## Set-theory

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 ______.

7 |

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 Π_{1} is the following partition of A

Π_{1} = {{a,b,c}{d}}

(a) List the ordered pairs of the equivalence relations induced by Π_{1}.

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

(c) Let Π_{2} = {{a},{b},{C},{d}}

Π_{3} = {{a,b,c,d}}

and Π_{4} = {{a,b},{c,d}}

Draw a Poset diagram of the poset,

({Π_{1},Π_{2},Π_{3},Π_{4}}, refines⟩

Theory Explanation. |

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 |

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 (x

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:

2 | |

4 | |

8 | |

None of the above |

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 __________.

2 ^{n-1} |

^{n}.

And so, no. of subsets with odd cardinality is half of total no. of subsets = 2

^{n}/n = 2

^{n-1}

Question 6 |

Let S be an infinite set and S_{1} ..., S_{n} be sets such that S_{1} ∪ S_{2} ∪ ... ∪ S_{n} = S. Then,

at least one of the set S _{i} is a finite set | |

not more than one of the set S _{i} can be finite | |

at least one of the sets S _{i} is an infinite set | |

not more than one of the sets S _{i} can be infinite | |

None of the above |

Question 7 |

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

2 ^{2n} | |

2 ^{n2} | |

(2 ^{n})^{2} | |

(2 ^{2})^{n} | |

None of the above |

^{2}

A = {1,2}

|A| = {∅, {1}, {2}, {1,2}}

Cardinality of power set of A = 2

^{n2}

A × A = {1,2} × {1,2}

= {(1,1), (1,2), (2,1), (2,2)}

Cardinality of A × A = n

^{2}

Cardinality of power set of A × A = 2

^{n2}

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 |

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 S_{1} and S_{2} in C, either S_{1 }⊂ S_{2} or S_{2 }⊂ S_{1}. What is the maximum cardinality of C?

n | |

n + 1 | |

2 ^{(n-1)} + 1 | |

n! |

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 = p^{2}q, 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) | |

pq | |

(p ^{2} - 1)(q - 1) | |

p(p - 1)(q - 1) |

^{2}q, [p, q are prime numbers]

→ No. of multiples of p in n = pq [n = p⋅p⋅q]

→ No. of multiples of q in n = p

^{2}[n = p

^{2}q]

→ 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 - p

^{2}- pq + p

= p

^{2}q - p

^{2}- pq + p

= p(pq - p - q + 1)

= p(p - 1)(q - 1)

Question 11 |

59049 |

Question 12 |

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. |

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.

disjoint |

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 |

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 15 |

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

2046 | |

2047 | |

2048 | |

2049 |

^{11}i.e., 2048.

Question 16 |

*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. |

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 17 |

*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?

Only I | |

Only II | |

Only III | |

None |

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

(d

_{i}= 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 |

*f:N*, deﬁned on the set of positive integers N

^{+}→ 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 __________.

2 | |

3 | |

4 | |

5 |

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 _________.

0 | |

1 | |

2 | |

3 |

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 |

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?

R _{2} only | |

R _{1} and R_{2} | |

Neither R _{1} and R_{2} | |

R _{1} only |

Consider Statement R

_{1}:

Reflexive:

aR

_{1}a

⇒ a = g

^{-1}ag

Left multiply both sides by g

⇒ ga = gg

^{-1}ag

Right multiply both sides by g

^{-1}

⇒ gag

^{-1}= gg

^{-1}agg

^{-1}

⇒ gag

^{-1}= a [∴ The relation is reflexive]

Symmetric:

If aR

_{1}b, then ∃g ∈ G such that gag

^{-1}= b then a = g

^{-1}bg, which is Correct.

⇒ So, given relation is symmetric.

Transitive:

The given relation is Transitive.

So, the given relation R

_{1}is equivalence.

R

_{2}:

The given relation is not reflexive.

So, which is not equivalence relation.

Such that a ≠ a

^{-1}.

So, only R

_{1}is true.

Question 21 |

Which of the above statements is/are TRUE?

Only II | |

Only I | |

Neither I nor II | |

Both I and II |

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 L^{3} = {(x,y,z): x, y, z ∈ L} be the set of all ordered triplets of the elements of L. Let p_{r} be the probability that an element (x,y,z) ∈ L^{3} chosen equiprobably satisfies x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z). Then

pr = 0 | |

pr = 1 | |

0 < pr ≤ 1/5 | |

1/5 < pr < 1 |

^{3}are 5×5×5 i.e., 125 n(s) = 125

Let A be the event that an element (x,y,z)∈ L

^{3}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 |

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 |

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. |

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') |

(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 |

∴(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 _________.

41 | |

42 | |

43 | |

44 |

__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 28 |

A binary operation ⊕ on a set of integers is defined as x ⊕ y = x^{2 }+ y^{2}. 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 |

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 = x

^{2}+ y

^{2}--------(1)

Replace x, y in (1)

y⊕x = y

^{2}+ x

^{2}which is same as (1), so this is cumulative

(x⊕y)⊕z = (x

^{2}+ y

^{2}) ⊕ z

= (x

^{2}+ y

^{2}) + z

^{2}

= x

^{2}+ y

^{2}+ z

^{2}+ 2x

^{2}y

^{2}----------(2)

x⊕(y ⊕ z) = x ⊕ (y

^{2}+ z

^{2})

= x

^{2}+ (y

^{2}+ z

^{2})

^{2}

= x

^{2}+ y

^{2}+ z

^{2}+ 2y

^{2z2 ----------- (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 |

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) |

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