## Set-Theory

 Question 1
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?
 A R2 only B R1 and R2 C Neither R1 and R2 D R1 only
Engineering-Mathematics       Set-Theory       GATE 2019       Video-Explanation
Question 1 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:
Reflexive:
aR1a
⇒ 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]
Symmetric:
If aR1b, then ∃g ∈ G such that gag-1 = b then a = g-1bg, which is Correct.
⇒ So, given relation is symmetric.
Transitive:
The given relation is Transitive.
So, the given relation R1 is equivalence.
R2:
The given relation is not reflexive.
So, which is not equivalence relation.
Such that a ≠ a-1.
So, only R1 is true.
 Question 2
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?
 A Only II B Only I C Neither I nor II D Both I and II
Engineering-Mathematics       Set-Theory       GATE 2019       Video-Explanation
Question 2 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.
Video Explanation
 Question 3

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

 A 41 B 42 C 43 D 44
Engineering-Mathematics       Set-Theory       GATE 2018       Video-Explanation
Question 3 Explanation:
Lagranges Theorem:
For any group ‘G’ with order ‘n’, every subgroup ‘H’ has order ‘k’ such that ‘n’ is divisible by ‘k’.
Solution:
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 4

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 _________.

 A 0 B 1 C 2 D 3
Engineering-Mathematics       Set-Theory       GATE 2017 [Set-2]       Video-Explanation
Question 4 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 5
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 __________.
 A 2 B 3 C 4 D 5
Engineering-Mathematics       Set-Theory       GATE 2016 [Set-1]       Video-Explanation
Question 5 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 6
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?
 A Both P and Q are true. B P is true and Q is false. C P is false and Q is true. D Both P and Q are false.
Engineering-Mathematics       Set-Theory       GATE 2016 [Set-2]       Video-Explanation
Question 6 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)
eg:
(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 7
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.
ALWAYS TRUE?
 A Only I B Only II C Only III D None
Engineering-Mathematics       Set-Theory       GATE 2016 [Set-2]       Video-Explanation
Question 7 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 8

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

 A pr = 0 B pr = 1 C 0 < pr ≤ 1/5 D 1/5 < pr < 1
Engineering-Mathematics       Set-Theory       GATE 2015 [Set-1]
Question 8 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 9

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?

 A Both {f} and {g} are functionally complete B Only {f} is functionally complete C Only {g} is functionally complete D Neither {f} nor {g} is functionally complete
Engineering-Mathematics       Set-Theory       GATE 2015 [Set-1]
Question 9 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 10

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?

 A R is symmetric and reflexive but not transitive B R is reflexive but not symmetric and not transitive C R is transitive but not reflexive and not symmetric D R is symmetric but not reflexive and not transitive
Engineering-Mathematics       Set-Theory       GATE 2015 [Set-2]
Question 10 Explanation:
Reflexive:
In aRb, 'a' and 'b' are distinct. So it can never be reflexive.
Symmetric:
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.
Transitive:
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 11

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

 A 2046 B 2047 C 2048 D 2049
Engineering-Mathematics       Set-Theory       GATE 2015 [Set-2]
Question 11 Explanation:
Cardinality of the power set of {0, 1, 2, … , 10} is 211 i.e., 2048.
 Question 12

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?

 A ∀X ∈ U (|X| = |X'|) B ∃X ∈ U ∃Y ∈ U (|X| = 5, |Y| = 5 and X ∩ Y = ∅) C ∀X ∈ U ∀Y ∈ U (|X| = 2, |Y| = 3 and X \ Y = ∅) D ∀X ∈ U ∀Y ∈ U (X \ Y = Y' \ X')
Engineering-Mathematics       Set-Theory       GATE 2015 [Set-3]
Question 12 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 13

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?

 A Both reflexive and symmetric B Reflexive but not symmetric C Not reflexive but symmetric D Neither reflexive nor symmetric
Engineering-Mathematics       Set-Theory       GATE 2015 [Set-3]
Question 13 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 14

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?

 A Both S1 and S2 are true B S1 is true and S2 is false C S2 is true and S1 is false D Neither S1 nor S2 is true
Engineering-Mathematics       Set-Theory       GATE 2014 [Set-2]
Question 14 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 15

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

 A For any subsets A and B of X, |f(A ∪ B)| = |f(A)|+|f(B)| B For any subsets A and B of X, f(A ∩ B) = f(A) ∩ f(B) C For any subsets A and B of X, |f(A ∩ B)| = min{ |f(A)|,f|(B)|} D 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]
Question 15 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
 Question 16

Let G be a group with 15 elements. Let L be a subgroup of G. It is known that L ≠ G and that the size of L is at least 4. The size of L  is __________.

 A 5 B 6 C 7 D 8
Engineering-Mathematics       Set-Theory       GATE 2014 [Set-3]
Question 16 Explanation:
Lagrange's theorem, in the mathematics of group theory, states that for any finite group G, the order (number of elements) of every subgroup H of G divides the order of G.
So, 15 is divided by {1, 3, 5, 15}.
As minimum is 4 and total is 15, we eliminate 1,3,15.
 Question 17

If V1 and V2 are 4-dimensional subspace of a 6-dimensional vector space V, then the smallest possible dimension of V1∩V2   is ______.

 A 2 B 3 C 4 D 5
Engineering-Mathematics       Set-Theory       GATE 2014 [Set-3]
Question 17 Explanation:
In a 6 dimensional vector space, sub space of 4 dimensional subspace V1, V2 are provided. Then the V1∩V2?
For eg: a two dimensional vector space have x, y axis. For dimensional vector space, it have x, y, z axis.
In the same manner, 6 dimensional vector space has x, y, z, p, q, r (assume).
Any subspace of it, with 4 dimensional subspace consists any 4 of the above. Then their intersection will be atmost 2.
[{x,y,z,p} ∩ {r,q,p,z}] = #2
V1 ∩ V2 = V1 + V2 - V1 ∪ V2 = 4 + 4 + (-6) = 2
 Question 18

Consider the set of all functions f: {0,1, … ,2014} → {0,1, … ,2014} such that f(f(i)) = i, for all 0 ≤ i ≤ 2014. Consider the following statements:

P. For each such function it must be the case that for every i, f(i) = i.
Q. For each such function it must be the case that for some i, f(i) = i.
R. Each such function must be onto.

Which one of the following is CORRECT?

 A P, Q and R are true B Only Q and R are true C Only P and Q are true D Only R is true
Engineering-Mathematics       Set-Theory       GATE 2014 [Set-3]
Question 18 Explanation:
f: {0,1,…,2014} → {0,1,…,2014} and f(f(i)) = i So f(i)should be resulting only {0, 1, …2014}
So, every element in range has a result value to domain. This is onto. (Option R is correct)
We have ‘2015’ elements in domain.
So atleast one element can have f(i) = i,
so option ‘Q’ is also True.
∴ Q, R are correct.
 Question 19

Let δ denote the minimum degree of a vertex in a graph. For all planar graphs on n vertices with δ ≥ 3, which one of the following is TRUE?

 A In any planar embedding, the number of faces is at least n/2 + 2 B In any planar embedding, the number of faces is less than n/2 + 2 C There is a planar embedding in which the number of faces is less than n/2 + 2 D There is a planar embedding in which the number of faces is at most n/(δ+1)
Engineering-Mathematics       Set-Theory       GATE 2014 [Set-3]
Question 19 Explanation:
Euler’s formula:
v – e + f = 2 →①
Point ① degree of each vertex is minimum ‘3’. 3×n ≥ 2e
e ≤ 3n/2
From ① :
n-3n/2+f = 2 ⇒ Question 20

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

 A Commutative but not associative B Both commutative and associative C Associative but not commutative D Neither commutative nor associative
Engineering-Mathematics       Set-Theory       GATE 2013
Question 20 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 21

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:

 A Neither a Partial Order nor an Equivalence Relation B A Partial Order but not a Total Order C A Total Order D An Equivalence Relation
Engineering-Mathematics       Set-Theory       GATE 2006
Question 21 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 22

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⟩

 A Theory Explanation.
Engineering-Mathematics       Set-Theory       GATE 1998
 Question 23

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

 A 2 B 4 C 8 D None of the above
Engineering-Mathematics       Set-Theory       GATE 1995
Question 23 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 24

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

 A 2n-1
Engineering-Mathematics       Set-Theory       GATE 1994
Question 24 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 25

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

 A 7
Engineering-Mathematics       Set-Theory       GATE 2020       Video-Explanation
Question 25 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 26

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

 A at least one of the set Si is a finite set B not more than one of the set Si can be finite C at least one of the sets Si is an infinite set D not more than one of the sets Si can be infinite E None of the above
Engineering-Mathematics       Set-Theory       GATE 1993
Question 26 Explanation:
Given sets are finite union of sets. One set must be infinite to make whole thing to be infinite.
 Question 27

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

 A 22n B 2n2 C (2n)2 D (22)n E None of the above
Engineering-Mathematics       Set-Theory       GATE 1993
Question 27 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 28

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

 A disjoint
Engineering-Mathematics       Set-Theory       GATE 1991       Video-Explanation
Question 28 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 29

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?

 A g is onto ⇒ h is onto B h is onto ⇒ f is onto C h is onto ⇒ g is onto D h is onto ⇒ f and g are onto
Engineering-Mathematics       Set-Theory       GATE 2005-IT
Question 29 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.
There are 29 questions to complete.

Register Now