Question 1

Let G be a group of 35 elements. Then the largest possible size of a subgroup of G other than G itself is ______.

       Engineering-Mathematics       Set-Theory       GATE 2020       Video-Explanation
Question 1 Explanation: 
Lagrange’s Theorem:
If ‘H” is a subgroup of finite group (G,*) then O(H) is the divisor of O(G).
Given that the order of group is 35. Its divisors are 1,5,7,35.
It is asked that the size of largest possible subgroup other than G itself will be 7.
Question 2

Suppose A = {a,b,c,d} and Π1 is the following partition of A
Π1 = {{a,b,c}{d}}

(a) List the ordered pairs of the equivalence relations induced by Π1.

(b) Draw the graph of the above equivalence relation.

(c) Let Π2 = {{a},{b},{C},{d}}
Π3 = {{a,b,c,d}}
and Π4 = {{a,b},{c,d}}

Draw a Poset diagram of the poset,
({Π1234}, refines⟩

Theory Explanation.
       Engineering-Mathematics       Set-Theory       GATE 1998
Question 3

A relation R is defined on ordered pairs of integers as follows: (x,y) R(u,v) if x < u and y > v. Then R is: Then R is:

Neither a Partial Order nor an Equivalence Relation
A Partial Order but not a Total Order
A Total Order
An Equivalence Relation
       Engineering-Mathematics       Set-Theory       GATE 2006
Question 3 Explanation: 
If a relation is equivalence then it must be
i) Symmetric
ii) Reflexive
iii) Transitive
If a relation is partial order relation then it must be
i) Reflexive
ii) Anti-symmetric
iii) Transitive
If a relation is total order relation then it must be
i) Reflexive
ii) Symmetric
iii) Transitive
iv) Comparability
Given ordered pairs are (x,y)R(u,v) if (xv).
Here <, > are using while using these symbol between (x,y) and (y,v) then they are not satisfy the reflexive relation. If they uses (x<=u) and (y>=u) then reflexive relation can satisfies.
So, given relation cannot be a Equivalence. Total order relation or partial order relation.
Question 4

The number of elements in he power set P (S) of the set S = {(φ), 1, (2, 3)} is:

None of the above
       Engineering-Mathematics       Set-Theory       GATE 1995
Question 4 Explanation: 
S = {(φ), 1, (2, 3)}
P(S) = {φ, {{φ}}, {1}, {{2, 3}}, {{φ}, 1}, {1, {2, 3}}, {{φ}, 1, {2, 3}}}
In P(S) it contains 8 elements.
Question 5

The number of subsets {1, 2, ... n} with odd cardinality is __________.

       Engineering-Mathematics       Set-Theory       GATE 1994
Question 5 Explanation: 
Total no. of subsets with n elements is 2n.
And so, no. of subsets with odd cardinality is half of total no. of subsets = 2n /n = 2n-1
Question 6

Let S be an infinite set and S1 ..., Sn be sets such that S1 ∪ S2 ∪ ... ∪ Sn = S. Then,

at least one of the set Si is a finite set
not more than one of the set Si can be finite
at least one of the sets Si is an infinite set
not more than one of the sets Si can be infinite
None of the above
       Engineering-Mathematics       Set-Theory       GATE 1993
Question 6 Explanation: 
Given sets are finite union of sets. One set must be infinite to make whole thing to be infinite.
Question 7

Let A be a finite set of size n. The number of elements in the power set of A × A is:

None of the above
       Engineering-Mathematics       Set-Theory       GATE 1993
Question 7 Explanation: 
Cardinality of A = n × n = n2
A = {1,2}
|A| = {∅, {1}, {2}, {1,2}}
Cardinality of power set of A = 2n2
A × A = {1,2} × {1,2}
= {(1,1), (1,2), (2,1), (2,2)}
Cardinality of A × A = n2
Cardinality of power set of A × A = 2n2
Question 8

Let f be a function from a set A to a set B, g a function from B to C, and h a function from A to C, such that h(a) = g(f(a)) for all a ∈ A. Which of the following statements is always true for all such functions f and g?

g is onto ⇒ h is onto
h is onto ⇒ f is onto
h is onto ⇒ g is onto
h is onto ⇒ f and g are onto
       Engineering-Mathematics       Set-Theory       GATE 2005-IT
Question 8 Explanation: 
g(f(a)) is a composition function which is A→B→C.
If h: A→C is a onto function, the composition must be onto, but the first function in the composition need to be onto.
So, B→C is must be onto.
Question 9

Let A be a set with n elements. Let C be a collection of distinct subsets of A such that for any two subsets S1 and S2 in C, either S⊂ S2 or S⊂ S1. What is the maximum cardinality of C?

n + 1
2(n-1) + 1
       Engineering-Mathematics       Set-Theory       GATE 2005-IT
Question 9 Explanation: 
Consider an example:
Let A = {a, b, c}, here n = 3
Now, P(A) = {Ø, {a}, {b}, {c}, {a,b}, {b,c}, {{a}, {a,b,c}}
Now C will be contain Ø (empty set) and {a,b,c} (set itself) as Ø is the subset of every set. And every other subset is the subset of {a,b,c}.
Now taking the subset of cardinality, we an take any 1 of {a}, {b}, {c} as none of the set is subset of other.
Let's take {2}
→ Now taking the sets of cardinality 2 -{a,b}, {b,c}
→ {b} ⊂ {a,b} and {b,c} but we can't take both as none of the 2 is subset of the other.
→ So let's take {c,a}.
So, C = {Ø, {b}, {b,c}, {a,b,c}}
→ So, if we observe carefully, we can see that we can select only 1 set from the subsets of each cardinality 1 to n
i.e., total n subsets + Ø = n + 1 subsets of A can be there in C.
→ So, even though we can have different combinations of subsets in C but maximum cardinality of C will be n+1 only.
Question 10

Let n = p2q, where p and q are distinct prime numbers. How many numbers m satisfy 1 ≤ m ≤ n and gcd (m, n) = 1? Note that gcd (m, n) is the greatest common divisor of m and n.

p(p - 1)
(p2 - 1)(q - 1)
p(p - 1)(q - 1)
       Engineering-Mathematics       Set-Theory       GATE 2005-IT
Question 10 Explanation: 
n = p2q, [p, q are prime numbers]
→ No. of multiples of p in n = pq [n = p⋅p⋅q]
→ No. of multiples of q in n = p2 [n = p2q]
→ Prime factorization of n contains only p & q.
→ gcd(m,n) is to be multiple of p and (or) 1.
→ So, no. of possible m such that gcd(m,n) is 1 will be
n - number of multiples of either p (or) q
= n - p2 - pq + p
= p2q - p2 - pq + p
= p(pq - p - q + 1)
= p(p - 1)(q - 1)
Question 11
 Let S be a set consisting of 10 elements. The number of tuples of the form (A, B) such that A and B are subsets of S, and A ⊆ B is _________.
       Engineering-Mathematics       Set-Theory       GATE 2021 CS-Set-2       Video-Explanation
Question 11 Explanation: 
Question 12
A relation R is said to be circular of aRb and bRc together imply cRa. Which of the following options is/are correct?
If a relation S is reflexive and circular, then S is an equivalence relation.
If a relation S is transitive and circular, then S is an equivalence relation.
If a relation S is circular and symmetric, then S is an equivalence relation.
If a relation S is reflexive and symmetric, then S is an equivalence relation.
       Engineering-Mathematics       Set-Theory       GATE 2021 CS-Set-1       Video-Explanation
Question 12 Explanation: 

Theorem: A relation R on a set A  is an equivalence relation if and only if it is reflexive and circular.


For symmetry, assume that x, y ∈ A so that xRy, lets check for  yRx. 

Since R is reflexive and y ∈ A, we know that yRy. Since R is circular and xRy and yRy, we know that yRx. Thus R is symmetric. 


For transitivity, assume that x, y, z ∈ A so that xRy and yRz. Check for  xRz. Since R is circular and xRy and yRz, we know that zRx. Since we already proved that R is symmetric, zRx implies that xRz. Thus R is transitive.

Question 13

If the longest chain in a partial order is of length n then the partial order can be written as a ______ of n antichains.

       Engineering-Mathematics       Set-Theory       GATE 1991       Video-Explanation
Question 13 Explanation: 
Suppose the length of the longest chain in a partial order is n. Then the elements in the poset can be partitioned into a disjoint antichains.
Question 14

Let R be the relation on the set of positive integers such that aRb if and only if a and b are distinct and have a common divisor other than 1. Which one of the following statements about R is true?

R is symmetric and reflexive but not transitive
R is reflexive but not symmetric and not transitive
R is transitive but not reflexive and not symmetric
R is symmetric but not reflexive and not transitive
       Engineering-Mathematics       Set-Theory       GATE 2015 [Set-2]       Video-Explanation
Question 14 Explanation: 
In aRb, 'a' and 'b' are distinct. So it can never be reflexive.
In aRb, if 'a' and 'b' have common divisor other than 1, then bRa, i.e., 'b' and 'a' also will have common divisor other than 1. So, yes symmetric.
Take (3, 6) and (6, 2) elements of R. For transitivity (3, 2) must be the element of R, but 3 and 2 don't have a common divisor. So not transitive.
Question 15

The cardinality of the power set of {0, 1, 2, … 10} is _________.

       Engineering-Mathematics       Set-Theory       GATE 2015 [Set-2]       Video-Explanation
Question 15 Explanation: 
Cardinality of the power set of {0, 1, 2, … , 10} is 211 i.e., 2048.
Question 16
A binary relation R on ℕ × ℕ is defined as follows: (a,b)R(c,d) if a≤c or b≤d. Consider the following propositions:
P:R is reflexive
Q:R is transitive
Which one of the following statements is TRUE?
Both P and Q are true.
P is true and Q is false.
P is false and Q is true.
Both P and Q are false.
       Engineering-Mathematics       Set-Theory       GATE 2016 [Set-2]       Video-Explanation
Question 16 Explanation: 
For every a,b ∈ N,
a≤c ∨ b≤d
Let a≤a ∨ b≤b is true for all a,b ∈ N
So there exists (a,a) ∀ a∈N.
It is Reflexive relation.
Consider an example
c = (a,b)R(c,d) and (c,d)R(e,f) then (a,b)R(e,f)
This does not hold for any (a>e) or (b>f)
(2,2)R(1,2) as 2≤2
(1,2)R(1,1) as 1≤1
but (2,2) R (1,1) is False
So, Not transitive.
Question 17
Consider a set U of 23 different compounds in a Chemistry lab. There is a subset S of U of 9 compounds, each of which reacts with exactly 3 compounds of U. Consider the following statements:
I. Each compound in U\S reacts with an odd number of compounds.
U\S reacts with an odd number of compounds.
U\S reacts with an even number of compounds.
Only I
Only II
Only III
       Engineering-Mathematics       Set-Theory       GATE 2016 [Set-2]       Video-Explanation
Question 17 Explanation: 
There are set of ‘23’ different compounds.
U = 23
∃S ∋ (S⊂U)
Each component in ‘S’ reacts with exactly ‘3’ compounds of U,

If a component ‘a’ reacts with ‘b’, then it is obvious that ‘b’ also reacts with ‘a’.
It’s a kind of symmetric relation.>br> If we connect the react able compounds, it will be an undirected graph.
The sum of degree of vertices = 9 × 3 = 27
But, in the graph of ‘23’ vertices the sum of degree of vertices should be even because
(di = degree of vertex i.e., = no. of edges)
But ‘27’ is not an even number.
To make it an even, one odd number should be added.
So, there exists atleast one compound in U/S reacts with an odd number of compounds.
Question 18
A function f:N+ → N+, defined on the set of positive integers N+, satisfies the following properties:
f(n) = f(n/2)           if n is even
f(n) = f(n+5)           if n is odd
Let R = {i|∃j: f(j)=i} be the set of distinct values that f takes. The maximum possible size of R is __________.
       Engineering-Mathematics       Set-Theory       GATE 2016 [Set-1]       Video-Explanation
Question 18 Explanation: 
f(n)= f(n⁄2)          if n is even
f(n)= f(n+5)        if n is odd

We can observe that

and f(5) = f(10) = f(15) = f(20)
Observe that f(11) = f(8)
f(12) = f(6) = f(3)
f(13) = f(9) = f(14) = f(7) = f(12) = f(6) = f(3)
f(14) = f(9) = f(12) = f(6) = f(3)
f(16) = f(8) = f(4) = f(2) = f(1) [repeating]
So, we can conclude that
‘R’ can have size only ‘two’ [one: multiple of 5’s, other: other than 5 multiples]
Question 19

Consider the set X = {a,b,c,d e} under the partial ordering

    R = {(a,a),(a,b),(a,c),(a,d),(a,e),(b,b),(b,c),(b,e),(c,c),(c,e),(d,d),(d,e),(e,e)}.

The Hasse diagram of the partial order (X,R) is shown below.

The minimum number of ordered pairs that need to be added to R to make (X,R) a lattice is _________.

       Engineering-Mathematics       Set-Theory       GATE 2017 [Set-2]       Video-Explanation
Question 19 Explanation: 
R = {(a,a), (a,b), (a,c), (a,d), (a,e), (b,b), (b,c), (b,e), (c,c), (c,e), (d,d), (d,e), (e,e)}
As per the definition of lattice, each pair should have GLB, LUB.
The given ‘R’ has GLB, LUB for each and every pair.
So, no need to add extra pair.
Thus no. of required pair such that Hasse diagram to become lattice is “0”.
Question 20
Let G be an arbitrary group. Consider the following relations on G:
R: ∀a,b ∈ G, aRb if and only if ∃g ∈ G such that a = gbg
R: ∀a,b ∈ G, aRb if and only if a = b-1
Which of the above is/are equivalence relation/relations?
R2 only
R1 and R2
Neither R1 and R2
R1 only
       Engineering-Mathematics       Set-Theory       GATE 2019       Video-Explanation
Question 20 Explanation: 
A relation between the elements of a set is symmetric, reflexive and transitive then such relation is called as equivalence relation.
Consider Statement R1:
⇒ a = g-1ag
Left multiply both sides by g
⇒ ga = gg-1ag
Right multiply both sides by g-1
⇒ gag-1 = gg-1agg-1
⇒ gag-1 = a [∴ The relation is reflexive]
If aR1b, then ∃g ∈ G such that gag-1 = b then a = g-1bg, which is Correct.
⇒ So, given relation is symmetric.
The given relation is Transitive.
So, the given relation R1 is equivalence.
The given relation is not reflexive.
So, which is not equivalence relation.
Such that a ≠ a-1.
So, only R1 is true.
Question 21
Let U = {1,2,...,n}. Let A = {(x,X)|x ∈ X, X ⊆ U}. Consider the following two statements on |A|.

Which of the above statements is/are TRUE?
Only II
Only I
Neither I nor II
Both I and II
       Engineering-Mathematics       Set-Theory       GATE 2019       Video-Explanation
Question 21 Explanation: 
Let us consider U = {1, 2}
and given A = {(x, X), x∈X and X⊆U}
Possible sets for U = {Φ, {1}, {2}, {1, 2}}
if x=1 then no. of possible sets = 2
x=2 then no. of possible sets = 2
⇒ No. of possible sets for A = (no. of sets at x=1) + (no. of sets at x=2) = 2 + 2 = 4
Consider statement (i) & (ii) and put n=2

Statement (i) is true

Statement (i) and (ii) both are true.
Answer: (C)
Video Explanation
Question 22

Suppose L = {p, q, r, s, t} is a lattice represented by the following Hasse diagram:

For any x, y ∈ L, not necessarily distinct, x ∨ y and x ∧ y are join and meet of x, y respectively. Let L3 = {(x,y,z): x, y, z ∈ L} be the set of all ordered triplets of the elements of L. Let pr be the probability that an element (x,y,z) ∈ L3 chosen equiprobably satisfies x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z). Then

pr = 0
pr = 1
0 < pr ≤ 1/5
1/5 < pr < 1
       Engineering-Mathematics       Set-Theory       GATE 2015 [Set-1]       Video-Explanation
Question 22 Explanation: 
Total number of elements i.e., ordered triplets (x,y,z) in L3 are 5×5×5 i.e., 125 n(s) = 125
Let A be the event that an element (x,y,z)∈ L3 satisfies x ∨(y∧z) = (x∨y) ∧ (x∨z) Since q∨(r∧s) = q∨p = q
and (q∨r)∧(q∨s) = t∧t = t q∨(r∧s) ≠ (q∨r)∧(q∨s)
Therefore, (x,y,z) = (q,r,s),(q,s,r),(r,q,s),(r,s,q),(s,r,q),(s,q,r)
i.e., 3! = 6 elements will not satisfy distributive law and all other (x,y,z) choices satisfy distributive law
n(A) = 125-6 = 119
∴ required probability is 119/125
⇒ 1/5
Question 23

Consider the operations f(X, Y, Z) = X'YZ + XY' + Y'Z'  and  g(X′, Y, Z) = X′YZ + X′YZ′ + XY Which one of the following is correct?

Both {f} and {g} are functionally complete
Only {f} is functionally complete
Only {g} is functionally complete
Neither {f} nor {g} is functionally complete
       Engineering-Mathematics       Set-Theory       GATE 2015 [Set-1]       Video-Explanation
Question 23 Explanation: 
A function is functionally complete if (OR, NOT) or (AND, NOT) are implemented by it.
f(X,X,X) = X'XX'+XX'+X'X'
= 0+0+X'
= X'
Similarly, f(Y,Y,Y) = Y' and f(X,Z,Z) = Z'
f(Y',Y',Z') = (Y')'Y'Z'+Y'(Y')'+(Y')'(Z')'
= YY'Z'+Y'Y+YZ
= 0+0+YZ
= YZ
We have derived NOT and AND. So f(X,Y,Z) is functionally complete.
g(X,Y,Z) = X'YZ+X'YZ'+XY
g(X,X,X) = X'XX+X'XZ'+XX
= 0+0+X
= X
Similarly, g(Y,Y,Y) = Y and g(Z,Z,Z) = Z
NOT is not derived. Hence, g is not functionally complete.
Question 24
Let X be a set and 2x denote the powerset of X. Define a binary operation Δ on 2x as follows:
AΔB = (A − B) ∪ (B − A) .
Let H = (2x, Δ). Which of the following statements about H is/are correct?
H is a group.
Every element in H has an inverse, but H is NOT a group.
For every A ε 2x, the inverse of A is the complement of A.
For every A ε 2x, the inverse of A is A.
       Engineering-Mathematics       Set-Theory       GATE 2023       Video-Explanation
Question 25

Suppose U is the power set of the set S = {1, 2, 3, 4, 5, 6}. For any T ∈ U, let |T| denote the number of element in T and T' denote the complement of T. For any T, R ∈ U, let TR be the set of all elements in T which are not in R. Which one of the following is true?

∀X ∈ U (|X| = |X'|)
∃X ∈ U ∃Y ∈ U (|X| = 5, |Y| = 5 and X ∩ Y = ∅)
∀X ∈ U ∀Y ∈ U (|X| = 2, |Y| = 3 and X \ Y = ∅)
∀X ∈ U ∀Y ∈ U (X \ Y = Y' \ X')
       Engineering-Mathematics       Set-Theory       GATE 2015 [Set-3]       Video-Explanation
Question 25 Explanation: 
As X and Y are elements of U, so X and Y are subsets of S.
(A) False. Consider X = {1,2}. Therefore, X' = {3,4,5,6}, |X| = 2 and |X'| = 4.
(B) False. Because for any two possible subsets of S with 5 elements should have atleast 4 elements in common. Hence X∩Y cannot be null.
(C) False. Consider X = {1,4}, Y= {1,2,3} then X\Y = {4} which is not null.
(D) True. Take any possible cases.
Question 26

Let R be a relation on the set of ordered pairs of positive integers such that ((p,q),(r,s)) ∈ R if and only if p - s = q - r. Which one of the following is true about R?

Both reflexive and symmetric
Reflexive but not symmetric
Not reflexive but symmetric
Neither reflexive nor symmetric
       Engineering-Mathematics       Set-Theory       GATE 2015 [Set-3]       Video-Explanation
Question 26 Explanation: 
Since p-q ≠ q-p
∴(p,q) R (p,q)
⇒ R is not reflexive.
Let (p,q) R (r,s) then p-s = q-r
⇒ r-q = s-p
⇒ (r,s) R (p,q)
⇒ R is symmetric.
Question 27

Let G be a finite group on 84 elements. The size of a largest possible proper subgroup of G is _________.

       Engineering-Mathematics       Set-Theory       GATE 2018       Video-Explanation
Question 27 Explanation: 
Lagranges Theorem:
For any group ‘G’ with order ‘n’, every subgroup ‘H’ has order ‘k’ such that ‘n’ is divisible by ‘k’.
Given order n = 84
Then the order of subgroups = {1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84}
As per the proper subgroup definition, it should be “42”.
Question 28

A binary operation ⊕ on a set of integers is defined as x ⊕ y = x+ y2. Which one of the following statements is TRUE about ⊕?

Commutative but not associative
Both commutative and associative
Associative but not commutative
Neither commutative nor associative
       Engineering-Mathematics       Set-Theory       GATE 2013       Video-Explanation
Question 28 Explanation: 
Cumulative property:
A binary relation on a set S is called cumulative if a*b = b*a ∀ x,y∈S.
Associative property:
A binary relation on set is called associative if (a*b)*c = a*(b*c) ∀ x,y∈S.
Given x⊕y = x2 + y2 --------(1)
Replace x, y in (1)
y⊕x = y2 + x2 which is same as (1), so this is cumulative
(x⊕y)⊕z = (x2 + y2) ⊕ z
= (x2 + y2) + z2
= x2 + y2 + z2 + 2x2y2 ----------(2)
x⊕(y ⊕ z) = x ⊕ (y2 + z2)
= x2 + (y2 + z2)2
= x2 + y2 + z2 + 2y2z2 ----------- (3)
(2) & (3) are not same so this is not associative.
Question 29

Consider the following relation on subsets of the set S of integers between 1 and 2014. For two distinct subsets U and V of S we say U < V if the minimum element in the symmetric difference of the two sets is in U. Consider the following two statements:

S1: There is a subset of S that is larger than every other subset.
S2: There is a subset of S that is smaller than every other subset.

Which one of the following is CORRECT?

Both S1 and S2 are true
S1 is true and S2 is false
S2 is true and S1 is false
Neither S1 nor S2 is true
       Engineering-Mathematics       Set-Theory       GATE 2014 [Set-2]       Video-Explanation
Question 29 Explanation: 
Given: S = {1, 2, 3, …, 2014}
U⊂S, V⊂S
Let U = {1, 2, 3}
V = {2, 3, 4}
Symmetric difference:
(U – V) ∪ (V – U) = {1} ∪ {4} = {1, 4}
The minimum element in the symmetric difference is 1 and 1∈U.
S1: Let S = Universal set = {1, 2, … 2014}
This universal set is larger than every other subset.
S2: Null set is smaller than every other set.
Let U = { }, V = {1}
Symmetric difference = ({ } – {1}) ∪ ({1} – { }) = { } ∪ {1} = {1}
So, U < V because { } ∈ U.
Question 30

Let X and Y be finite sets and f: X→Y be a function. Which one of the following statements is TRUE?

For any subsets A and B of X, |f(A ∪ B)| = |f(A)|+|f(B)|
For any subsets A and B of X, f(A ∩ B) = f(A) ∩ f(B)
For any subsets A and B of X, |f(A ∩ B)| = min{ |f(A)|,f|(B)|}
For any subsets S and T of Y, f -1 (S ∩ T) = f -1 (S) ∩ f -1 (T)
       Engineering-Mathematics       Set-Theory       GATE 2014 [Set-3]       Video-Explanation
Question 30 Explanation: 
The function f: x→y.
We need to consider subsets of 'x', which are A & B (A, B can have common elements are exclusive).
Similarly S, T are subsets of 'y'.

To be a function, each element should be mapped with only one element.
(a) |f(A∪B)| = |f(A)|+|f(B)|
|{a,b,c}|∪|{c,d,e}| = |{a,b,c}| + |{c,d,e}|
|{a,b,c,d,e}| = 3+3
5 = 6 FALSE
(d) To get inverse, the function should be one-one & onto.
The above diagram fulfills it. So we can proceed with inverse.
f-1 (S∩T ) = f-1 (S)∩f-1 (T)
f-1 (c) = f-1 ({a,b,c})∩f-1 ({c,d,e})
2 = {1,2,3}∩{2,4,5}
2 = 2 TRUE
There are 30 questions to complete.

Access quiz wise question and answers by becoming as a solutions adda PRO SUBSCRIBER with Ad-Free content

Register Now

If you have registered and made your payment please contact to get access