Surjectivity: Not surjective, since for any y∈N, there is no x∈N such that f(x)=y.
(v) f:Z→Z given by f(x)=x^3 Injectivity: Not injective, since f(2)=f(−2)=−8. Surjectivity: Not surjective, since for any y∈Z, there is no x∈Z such that f(x)=y.
Question:
Relation R in the set A of human beings in a town at a particular time given by R={(x,y):xandyliveinthesamelocality} enter 1reflexive and transitive but not symmetric 2reflexive only 3Transitive only 4Equivalence 5None
Answer:
1Reflexive and transitive but not symmetric: This is not an equivalence relation.
2Reflexive only: This is not an equivalence relation.
3Transitive only: This is not an equivalence relation.
4Equivalence: This is an equivalence relation.
5None: None of the above.
Question:
Let f,g and h be functions from R to R. Show that (i) (f+g)oh=foh+goh (ii) (f.g)oh=(foh).(goh)
Answer:
(i) (f+g)oh=foh+goh
Proof:
Let x be an element of R.
Then, (f+g)oh(x) = (f+g)(h(x)) = f(h(x)) + g(h(x)) = foh(x) + goh(x) = foh + goh
Therefore, (f+g)oh=foh+goh
(ii) (f.g)oh=(foh).(goh)
Proof:
Let x be an element of R.
Then, (f.g)oh(x) = (f.g)(h(x)) = f(h(x)) . g(h(x)) = foh(x) . goh(x) = foh . goh
Therefore, (f.g)oh=(foh).(goh)
Question:
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.
Answer:
(i) R={(a,b):ab is a multiple of 4}
To show that R is an equivalence relation, we must show that it is reflexive, symmetric, and transitive.
Reflexive: For any element a in A, (a,a) must be in R. Since aa is 0, which is a multiple of 4, (a,a) is in R.
Symmetric: For any elements a and b in A, if (a,b) is in R, then (b,a) must also be in R. Since ab is a multiple of 4, then ba is also a multiple of 4, so (b,a) is in R.
Transitive: For any elements a, b, and c in A, if (a,b) and (b,c) are in R, then (a,c) must also be in R. Since ab and bc are both multiples of 4, then ac must also be a multiple of 4, so (a,c) is in R.
Therefore, R is an equivalence relation.
The set of all elements related to 1 in R is {3, 5, 9}.
(ii) R={(a,b):a=b}
To show that R is an equivalence relation, we must show that it is reflexive, symmetric, and transitive.
Reflexive: For any element a in A, (a,a) must be in R. Since a=a, (a,a) is in R.
Symmetric: For any elements a and b in A, if (a,b) is in R, then (b,a) must also be in R. Since a=b, then b=a, so (b,a) is in R.
Transitive: For any elements a, b, and c in A, if (a,b) and (b,c) are in R, then (a,c) must also be in R. Since a=b and b=c, then a=c, so (a,c) is in R.
Therefore, R is an equivalence relation.
The set of all elements related to 1 in R is {1}.
Question:
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.
Answer:
(i) Symmetric but neither reflexive nor transitive: A relation R on a set A = {1, 2, 3, 4, 5} can be given as R = {(1, 4), (4, 1), (2, 3), (3, 2)}. This relation is symmetric since for every (x, y) ∈ R, (y, x) ∈ R. However, it is neither reflexive nor transitive since (1, 1) and (1, 2) are not present in R.
(ii) Transitive but neither reflexive nor symmetric: A relation R on a set A = {1, 2, 3, 4, 5} can be given as R = {(1, 2), (2, 3), (3, 4), (4, 5)}. This relation is transitive since for every (x, y) and (y, z) ∈ R, (x, z) ∈ R. However, it is neither reflexive nor symmetric since (1, 1) and (2, 1) are not present in R.
(iii) Reflexive and symmetric but not transitive: A relation R on a set A = {1, 2, 3, 4, 5} can be given as R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (1, 2), (2, 1), (3, 4), (4, 3)}. This relation is reflexive since for every x ∈ A, (x, x) ∈ R. It is also symmetric since for every (x, y) ∈ R, (y, x) ∈ R. However, it is not transitive since for every (x, y) and (y, z) ∈ R, (x, z) is not necessarily present in R.
(iv) Reflexive and transitive but not symmetric: A relation R on a set A = {1, 2, 3, 4, 5} can be given as R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (1, 2), (2, 3), (3, 4), (4, 5)}. This relation is reflexive since for every x ∈ A, (x, x) ∈ R. It is also transitive since for every (x, y) and (y, z) ∈ R, (x, z) ∈ R. However, it is not symmetric since for every (x, y) ∈ R, (y, x) is not necessarily present in R.
(v) Symmetric and transitive but not reflexive: A relation R on a set A = {1, 2, 3, 4, 5} can be given as R = {(1, 4), (4, 1), (2, 3), (3, 2), (1, 2), (2, 3), (3, 4), (4, 5)}. This relation is symmetric since for every (x, y) ∈ R, (y, x) ∈ R. It is also transitive since for every (x, y) and (y, z) ∈ R, (x, z) ∈ R. However, it is not reflexive since for every x ∈ A, (x, x) is not necessarily present in R.
Question:
Show that each of the relation R in the set A={x∈Z:0≤x≤12}, given by is an equivalence relation. Find the set of all elements related to 1 in each case. R={(a,b):a=b}
Answer:

Show that R is reflexive: For any element x in the set A={x∈Z:0≤x≤12}, we have that (x,x)∈R, since a=b. Therefore, R is reflexive.

Show that R is symmetric: For any elements a and b in the set A={x∈Z:0≤x≤12}, if (a,b)∈R, then a=b. Therefore, (b,a)∈R, since b=a. Therefore, R is symmetric.

Show that R is transitive: For any elements a, b, and c in the set A={x∈Z:0≤x≤12}, if (a,b)∈R and (b,c)∈R, then a=b and b=c. Therefore, a=c. Therefore, (a,c)∈R. Therefore, R is transitive.

Find the set of all elements related to 1 in R: The set of all elements related to 1 in R is {1}, since (1,1)∈R and a=b.
Question:
Relation R in the set A of human beings in a town at a particular time given by R={(x,y):xisexactly7cmtallerthany} enter 1reflexive and transitive but not symmetric 2reflexive only 3Transitive only 4Equivalence 5Neither reflexive, nor symmetric, nor transitive
Answer:
Answer: 4Equivalence
Question:
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 as
R={(x,y):3x−y=0}
(ii) Relative R in the set N of natural numbers defined as
R={(x,y):y=x+5 andx<4}
(iii) Relation R in the set A={1,2,3,4,5,6} as
R={(x,y):yis divisible byx}
(iv) Relative R in the set Z of all integers defined as
R={(x,y):x−y is an integer}
(v) Relation R in the set A of human beings in a town at a particular time given by
(a)R={(x,y):xandywork at the same place }
(b)R={(x,y):xandylive in the same locality}
(c)R={(x,y):xis exactly7cm taller thany}
(d)R={(x,y):xis wife ofy}
(e)R={(x,y):x is father ofy}
Answer:
(i) Reflexive: No, Symmetric: No, Transitive: No
(ii) Reflexive: No, Symmetric: No, Transitive: No
(iii) Reflexive: Yes, Symmetric: No, Transitive: Yes
(iv) Reflexive: Yes, Symmetric: Yes, Transitive: Yes
(v) (a) Reflexive: No, Symmetric: No, Transitive: No (b) Reflexive: No, Symmetric: No, Transitive: No (c) Reflexive: No, Symmetric: No, Transitive: No (d) Reflexive: Yes, Symmetric: Yes, Transitive: No (e) Reflexive: No, Symmetric: No, Transitive: Yes
Question:
Given an example of a relation. Which is Reflexive and symmetric but not transitive.
Answer:
Answer: A relation is a set of ordered pairs. An example of a relation that is reflexive and symmetric but not transitive is the relation {(1, 1), (2, 2), (1, 2)}.
This relation is reflexive because it contains the ordered pair (1,1) and (2,2), which means that each element is related to itself.
This relation is symmetric because the ordered pair (1,2) implies that the ordered pair (2,1) is also in the relation.
This relation is not transitive because there is no ordered pair (2,1) that would imply the ordered pair (1,2).
Question:
Check the injectivity and surjectivity of the following functions:f:Z→Z given by f(x)=x^{2}
Answer:
Step 1: Determine the domain and range of the function.
The domain of the function is all real numbers, and the range is all nonnegative real numbers.
Step 2: Determine if the function is injective.
To determine if the function is injective, we need to see if there are any xvalues that produce the same yvalue. In this case, there are two xvalues that produce the same yvalue: x=1 and x=1. Therefore, the function is not injective.
Step 3: Determine if the function is surjective.
To determine if the function is surjective, we need to see if all yvalues in the range can be produced by an xvalue in the domain. Since all nonnegative real numbers can be produced by an xvalue in the domain, the function is surjective.
Question:
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):yis divisible byx} enter 1reflexive and transitive but not symmetric 2reflexive only 3Transitive only 4Equivalence 5None
Answer:
5None
Question:
Relation R in the set A of human beings in a town at a particular time given by R={(x,y):xiswifeofy} enter 1reflexive and transitive but not symmetric 2reflexive only 3Transitive only 4Equivalence 5Neither reflexive, nor symmetric, nor transitive
Answer:
1 Reflexive and transitive but not symmetric: A relation R is said to be reflexive if (x,x)∈R for all x∈A. It is said to be transitive if (x,y)∈R and (y,z)∈R implies (x,z)∈R. It is said to be symmetric if (x,y)∈R implies (y,x)∈R.
2 Reflexive only: A relation R is said to be reflexive if (x,x)∈R for all x∈A.
3 Transitive only: A relation R is said to be transitive if (x,y)∈R and (y,z)∈R implies (x,z)∈R.
4 Equivalence: A relation R is said to be an equivalence relation if it is reflexive, symmetric, and transitive.
5 Neither reflexive, nor symmetric, nor transitive: A relation R is said to be neither reflexive, nor symmetric, nor transitive if it does not satisfy any of the properties of reflexivity, symmetry, or transitivity.
Question:
Relation R in the set Z of all integers defined as R={(x,y):(x−y)isaninteger} enter 1reflexive and transitive but not symmetric 2reflexive only 3Transitive only 4Equivalence 5None
Answer:
Answer: 4Equivalence
Question:
Determine whether the following relation is reflexive, symmetric and transitive Relative R in the set N of natural numbers defined as R={(x,y):y=x+5andx<4} enter 1Reflexive 2Symmetric 3Transitive 4Equivalence 5None
Answer:
1 Not Reflexive 2 Not Symmetric 3 Not Transitive 4 Not Equivalence 5 None
Question:
Symmetric and transitive but not reflexive.
Answer:

Symmetric: This means that if (a, b) is an element of the relation, then (b, a) is also an element of the relation.

Transitive: This means that if (a, b) and (b, c) are elements of the relation, then (a, c) is also an element of the relation.

Not Reflexive: This means that the relation does not contain (a, a) for any element a.
Question:
Relation R in the set A of human beings in a town at a particular time given by R={(x,y):xandyworkatthesameplace} enter 1reflexive and transitive but not symmetric 2reflexive only 3Transitive only 4Equivalence 5None
Answer:
1Reflexive and transitive but not symmetric: This is not an equivalence relation because it is not symmetric.
2Reflexive only: This is not an equivalence relation because it is not transitive.
3Transitive only: This is not an equivalence relation because it is not reflexive.
4Equivalence: This is an equivalence relation because it is reflexive, transitive, and symmetric.
5None: This is not an equivalence relation because it is neither reflexive, transitive, nor symmetric.
Question:
In each of the following cases, state whether the function is oneone, onto or bijective. Justify your answer. (i) f:R→R defined by f(x)=3−4x (ii) f:R→R defined by f(x)=1+x^{2}
Answer:
(i) f:R→R defined by f(x)=3−4x
This function is not oneone, onto or bijective. This is because the function is not injective, as f(1) = f(1) = 1. It is also not surjective, as there is no value of x for which f(x) = 0. Therefore, the function is not bijective.
(ii) f:R→R defined by f(x)=1+x^{2}
This function is not oneone, onto or bijective. This is because the function is not injective, as f(1) = f(1) = 2. It is also not surjective, as there is no value of x for which f(x) = 0. Therefore, the function is not bijective.
Question:
Relation R in the set A of human beings in a town at a particular time given by R={(x,y):xisfatherofy} enter 1reflexive and transitive but not symmetric 2reflexive only 3Transitive only 4Equivalence 5Neither reflexive, nor symmetric, nor transitive
Answer:
1 Reflexive: xRx for all x in A Transitive: if xRy and yRz then xRz for all x, y, z in A Symmetric: if xRy then yRx for all x, y in A
Answer: 5Neither reflexive, nor symmetric, nor transitive