Set-Theory

 Question 1
A relation R is said to be circular of aRb and bRc together imply cRa. Which of the following options is/are correct?
 A If a relation S is reflexive and circular, then S is an equivalence relation. B If a relation S is transitive and circular, then S is an equivalence relation. C If a relation S is circular and symmetric, then S is an equivalence relation. D If a relation S is reflexive and symmetric, then S is an equivalence relation.
Question 1 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 2

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

 A 2n-1
Question 2 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 3

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
Question 3 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 4

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
Question 4 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 5

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.
 Question 6

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
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:

 A 22n B 2n2 C (2n)2 D (22)n E None of the above
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

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
Question 8 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 9
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 _________.
 A 59049
Question 9 Explanation:
 Question 10

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
Question 10 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 11

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?

 A n B n + 1 C 2(n-1) + 1 D n!
Question 11 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 12

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.

 A p(p - 1) B pq C (p2 - 1)(q - 1) D p(p - 1)(q - 1)
Question 12 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 13

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)
Question 13 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 14

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
Question 14 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 15

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
Question 15 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 16

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
Question 16 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 17

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
Question 17 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 18

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

 A 2046 B 2047 C 2048 D 2049
Question 18 Explanation:
Cardinality of the power set of {0, 1, 2, … , 10} is 211 i.e., 2048.
 Question 19

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
Question 19 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 20

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
Question 20 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 21
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.
Question 21 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 22
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
Question 22 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 23
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?
 A H is a group. B Every element in H has an inverse, but H is NOT a group. C For every A ε 2x, the inverse of A is the complement of A. D For every A ε 2x, the inverse of A is A.
 Question 24
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
Question 24 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 25

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
Question 25 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 26

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
Question 26 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 27
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
Question 27 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 28
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
Question 28 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 29

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')
Question 29 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 30

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
Question 30 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.
There are 30 questions to complete.

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