Question 1
A binary operation \oplus on a set of integers is defined as x \oplus y = x2 + y2. Which one of the following statements is TRUE about \oplus?
A
Commutative but not associative
B
Both commutative and associative
C
Associative but not commutative
D
Neither commutative nor associative
GATE CS 2013    Set Theory & Algebra    
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Question 1 Explanation: 
Associativity: A binary operation ∗ on a set S is said to be associative if it satisfies the associative law: a ∗ (b ∗c) = (a ∗b) ∗c for all a, b, c ∈S. Commutativity: A binary operation ∗ on a set S is said to be commutative if it satisfies the condition: a ∗b=b ∗a for all a, b, ∈S. In this case, the order in which elements are combined does not matter. Solution: Here a binary operation on a set of integers is defined as x⊕ y = x2 + y2. for Commutativity: x ⊕y= y ⊕x. LHS=> x ⊕y= x^2+ y^2 RHS=> y ⊕x= y^2+x^2 LHS = RHS. hence commutative. for Associativity: x ⊕ (y ⊕ z) =(x ⊕ y) ⊕ z LHS=> x ⊕ (y⊕ z) = x ⊕ ( y^2+z^2)= x^2+(y^2+z^2)^2 RHS=> (x ⊕y) ⊕z= ( x^2+y^2) ⊕z=(x^2+y^2)^2+z^2 So, LHS ≠ RHS, hence not associative. Reference: http://faculty.atu.edu/mfinan/4033/absalg3.pdf This solution is contributed by Nitika Bansal Another Solution : [Tex]\oplus[/Tex] commutative as x[Tex]\oplus[/Tex]y is always same as y[Tex]\oplus[/Tex]x. [Tex]\oplus[/Tex] is not associative as (x[Tex]\oplus[/Tex]y)[Tex]\oplus[/Tex]z is (x^2 + y^2)^2 + z^2, but x[Tex]\oplus[/Tex](y[Tex]\oplus[/Tex]z) is x^2 + (y^2 + z^2)^2.
Question 2
Consider the set S = {1, ω, ω2}, where ω and w2 are cube roots of unity. If * denotes the multiplication operation, the structure (S, *) forms
A
A group
B
A ring
C
An integral domain
D
A field
GATE CS 2010    Set Theory & Algebra    
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Question 2 Explanation: 
A group is a set of elements together with an operation that combines any two of its elements to form a third element also in the set while satisfying four conditions called the group axioms, namely closureassociativityidentity and invertibility. (Source: http://en.wikipedia.org/wiki/Group_(mathematics) (S, *)  is a group with identity as 1    
Question 3
Which one of the following in NOT necessarily a property of a Group?
A
Commutativity
B
Associativity
C
Existence of inverse for every element
D
Existence of identity
GATE-CS-2009    Set Theory & Algebra    
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Question 3 Explanation: 
A group is a set, G, together with an operation • (called the group law of G) that combines any two elements a and b to form another element, denoted a • b or ab. To qualify as a group, the set and operation, (G, •), must satisfy four requirements known as the group axioms: Closure For all a, b in G, the result of the operation, a • b, is also in G.b Associativity For all a, b and c in G, (a • b) • c = a • (b • c). Identity element There exists an element e in G, such that for every element a in G, the equation e • a = a • e = a holds. Such an element is unique (see below), and thus one speaks of the identity element. Inverse element For each a in G, there exists an element b in G such that a • b = b • a = e, where e is the identity element. The result of an operation may depend on the order of the operands. In other words, the result of combining element a with element b need not yield the same result as combining element b with element a; the equation a • b = b • a may not always be true. This equation always holds in the group of integers under addition, because a + b = b + a for any two integers (commutativity of addition). Groups for which the commutativity equation a • b = b • a always holds are called abelian groups (in honor of Niels Abel) Source: http://en.wikipedia.org/wiki/Group_(mathematics)
Question 4
Consider the binary relation R = {(x, y), (x, z), (z, x), (z, y)} on the set {x, y, z}. Which one of the following is TRUE?
A
R is symmetric but NOT antisymmetric
B
R is NOT symmetric but antisymmetric
C
R is both symmetric and antisymmetric
D
R is neither symmetric nor antisymmetric
GATE-CS-2009    Set Theory & Algebra    
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Question 4 Explanation: 
R is not symmetric as (x, y) is present, but (y, x) is not present in R. R is also not antisymmetric as both (x, z) and (z, x) are present in R.
Question 5
For the composition table of a cyclic group shown below

*

a

b

c

d

a

a

b

c

d

b

b

a

d

c

c

c

d

b

a

d

d

c

a

b

 
Which one of the following choices is correct?
A
a, b are generators
B
b, c are generators
C
c, d are generators
D
d, a are generators
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Question 5 Explanation: 
Check for all:-
a1 = a ,
a2 = a * a = a
a3 = a2 * a = a * a = a
a is not the generator since we are not able to express 
other members of the group in powers of a

Check for c -
c1 = c
c2 = c * c = b
c3 = c2 * c = b * c = d
c4 = c2 * c2 = b * b = a
We are able to generate all the members of the group from c ,
Hence c is the generator

Similarly check for d
Question 6
If P, Q, R are subsets of the universal set U, then 2 
A
Qc U Rc
B
P U Qc U Rc
C
Pc U Qc U Rc
D
U
Set Theory & Algebra    GATE CS 2008    
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Question 6 Explanation: 
104
Question 7
Let S be a set of nelements. The number of ordered pairs in the largest and the smallest equivalence relations on S are:
A
n and n
B
n2 and n
C
n2 and 0
D
n and 1
Set Theory & Algebra    GATE-CS-2007    
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Question 7 Explanation: 
Consider an example set, S = (1,2,3)

Equivalence property follows, reflexive, symmetric
and transitive

Largest ordered set are s x s = 
{ (1,1) (1,2) (1,3) (2,1) (2,2) (2,3) (3,1) (3,2) 
(3,3) } which are 9 which equal to 3^2 = n^2

Smallest ordered set are { (1,1) (2,2) ( 3,3)}
which are 3 and equals to n. number of elements.
Question 8
How many different non-isomorphic Abelian groups of order 4 are there
A
2
B
3
C
4
D
5
Set Theory & Algebra    GATE-CS-2007    
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Question 8 Explanation: 
2 can be written as 2 power 2.
Number of partitioning of 2 = no. of non isomorphic
                              abelian groups
2 can be partitioned as {(2),(1,1)}
Question 9
gatecs200726 gatecs2007261
A
A
B
B
C
C
D
D
Set Theory & Algebra    GATE-CS-2007    
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Question 10
gatecs200727
A
A
B
B
C
C
D
D
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Question 10 Explanation: 
To be basis of subspace x, 2 conditions are to be fulfilled 1) They must span x 2) The vectors have to be linearly independent 1)the general solution of x1+x2+x3=0 is [-x2-x3 , x2 , x3]^T (Transpose) Which gives two linearly independent solutions by by assuming x2 = 1 and x3 = 0 and next x3 = 1 and x2 = 0 gives [-1,1,0]^T and [-1,0,1]^T respectively. Since both of these can be generated by linear combinations of [1,-1,0]^T & [-1,0,1]^T given in question, it span x. 2) Above set of column vector is linearly independent because one cannot be obtained from another by scalar multiplication (second method rank is 2..that is why linearly independent)
Question 11
Let X, Y, Z be sets of sizes x, y and z respectively. Let W = X x Y. Let E be the set of all subsets of W. The number of functions from Z to E is:
A
z2xy
B
z x 2 xy
C
z2x + y
D
2xyz
Set Theory & Algebra    GATE-CS-2006    
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Question 11 Explanation: 
Number of functions from a set A of size m to set B of size n is nm, because each of the m elements of A has n choices for mapping. Now here m=|Z|=z, and n=|E|=2xy because number of subsets of a set of size n is 2n, and here set W has size of xy.
So number of functions from Z to E = (2xy)z=2xyz. So option (D) is correct. Source: http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2006.html
Question 12
The set {1, 2, 3, 5, 7, 8, 9} under multiplication modulo 10 is not a group. Given below are four plausible reasons. Which one of them is false?
A
It is not closed
B
2 does not have an inverse
C
3 does not have an inverse
D
8 does not have an inverse
Set Theory & Algebra    GATE-CS-2006    
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Question 12 Explanation: 
A is not closed under multiplication as we may get 0 after multiplication and 0 is not present in set. 2 doesn't have an inverse as there is no x such that (2*x) mod 10 is 1. 3 has an inverse as (3*7) mod 10 is 1. 8 doesn't have an inverse as there is no x such that (2*x) mod 10 is 1.
Question 13
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
Set Theory & Algebra    GATE-CS-2006    
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Question 13 Explanation: 
  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

Solution:

An equivalence relation on a set x is a subset of x*x, i.e., a collection R of ordered pairs of elements of x, satisfying certain properties. Write “x R y" to mean (x,y) is an element of R, and we say "x is related to y," then the properties are 1. Reflexive: a R a for all a Є R, 2. Symmetric: a R b implies that b R a for all a,b Є R 3. Transitive: a R b and b R c imply a R c for all a,b,c Є R.

An partial order relation on a set x is a subset of x*x, i.e., a collection R of ordered pairs of elements of x, satisfying certain properties. Write “x R y" to mean (x,y) is an element of R, and we say "x is related to y," then the properties are

1. Reflexive: a R a for all a Є R, 2. Anti-Symmetric: a R b and b R a implies that for all a,b Є R 3. Transitive: a R b and b R c imply a R c for all a,b,c Є R.

An total order relation a set x is a subset of x*x, i.e., a collection R of ordered pairs of elements of x, satisfying certain properties. Write “x R y" to mean (x,y) is an element of R, and we say "x is related to y," then the properties are

1. Reflexive: a R a for all a Є R, 2. Symmetric: a R b implies that b R a for all a,b Є R 3. Transitive: a R b and b R c imply a R c for all a,b,c Є R. 4. Comparability : either a R b or b R a for all a,b Є R.

As given in question, a relation R is defined on ordered pairs of integers as follows: (x,y) R(u,v) if x < u and y > v , reflexive property is not satisfied here , because there is > or < relationship between (x ,y) pair set and (u,v) pair set . Other way , if there would have been x <= u and y>= v (or x=u and y=v) kind of relation amongs elements of sets then reflexive property could have been satisfied. Since reflexive property in not satisfied here , so given realtion can not be equivalence ,partial order or total order relation.So ,Answer (A) is true .

This solution is contributed by Nirmal Bharadwaj.

Question 14
Let S denote the set of all functions f: {0,1}4 -> {0,1}. Denote by N the number of functions from S to the set {0,1}. The value of Log2Log2N is ______.
A
12
B
13
C
15
D
16
Set Theory & Algebra    GATE-CS-2014-(Set-1)    
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Question 14 Explanation: 
The given mapping S is defined by f:{0,1}^4 -> {0,1} .
So, number of functions from S will be 2^16.
Now N is defined by f : S-> {0,1}.
So Number of functions from S to {0,1} will be 2^S.
Hence log2log2N = log2S = 16 
Question 15
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
Set Theory & Algebra    GATE-CS-2014-(Set-2)    
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Question 15 Explanation: 
According to the given information :
S1 is true because NULL set is smaller than every other set.
S2 is true because the UNIVERSAL set {1, 2, ..., 2014} is larger than every other set.
 
Thus, both S1 and S2 are true.
 
Please comment below if you find anything wrong in the above post.
Question 16
Let X and Y be finite sets and f: X -> Y be a function. Which one of the following statements is TRUE? GATECS2014Q11
A
A
B
B
C
C
D
D
Set Theory & Algebra    GATE-CS-2014-(Set-3)    
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Question 16 Explanation: 
Let x = {a, b, c} and y = {1, 2}
A Function f maps each element of x to 1 in y.
f(a)=1 , f(b)=1 , f(c) =1
A = {a, b} B = {b, c}
----------------------------------------------
A ]
| f(A u B) | = |f({a, b, c})| = 3
| f(A)|+|f(B)| = 2 + 2 = 4 , LHS != RHS.
----------------------------------------------
B ]
f(A ∩ B) = f({b}) = { 1 }
f(A) ∩ f(B) = {1, 1} ∩ {1, 1} = {1, 1}
LHS != RHS
-----------------------------------------------
C ]
|f(A ∩ B)| = |f({b})| = |{ 1 }| = 1
min{|f(A)|,|f(B)|} = min(2,2) = 2
LHS != RHS
-----------------------------------------------
D ] In a function a value can be mapped only to one value.
Question 17
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
3
B
5
C
7
D
9
Set Theory & Algebra    GATE-CS-2014-(Set-3)    
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Question 17 Explanation: 
This is according to Lagrange's theorem that states that the order of subgroup should divide the order of group. Subgroups of G can be of order 1,3,5 or 15 and according to the conditions and choices, B is the right answer.
Question 18
If V1 and V2 are 4-dimensional subspaces of a 6-dimensional vector space V, then the smallest possible dimension of V1 ∩ V2 is ______.
A
1
B
2
C
3
D
4
Set Theory & Algebra    GATE-CS-2014-(Set-3)    
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Question 18 Explanation: 
First, note that V1+V2 is still in V, so dim(V1+V2)≤ 6. We know that dim(V1+V2)=dimV1+dimV2−dim(V1∩V2). So 6≥dim(V1+V2)=dimV1+dimV2−dim(V1∩V2) dim(V1∩V2)≥4+4−6=2. The answer is B.
Question 19
There are two elements x, y in a group (G,∗) such that every element in the group can be written as a product of some number of x's and y's in some order. It is known that
  x ∗ x = y ∗ y = x ∗ y ∗ x ∗ y = y ∗ x ∗ y ∗ x = e
where e is the identity element. The maximum number of elements in such a group is __________.
A
2
B
3
C
4
D
5
Set Theory & Algebra    GATE-CS-2014-(Set-3)    
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Question 19 Explanation: 
x * x = e, x is its own inverse
y * y = e, y is its own inverse

(x*y) * (x* y) = e, x*y is its own inverse

(y*x) * (y*x) = e, y*x is its own inverse
also x*x*e = e*e can be rewritten as follows

x*y*y*x = e*y*y*e = e, (Since y *y = e)

(x*y) * (y*x) = e shows that (x *y) and (y *x)
are each other’s inverse and we already know that
(x*y) and (y*x) are inverse of its own.

As per (G,*) to be group any element should have
only one inverse element (unique)

This implies x*y = y*x (is one element)

So the elements of such group are 4 which are
{x, y, e, x*y}.
See following definition of group from wikipedia. A group is a set, G, together with an operation • (called the group law of G) that combines any two elements a and b to form another element, denoted a • b or ab. To qualify as a group, the set and operation, (G, •), must satisfy four requirements known as the group axioms:[5] Closure For all a, b in G, the result of the operation, a • b, is also in G.b[›] Associativity For all a, b and c in G, (a • b) • c = a • (b • c). Identity element There exists an element e in G, such that for every element a in G, the equation e • a = a • e = a holds. Such an element is unique (see below), and thus one speaks of the identity element. Inverse element For each a in G, there exists an element b in G such that a • b = b • a = e, where e is the identity element. Source: https://en.wikipedia.org/wiki/Group_%28mathematics%29
Question 20
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
Set Theory & Algebra    GATE-CS-2014-(Set-3)    
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Question 20 Explanation: 
This kind of functions are called identity functions. We assume f(i) = k. So, f(k) = i. Now, since the values of ' i ' and ' j ' would be same for atleast some values if the domain and co - domain intersect, which is true for the given question, Q is definitely true. But this might not happen for all the values of ' i ', hence, P is not always true. Now, ' i ' ranges from 0 to 2014, so, it takes 2015 possible values. From the definition of a function, we know that for each input to the function, we have a unique output. Also, in the given question, domain and co - domain are exactly same. Therefore, the function is onto and hence, R is definitely true.   Thus, the correct option is B.
Question 21
Let E, F and G be finite sets. Let X = (E ∩ F) - (F ∩ G) and Y = (E - (E ∩ G)) - (E - F). Which one of the following is true?
A
X ⊂ Y
B
X ⊃ Y
C
X = Y
D
X - Y ≠ φ and Y - X ≠ φ
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Question 21 Explanation: 
If we draw the venn diagrams of both X and Y, we find that both cover exactly same region (shown in gigure below). 2006_22 So option (C) is correct. Source: http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2006.html
Question 22
Given a set of elements N = {1, 2, ..., n} and two arbitrary subsets A⊆N and B⊆N, how many of the n! permutations π from N to N satisfy min(π(A)) = min(π(B)), where min(S) is the smallest integer in the set of integers S, and π(S) is the set of integers obtained by applying permutation π to each element of S?
A
(n - |A ∪ B|) |A| |B|
B
(|A|2+|B|2)n2
C
n! |A∩B| / |A∪B|
D
|A∩B|2nC|A∪B|
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Question 22 Explanation: 
First let us understand what question is asking. So π is a function from N to N, which just permutes the elements of N, so there will be n! such permutations. Now given a particular π i.e. given a particular permutation scheme, we have to find number of permutations out of these n! permuations in which minimum elements of A and B after applying π to them are same. So for example, if N = {1,2,3}, π is {2,3,1}, and if A is {1,3}, then π(A) = {2,1}. Now number of elements in A ∪ B is |A ∪ B|. We can choose permutations for A ∪ B in nC|A∪B| ways. Note that here we are just choosing elements for permutation, and not actually permuting. Let this chosen set be P. Now once we have chosen numbers for permutations, we have to select mapping from each element of A ∪ B to some element of P. So first of all, to achieve required condition specified in question, we have to map minimum number in P to any of the number in A ∩ B, so that min(π(A)) = min(π(B)). We can do this in |A∩B| ways, since we can choose any element of |A∩B| to be mapped to minimum number in P. Now we come to permutation. We can permute numbers in P in |A∪B-1|! ways, since one number (minimum) is already fixed. Moreover, we can also permute remaining n - |A∪B-1| in (n - |A∪B-1|)! ways, so total no. of ways = nC|A∪B|∗|A∩B|∗|A∪B−1|!∗(n−|A∪B−1|)!=n!|A∩B||A∪B| So option (C) is correct. Note: Some answer keys on web have shown answer as option (D), which is clearly incorrect. Suppose |A ∪ B| = 3, and |A ∩ B| = 1, and n = 4, then option (D) evaluates to 14=0.25, which doesn't make sense. Source: http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2006.html
Question 23
GATECS2006Q25
A
3m
B
3n
C
2m + 1
D
2n + 1
Set Theory & Algebra    GATE-CS-2006    
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Question 23 Explanation: 
First of all, number of subsets of S of size 3 is mC3 i.e. n=mC3. Now we count number of subsets in which a particular element i appears, that will be (m−1)C2, because 1 element is already known, and we have to choose 2 elements from remaining m-1 elements. [Tex]\sum\limits_{i=1}^{m} f(i) = m * ^{m-1}\mathrm{C}_2 = 3 * ^m\mathrm{C}_3 = 3n[/Tex]
Question 24
Let A, B and C be non-empty sets and let X = (A - B) - C and Y = (A - C) - (B - C). Which one of the following is TRUE?
A
X = Y
B
X ⊂ Y
C
Y ⊂ X
D
none of these
Set Theory & Algebra    GATE-CS-2005    
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Question 24 Explanation: 
We can solve it by making Venn diagram
Question 25
The following is the Hasse diagram of the poset [{a, b, c, d, e}, ≤] GATECS2005Q9 The poset is
A
not a lattice
B
a lattice but not a distributive lattice
C
a distributive lattice but not a Boolean algebra
D
a Boolean algebra
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Question 25 Explanation: 
It is a lattice but not a distributive lattice.

Table for Join Operation of above Hesse diagram

V |a b c d e
________________
a |a a a a a
b |a b a a b
c |a a c a c
d |a a a d d
e |a b c d e

Table for Meet Operation of above Hesse diagram

^ |a b c d e
_______________
a |a b c d e
b |b b e e e
c |c e c e e
d |d e e d e
e |e e e e e

Therefore for any two element p, q in the lattice (A,<=)

p <= p V q ; p^q <= p

This satisfies for all element (a,b,c,d,e).

which has 'a' as unique least upper bound and 'e' as unique 
greatest lower bound.

The given lattice doesn't obey distributive law, so it is 
not distributive lattice,

Note that for b,c,d we have distributive law

b^(cVd) = (b^c) V (b^d). From the diagram / tables given above 
we can verify as follows,

(i) L.H.S. = b ^ (c V d) = b ^ a = b

(ii) R.H.S. = (b^c) V (b^d) = e v e = e

b != e which contradict the distributive law. 
Hence it is not distributive lattice.

so, option (B) is correct. 
Question 26
The set {1, 2, 4, 7, 8, 11, 13, 14} is a group under multiplication modulo 15. The inverses of 4 and 7 are respectively
A
3 and 13
B
2 and 11
C
4 and 13
D
8 and 14
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Question 26 Explanation: 
We know that,
For a number 'n', (n) x (n') = I, where
n' = inverse of 'n' I = Identity element Now, the identity element for multiplication is 1. So, we need to find two numbers 'm' and 'n' such that (4 x m) % 15 = 1 and (7 x n) % 15 = 1,
where 'm' is the inverse of 4 and 'n' is the inverse of 7, and both 'm' and 'n' belong to the given set. Thus, from the given set, it can be easily identified by putting values that m = 4 and n = 13. So, C is the correct choice.
  Please comment below if you find anything wrong in the above post.
Question 27
Let R and S be any two equivalence relations on a non-empty set A. Which one of the following statements is TRUE?
A
R ∪ S, R ∩ S are both equivalence relations
B
R ∪ S is an equivalence relation
C
R ∩ S is an equivalence relation
D
Neither R ∪ S nor R ∩ S is an equivalence relation
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Question 28
Let f: B → C and g: A → B be two functions and let h = f o g. Given that h is an onto function. Which one of the following is TRUE?
A
f and g should both be onto functions.
B
f should be onto but g need not be onto
C
g should be onto but f need not be onto
D
both f and g need not be onto
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Question 28 Explanation: 
A function f: X → Y is called on-to function if for every value in set Y, there is a value in set X.
Given that, f: B → C and g: A → B and h = f o g.  

Note that the sign o represents composition. 

h is basically f(g(x)). So h is a function from set A
to set C.

It is also given that h is an onto function which means
for every value in C there is a value in A. 
We map from C to A using B. So for every value in C, there must be a value in B. It means f must be onto. But g may or may not be onto as there may be some values in B which don't map to A. Example :
Let us consider following sets
A : {a1, a2, a3}
B : {b1, b2}
C : {c1}

And following function values
f(b1) = c1
g(a1) = b1, g(a2) = b1, g(a3) = b1

Values of h() would be,
h(a1) = c1, h(a2) = c1, h(a3) = c1

Here h is onto, therefore f is onto, but g is 
onto as b2 is not mapped to any value in A.
Given that, f: B → C and g: A → B and h = f o g.
Question 29
What is the minimum number of ordered pairs of non-negative numbers that should be chosen to ensure that there are two pairs (a, b) and (c, d) in the chosen set such that "a ≡ c mod 3" and "b ≡ d mod 5"
A
4
B
6
C
16
D
24
Set Theory & Algebra    GATE-CS-2005    
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Question 29 Explanation: 

a = c mod 3 (given) Thus, ‘a’ can be any one of these values : 0, 1, 2
b = d mod 5 (given) Thus, ‘b’ can be any one of these values : 0, 1, 2, 3, 4
Thus, ordered pair for (a, b) are : (0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (2, 0), (2, 1), (2, 2), (2, 3), (2, 4)
Therefore, ordered pair (a, b) has 15 combinations and ordered pair (c, d) has 1 combination. Total combinations = 15 + 1 = 16
 
Hence, option (C) is correct.
 
Please comment below if you find anything wrong in the above post.
Question 30
Consider the binary relation:
S = {(x, y) | y = x+1 and x, y ∈ {0, 1, 2, ...}}
The reflexive transitive closure of S is
A
{(x, y) | y > x and x, y ∈ {0, 1, 2, ... }}
B
{(x, y) | y ≥ x and x, y ∈ {0, 1, 2, ... }}
C
{(x, y) | y < x and x, y ∈ {0, 1, 2, ... }}
D
{(x, y) | y ≤ x and x, y ∈ {0, 1, 2, ... }}
Set Theory & Algebra    GATE-CS-2004    
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Question 30 Explanation: 

Reflexive closure of a relation R on set S is the smallest reflexive relation which contains R.
If S = {(0, 1), (1, 2)} , we make it reflexive by taking its union with set {(0, 0), (1, 1), (2, 2)}. Thus, reflexive closure of S = {(0, 0), (0, 1), (1, 1), (1, 2), (2, 2)}.
Now transitive closure is defined as smallest transitive relation which contains S.
We check where does it violate property of transitivity then add appropriate pair. We have (0, 1) and (1, 2) but not (0, 2). So, S = {(0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2)} now.
 
Thus, option (B) matches the final set S.
 
Please comment below if you find anything wrong in the above post.
Question 31
The following is the incomplete operation table a 4-element group.
 *  e  a  b  c
 e  e  a  b  c
 a  a  b  c  e
 b
 c
The last row of the table is
A
c a e b
B
c b a e
C
c b e a
D
c e a b
Set Theory & Algebra    GATE-CS-2004    
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Question 31 Explanation: 
It is given that the given set of 4 elements is group. The element 'e' is clearly identity as the row corresponding to it has all values same as the other operand. Also, since a*c is e, c*a should also be e which is only the case in option D
Question 32
The inclusion of which of the following sets into
S = {{1, 2}, {1, 2, 3}, {1, 3, 5}, (1, 2, 4), (1, 2, 3, 4, 5}}
is necessary and sufficient to make S a complete lattice under the partial order defined by set containment ?
A
{1}
B
{1}, {2, 3}
C
{1}, {1, 3}
D
{1}, {1, 3}, (1, 2, 3, 4}, {1, 2, 3, 5)
Set Theory & Algebra    GATE-CS-2004    
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Question 32 Explanation: 
  • A partially ordered set L is called a complete lattice if every subset M of L has a least upper bound called as supremum and a greatest lower bound called as infimum.
  • We are given a set containment relation.
  • So, supremum element is union of all the subset and infimum element is intersection of all the subset.
  • Set S is not complete lattice because although it has a supremum for every subset, but some subsets have no infimum. We take subset {{1,3,5},{1,2,4}}.Intersection of these sets is {1}, which is not present in S. So we have to add set {1} in S to make it a complete lattice
Thus, option (A) is correct. Please comment below if you find anything wrong in the above post.
Question 33
How many graphs on n labeled vertices exist which have at least (n2 - 3n)/2 edges ? GATECS2004Q79
A
A
B
B
C
C
D
D
Set Theory & Algebra    GATE-CS-2004    
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Question 33 Explanation: 

Let there be total number of edges of a graph to be formed be e and total number of vertices be n.

So the total number of ways a graph can be formed having exactly e edges and v vertices is given by the total number of ways we can select e edges among v edges

i. e.

C (v, e)

Now, here e = (n2 – 3n)/2 and v = n(n-1)/2 (maximum number of edges in a simple graph).

So to select a edge we can do that in C (v, (n2 – 3n)/2) ways.

Since minimum no. of edges to be selected for a graph are (n2 – 3n)/2.

So total number of graphs possible will be :

C(v, e) + C(v, e+1) + C(v, e+2) +…………...+ C(v, v).

C(v, v-e) + C(v, v-(e+1)) + C(v, v-(e+2)) +…………...+ C(v, v-v).

Since v - e = n(n-1)/2 - (n2 – 3n)/2 = n

Therefore

C(v, n) + C(v, n-1) + C(v, n-2) +…………...+ C(v, 0).

Solving this we will get

 graph_90

This solution is contributed by Namita Singh.

Question 34
Consider the set ∑* of all strings over the alphabet ∑ = {0, 1}. ∑* with the concatenation operator for strings
A
does not form a group
B
forms a non-commutative group
C
does not have a right identity element
D
forms a group if the empty string is removed from ∑*
Set Theory & Algebra    GATE-CS-2003    
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Question 35
Let (5, ≤) be a partial order with two minimal elements a and b, and a maximum element c.
Let P : S → {True, False} be a predicate defined on S.
Suppose that P(a) = True, P(b) = False and 
P(x) ⇒ P(y) for all x, y ∈ S satisfying x ≤ y, 
where ⇒ stands for logical implication.
Which of the following statements CANNOT be true ?
A
P(x) = True for all x ∈ S such that x ≠ b
B
P(x) = False for all x ∈ S such that x ≠ a and x ≠ c
C
P(x) = False for all x ∈ S such that b ≤ x and x ≠ c
D
P(x) = False for all x ∈ S such that a ≤ x and b ≤ x
Set Theory & Algebra    GATE-CS-2003    
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Question 35 Explanation: 

'a' and 'b' are given as minimal elements. No other element in S is of lower order than either a or b. 'c' is given as maximum element. So, c is of higher order than any other element in S.
P(a) = True means all elements 'x' which have an edge from element 'a' have to be true. Since there is an edge from 'a', we have to satisfy formula P(a) => P(x), which can only be done by setting P(x) = True.
Elements which have an edge from b can be anything because formula P(b) => P(x) is satisfied as P(b) = False.
(A) This statement is true because making all elements true trivially satisfy formula P(x) => P(y).
(B) This statement is true if all elements are connected from b then all elements can be false.
(C) This statement is true because b<=x ensures x!=a and for all other elements P(x) can be false without violating the given implication.
(D) This statement is false. Since, P(a) = true , for all 'x' such that a<=x, P(x) must be true. We do have at least one such 'x', which is 'c' as it is the maximum element.
 
Thus, option (D) is the answer.
 
Please comment below if you find anything wrong in the above post.
Question 36
Let f : A → B be an injective (one-to-one) function.
Define g : 2A → 2B as :
g(C) = {f(x) | x ∈ C}, for all subsets C of A.
Define h : 2B → 2A as :
h(D) = {x | x ∈ A, f(x) ∈ D}, for all subsets D of B. 
Which of the following statements is always true ?
A
g(h(D)) ⊆ D
B
g(h(D)) ⊇ D
C
g(h(D)) ∩ D = ф
D
g(h(D)) ∩ (B - D) ≠ ф
Set Theory & Algebra    GATE-CS-2003    
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Question 37
Let ∑ = (a, b, c, d, e) be an alphabet. We define an encoding scheme as follows : g(a) = 3, g(b) = 5, g(c) = 7, g(d) = 9, g(e) = 11. GATECS2003Q39 Which of the following numbers is the encoding h of a non-empty sequence of strings ?
A
27 37 57
B
28 38 58
C
29 39 59
D
210 510 710
Set Theory & Algebra    GATE-CS-2003    
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Question 37 Explanation: 

Since, the answer is a product of three prime numbers 2, 3 and 5. So, we have three non-empty sequence of strings : "a", "a" and "a".
f(s) = 2x for some x Since, 7 and 9 are not multiple of 2. So, options (A) and (C) are eliminated.
f(a) = 23 = 8 h = 28 3858
 
Thus, option (B) is correct.
 
Please comment below if you find anything wrong in the above post.
Question 38
Which of the following is true?
A
The set of all rational negative numbers forms a group under multiplication.
B
The set of all non-singular matrices forms a group under multiplication.
C
The set of all matrices forms a group under multiplication.
D
Both (2) and (3) are true.
Set Theory & Algebra    GATE-CS-2002    
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Question 38 Explanation: 
A group is a set of elements such that any two elements of the group combine to form a third element of the same group. Also, a group must satisfy certain properties: Closure Property - Any two elements of the set when operated open by an operator form a third element that must also be in the set. Associative Property - For an expression with three or more operands having the same operator between them, the order of operation does not matter as long as the sequence of operands are not changed. For example, (a + b) + c = a + (b + c). Identity element Property - Each set must have an identity element, which is an element of the set such that when operated upon with another element of the set, it gives the element itself. For example, a + 0 = a. Here, 0 is the identity element. Invertibility Property - For each element of the set, inverse should exist.   Now, for the given statements, we have A is incorrect as it does not satisfies closure property. If we take two negative numbers and multiply them, we get a positive number which is not in the set. B is correct. The matrices in the set must be non - singular, i.e., their determinant should not be zero, for the inverse to exist (Invertibility Property). C is incorrect as the inverse of a singular (determinant = 0) matrix does not exist (Invertibility Property violated).   Thus, B is the correct option.     Please comment below if you find anything wrong in the above post.
Question 39
The binary relation S = ф (empty set) on set A = {1, 2, 3} is :
A
Neither reflexive nor symmetric
B
Symmetric and reflexive
C
Transitive and reflexive
D
Transitive and symmetric
Set Theory & Algebra    GATE-CS-2002    
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Question 39 Explanation: 
  • Reflexive : A relation is reflexive if every element of set is paired with itself. Here none of the element of A is paired with themselves, so S is not reflexive.
  • Symmetric : This property says that if there is a pair (a, b) in S, then there must be a pair (b, a) in S. Since there is no pair here in S, this is trivially true, so S is symmetric.
  • Transitive : This says that if there are pairs (a, b) and (b, c) in S, then there must be pair (a,c) in S. Again, this condition is trivially true, so S is transitive. Thus, option (D) is correct. Please comment below if you find anything wrong in the above post.
Question 40
Consider the following relations:
R1(a,b) iff (a+b) is even over the set of integers
R2(a,b) iff (a+b) is odd over the set of integers
R3(a,b) iff a.b > 0 over the set of non-zero rational numbers
R4(a,b) iff |a - b| <= 2 over the set of natural numbers
Which of the following statements is correct?
A
R1 and R2 are equivalence relations, R3 and R4 are not
B
R1 and R3 are equivalence relations, R2 and R4 are not
C
R1 and R4 are equivalence relations, R2 and R3 are not
D
R1, R2, R3 and R4 are all equivalence relations
Set Theory & Algebra    GATE-CS-2001    
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Question 40 Explanation: 
So basically, we have to tell whether these relations are equivalence or not.
  1. R1(a,b)
    • Reflexive : Yes, because (a+a) is even.
    • Symmetrix : Yes, (a+b) is even (b+a) is even.
    • Transitive : Yes, because (a+b) is even and (b+c) is even (a+c) is even.
    So R1 is equivalence relation.
  2. R2(a,b)
    • Reflexive : No, because (a+a) is even.
    So R2 is not equivalence relation.
  3. R3(a,b)
    • Reflexive : Yes, because a.a > 0.
    • Symmetrix : Yes, a.b > 0 b.a > 0.
    • Transitive : Yes, because a.b > 0 and b.c > 0 a.c > 0.
    So R3 is equivalence relation.
  4. R4(a,b)
    • Reflexive : Yes, because |a-a| 2.
    • Symmetrix : Yes, |a-b| 2 |b-a| 2.
    • Transitive : No, because |a-b| 2 and |b-c| 2 (a-c) is even.
    So R4 is not equivalence relation.
So option (b) is correct.. Source: http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2001.html
Question 41
Consider the following statements:
S1: There exists infinite sets A, B, C such that 
    A ∩ (B ∪ C) is finite.
S2: There exists two irrational numbers x and y such
    that (x+y) is rational.
Which of the following is true about S1 and S2?
A
Only S1 is correct
B
Only S2 is correct
C
Both S1 and S2 are correct
D
None of S1 and S2 is correct
Set Theory & Algebra    GATE-CS-2001    
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Question 41 Explanation: 
S1: A ∩ (B ∪ C) Here S1 is finite where A, B, C are infinite We’ll prove this by taking an example. Let A = {Set of all even numbers} = {2, 4, 6, 8, 10...} Let B = {Set of all odd numbers} = {1, 3, 5, 7...........} Let C = {Set of all prime numbers} = {2, 3, 5, 7, 11, 13......} B U C = {1, 2, 3, 5, 7, 9, 11, 13......} A ∩ (B ∪ C) Will be equals to: {2} which is finite. I.e. using A, B, C as infinite sets the statement S1 is finite. So, statement S1 is correct. S2: There exists two irrational numbers x, y such that (x+y) is rational To prove this statement as correct, we take an example. Let X = 2-Sqrt (3), Y = 2+Sqrt (3) => X, Y are irrational X+Y = 2+Sqrt (3) + 2-Sqrt (3) = 2+2 = 4 So, statement S2 is also correct. Answer is Option C Both Statements S1, S2 are correct.   This solution is contributed by Anil Saikrishna Devarasetty.
Question 42
A relation R is defined on the set of integers as xRy if f(x + y) is even. Which of the following state­ments is true?
A
R is not an equivalence relation
B
R is an equivalence relation having 1 equivalence class
C
R is an equivalence relation having 2 equivalence classes
D
R is an equivalence relation having 3 equivalence classes
Set Theory & Algebra    GATE-CS-2000    
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Question 42 Explanation: 
There are 2 2 equivalence classes. 1) All odd integers. (Reflexive as sum of two even is even, Symmetric and Transitive as + operation is Symmetric and Transitive) 2) All Even Integers. (Reflexive as sum of two odd is even, Symmetric and Transitive as + operation is Symmetric and Transitive)
Question 43
Let P(S) denotes the power set of set S. Which of the following is always true? GATECS2000!28
A
a
B
b
C
c
D
d
Set Theory & Algebra    GATE-CS-2000    
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Question 43 Explanation: 
  Ex:  let a set ‘s’ be, s={1,2} Power-set(s)=P(s)={{},{1},{2},{1,2}} Note: ‘{}’ denotes empty or NULL set or φ. In general, if a set ‘s’ contains ‘n’ elements,  power-set will contain 2^n elements, that denotes set of all subset of ‘s’. Solution: Option (a): P(P(S))=P(S) If, S={1} P(S)={ φ, {1}} P(P(S))={ φ,{ φ},{1},{ φ, {1}}} NOTE: LHS is set of set of subsets of ‘S’ and RHS is set of subsets of ‘S’. So, we can conclude this option is false. Option (b): P(P(S)) ∩P(S)={ φ}. If, S={1} P(S)={ φ, {1}} P(P(S))={ φ,{ φ},{1},{ φ, {1}}} P(P(S)) ∩P(S)= { φ}=RHS So, we can conclude this option is TRUE. Option (a): P(S) ∩S=P(S) If, S={1} P(S)={ φ, {1}} P(S) ∩S= φ NOTE: We can’t find anything common in set of elements (S) and set of sets(P(S)). So, we can conclude this option is false. Option (a): S ∉ P(S) If, S={1} P(S)={ φ, {1}} Clearly, S is subset of P(S). NOTE: By definition only, its clear that power-set is set of all subset that will contain S also. So, we can conclude this option is false.   This solution is contributed by Sandeep pandey.
Question 44
The binary operator ≠ is defined by the following truth table Q46 Which one of the following is true about the binary operator ≠?
A
Both commutative and associative
B
Commutative but not associative
C
Not commutative but associative
D
Neither commutative nor associative
Set Theory & Algebra    GATE-CS-2015 (Set 1)    
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Question 44 Explanation: 
The operation is basically XOR which is both commutative and associative.
Question 45
Suppose L = {p, q, r, s, t} is a lattice represented by the following Hasse diagram: Q48 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
Set Theory & Algebra    GATE-CS-2015 (Set 1)    
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Question 45 Explanation: 
Number of triplets in L3 = Number of ways in which 
                             we can choose 3 elements 
                             from 5 with repetition 

                              = 5 * 5 * 5 
                              = 125.

Now, when we take x = t, then the given condition for L is
satisfied for any y and z. Here, y and z can be taken in 
5 * 5 = 25 ways.

Take x = r, y = p, z = p. Here also, the given condition is
satisfied. So, pr > 25 / 125 > 1/5.

For x = q, y = r, z = s, the given condition is not satisfied
as q ⋁ (r ⋀ s) = q ⋁ p = q, while (q ⋁ r) ⋀ (q ⋁ s) = t ⋀ t = t.

So, pr ≠ 1.

Hence  D choice.
Question 46
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
Set Theory & Algebra    GATE-CS-2015 (Set 2)    
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Question 46 Explanation: 
R cannot be reflexive as 'a' and 'b' have to be distinct in aRb. R is symmetric if a and b have a common divisor, then b and a also have. R is not transitive as aRb and bRc doesn't mean aRc. For example 3 and 15 have common divisor, 15 and 5 have common divisor, but 3 and 5 don't have.
Question 47
The cardinality of the power set of {0, 1, 2 . . ., 10} is _________.
A
1024
B
1023
C
2048
D
2043
Set Theory & Algebra    GATE-CS-2015 (Set 2)    
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Question 47 Explanation: 
The power set has 2n elements. For n = 11, size of power set is 2048.
Question 48
Consider two relations R1(A, B) with the tuples (1, 5), (3, 7) and R1(A, C) = (1, 7), (4, 9). Assume that R(A,B,C) is the full natural outer join of R1 and R2. Consider the following tuples of the form (A,B,C)
  a = (1, 5, null),
  b = (1, null, 7), 
  c = (3, null, 9), 
  d = (4, 7, null), 
  e = (1, 5, 7), 
  f = (3, 7, null), 
  g = (4, null, 9). 
Which one of the following statements is correct?
A
R contains a, b, e, f, g but not c, d
B
R contains a, b, c, d, e, f, g
C
R contains e, f, g but not a, b
D
R contains e but not f, g
Set Theory & Algebra    GATE-CS-2015 (Set 2)    
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Question 48 Explanation: 
Below is R1 
  A   |   B
 -----------
  1   |   5
  3   |   7


Below is R2
  A   |   C
 -----------
  1   |   7
  4   |   9 


Full outer join of above two is 

  A   |   B   |   C
 -------------------
  1   |   5   |   7
  3   |   7   |  NULL
  4   |  NULL |   9
So the full outer join contains e = (1, 5, 7), f = (3, 7, null), g = (4, null, 9).
Question 49
The number of onto functions (surjective functions) from set X = {1, 2, 3, 4} to set Y = {a, b, c} is ________________
A
36
B
64
C
81
D
72
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Question 49 Explanation: 
A function f from X to Y is called onto if for all 'y' in Y there is an 'x' in X such that f(x) = y.          In onto functions, all elements in Y are used. Source: http://www.regentsprep.org/regents/math/algtrig/atp5/OntoFunctions.htm Every Surjective or Onto function sends two elements of {1, 2, 3, 4} to the same element of {a, b, c}. There are 4C2 = 6 such pairs of elements. The pairs are {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}. For a given pair {i, j} ⊂ {1, 2, 3, 4}, there are 3! sujective functions such that f(i) = f(j). Hence there are total 6 * 6 = 36 surjective functions.
Question 50
Let X and Y denote the sets containing 2 and 20 distinct objects respectively and F denote the set of all possible functions defined from X and Y. Let f be randomly chosen from F. The probability of f being one-to-one is _________
A
0.95
B
0.80
C
0.75
D
0.70
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Question 50 Explanation: 
X has 2 elements Y has 20 elements Number of functions from X to Y is 20*20 [Every element can take any of the 20 values] Number of one to one functions from X to Y is 20*19 [Every element takes a different value] So probability of a function being one to one = (20*19) / (20*20) = 380 / 400 = 0.95
Question 51
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
Set Theory & Algebra    GATE-CS-2015 (Set 3)    
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Question 51 Explanation: 
((p, q), (r, s)) ∈ R if and only if p–s = q–r

(p, q) is not related to (p, q)

as p-q is not same as q-p.

The relation is symmetric because if p–s = q–r, then s-q = s-p. 
Question 52
Let R1 be a relation from A = {1, 3, 5, 7} to B = {2, 4, 6, 8} and R2 be another relation from B to C = {1, 2, 3, 4} as defined below:
  1. An element x in A is related to an element y in B (under R1) if x + y is divisible by 3.
  2. An element x in B is related to an element y in C (under R2) if x + y is even but not divisible by 3.
Which is the composite relation R1R2 from A to C?  
A
R1R2 = {(1, 2), (1, 4), (3, 3), (5, 4), (7, 3)}
B
R1R2 = {(1, 2), (1, 3), (3, 2), (5, 2), (7, 3)}
C
R1R2 = {(1, 2), (3, 2), (3, 4), (5, 4), (7, 2)}
D
R1R2 = {(3, 2), (3, 4), (5, 1), (5, 3), (7, 1)}
Set Theory & Algebra    GATE-IT-2004    
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Question 52 Explanation: 

R1 is a relation from A = {1, 3, 5, 7} to B = {2, 4, 6, 8} . Under R1, an element x in A is related to an element y in B if x + y is divisible by 3.
Thus, R1 = {(1, 2), (1, 8), (3, 6), (5, 4), (7, 2), (7, 8)}
R2 is a relation from B = {2, 4, 6, 8} to C = {1, 2, 3, 4} Under R2, an element y in B is related to an element z in C if y + z is even but not divisible by 3.
Thus, R2 = {(2, 2), (4, 4), (6, 2), (6, 4), (8, 2)}
it
Thus, R1R2 = {(1, 2), (3, 2), (3, 4), (5, 4), (7, 2)}
 
Thus, option (C) is correct.
 
Please comment below if you find anything wrong in the above post.
Question 53
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
Set Theory & Algebra    Gate IT 2005    
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Question 54
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 S1 ⊂ S2 or S2⊂ S1. What is the maximum cardinality of C?  
A
n
B
n + 1
C
2n-1 + 1
D
n!
Set Theory & Algebra    Gate IT 2005    
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Question 54 Explanation: 
  As i already mentioned, even if there is small scope of getting solution by substitution method, Just go for it...!! Here let n=2 A = {1, 2} All subsets formed by A are: - {}, {1}, {2}, {1,2}. C is a collection of distinct subsets such that for any S1, S2 either S1⊂S2 or S2⊂S1. So for C, {} null set can be included always since it null. set is a subset of every set. We can choose one from either {1} or {2}, {1,2} can be included to maximise the cardinality. So, here 1) If {1} is chosen then C = {}, {1}, {1,2} here every set is subset of other. 2) If {2} is chosen then C = {}, {2}, {1,2} here also every set is subset of other. So, answer should be 2 but it incluedes empty set also therefore the maximum cardinality of C is 3.   This solution is contributed by Anil Saikrishna Devarasetty .
Question 55
GATECS201612 [This Question was originally a Fill-in-the-Blanks question]
A
1
B
2
C
3
D
4
Set Theory & Algebra    GATE-CS-2016 (Set 1)    
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Question 55 Explanation: 
Let us assume: f(1) = x. Then,
 f(2) = f(2/2) = f(1) = x
 f(3) = f(3+5) = f(8) = f(8/2) = f(4/2) = f(2/1) = f(1) = x
Similarly, f(4) = x
  f(5) = f(5+5) = f(10/2) = f(5) = y.
So it will have two values. All multiples of 5 will have value y and others will have value x. It will have 2 different values.
Question 56
A binary relation R on N x 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.
Set Theory & Algebra    GATE-CS-2016 (Set 2)    
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Question 56 Explanation: 
THEORY: REFLEXTIVE RELATION:   A relation ‘R’ on a set ‘A’ is said to be reflexive if, (xRx) for every x£A. Ex. If A= {1,2} And R, and P be a relation on AxA, defined as, R= {(2,2),(1,1)} => R is reflexive as it contains all pair of type (xRx). P= {(1,1)} => P is not reflexive relation on A, as it doesn’t contain (2,2).   TRANSITIVE RELATION: A relation ‘R’ on set ‘A’ is said to be transitive if (xRy) and (yRz), then (xRz) for every x,y,z £A. Ex: if A= {1,2} Let R be a relation on AxA, defined as, R= {(1,1),(1,2),(2,1)} => R is transitive.   SOLUTION: Given, (a, b) R (c, d) if a <= c or b <= d i.Check for reflexivity: if an element of set be (a,b) then, (a,b)R(a,b) should hold true. Here, a<=a or b<=b. So, (a,b)R(a,b) holds true. Hence, ‘R’ is reflexive. ii. Check for transitivity:  if elements of set be (2,3),(3,1) and(1,1) Then, (2,3)R(3,1) as 2<=3 And (3,1)R(1,1) as 1<=1 But (2,3)R(1,1) doesn’t hold true as 2>=1 and 3>=1. Hence, R is reflexive but not transitive.   This solution is contributed by Sandeep pandey.
Question 57
For the set N of natural numbers and a binary operation f : N x N → N, an element z ∊ N is called an identity for f, if f (a, z) = a = f(z, a), for all a ∊ N. Which of the following binary operations have an identity?
  1. f (x, y) = x + y - 3
  2. f (x, y) = max(x, y)
  3. f (x, y) = xy
 
A
I and II only
B
II and III only
C
I and III only
D
None of these
Set Theory & Algebra    GATE IT 2006    
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Question 57 Explanation: 
I f(x,y) = x+y-3 = x= y+x-3  =>  y=3 Here identity elements is 3 II f(x,y) = max(x,y)=x=max(y,x)  => y=1 Here identity elements is 1 (III f(x,y) =x^y is not same as f(y,x) = y^x. So no identity element.
Question 58
Given a boolean function f (x1, x2, ..., xn), which of the following equations is NOT true  
A
f (x1, x2, ..., xn) = x1'f(x1, x2, ..., xn) + x1f(x1, x2, ..., xn)
B
f (x1, x2, ..., xn) = x2f(x1, x2, …, xn) + x2'f(x1, x2, …,xn)
C
f (x1, x2, ..., xn) = xn'f(x1, x2, …, 0) + xnf(x1, x2, …,1)
D
f (x1, x2, ..., xn) = f(0, x2, …, xn) + f(1, x2, .., xn)
Functions    Set Theory & Algebra    GATE IT 2006    
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Question 58 Explanation: 
Option A: f (x1, x2, …, xn) = x1’f(x1, x2, …, xn) + x1f(x1, x2, …, xn) Case 1: taking x1=0 RHS = 1.f(x1, x2, …, xn) + 0.f(x1, x2, …, xn) RHS =f(x1, x2, …, xn). Case 2: taking x1=1 RHS = 0.f(x1, x2, …, xn) + 1.f(x1, x2, …, xn) RHS =f(x1, x2, …, xn). In both cases RHS=LHS, so, (A) is true Option B: f (x1, x2, …, xn) = x2f(x1, x2, …, xn) + x2’f(x1, x2, …, xn) Case 1: taking x2=0 RHS= 0.f(x1, x2, …, xn) + 1.f(x1, x2…,xn) RHS =f(x1, x2, …, xn). Case 2: taking x2=1 RHS = 1.f(x1, x2, …, xn) + 0.f(x1, x2, …, xn) RHS =f(x1, x2, …, xn). In both cases RHS=LHS, so, (B) is true. Option C: f (x1, x2, …, xn) = xn’f(x1, x2, …, 0) + xnf(x1, x2, …,1) Case 1: taking xn=0 RHS= 1.f(x1, x2, …, 0) + 0.f(x1, x2, …, 1) RHS =f(x1, x2, …, 0) Case 2: taking xn=1 RHS = 0.f(x1, x2, …, 0) + 1.f(x1, x2, …, 1) RHS =f(x1, x2, …, 1)In both cases RHS=LHS, so, (C) is true. Option D: f (x1, x2, …, xn) = f(0, x2, …, xn) + f(1, x2, .., xn) Here, no way to equate LHS and RHS so ‘NOT true’. NO term depends on value of ‘x1’.   This solution is contributed by Sandeep pandey.
Question 59
Consider the following first order logic formula in which R is a binary relation symbol. ∀x∀y (R(x, y)  => R(y, x)) The formula is  
A
satisfiable and valid
B
satisfiable and so is its negation
C
unsatisfiable but its negation is valid
D
satisfiable but its negation is unsatisfiable
Set Theory & Algebra    GATE IT 2006    
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Question 59 Explanation: 
  VxVy R(x,y) => R(y,x) The above given relation is symmetry But, we have both symmetric relastions possible and also possibility of anti symmetric relation  But neither of always holds for all possibilites of sets. => Both are satisfiable but not valid This solution is contributed by Anil Saikrishna Devarasetty.   One more solution : We are given a logical formula. So, to be valid it must be a symmetric relation. Hence, Option A is incorrect. Since, it is a logical formula => it is along with it's negation is satisfiable. Hence, option B is correct. This solution is contributed by Mohit Gupta .
Question 60
Let P, Q and R be sets let Δ denote the symmetric difference operator defined as PΔQ = (P U Q) - (P ∩ Q). Using Venn diagrams, determine which of the following is/are TRUE? PΔ (Q ∩ R) = (P Δ Q) ∩ (P Δ R) P ∩ (Q ∩ R) = (P ∩ Q) Δ (P Δ R)
A
I only
B
II only
C
Neither I nor II
D
Both I and II
Set Theory & Algebra    GATE IT 2006    
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Question 60 Explanation: 
2006_23
From above venn diagrams answer C
Question 61
What is the cardinality of the set of integers X defined below? X = {n | 1 ≤ n ≤ 123, n is not divisible by either 2, 3 or 5}
A
28
B
33
C
37
D
44
Set Theory & Algebra    GATE IT 2006    
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Question 61 Explanation: 
  anil_4   This solution is contributed by Anil Saikrishna Devarasetty.  
Question 62
Let A = {a, b, c, d }, B = { p, q, r, s } denote sets. R : A –> B, R is a function from A to B. Then which of the following relations are not functions ?
(i) { (a, p) (b, q) (c, r) }
(ii) { (a, p) (b, q) (c, s) (d, r) }
(iii) { (a, p) (b, s) (b, r) (c, q) }
A
(i) and (ii) only
B
(ii) and (iii) only
C
(i) and (iii) only
D
None of these
Set Theory & Algebra    GATE 2017 Mock    
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Question 62 Explanation: 
(i) is not a function since the element d (∈A) is not related to any element in B.
(iii) is not a function since b (∈A) is associated with two different elements of B.
Question 63
Let A = { 1,2,3,4,…….∞ } and a binary operation ‘+’ is defined by a + b = ab ∀ a,b ∈ A. Which of the following is true ?
A
( A, + ) is a semi group but not monoid
B
( A, + ) is a monoid but not group
C
( A, + ) is a group
D
( A, + ) is not a semi group
Set Theory & Algebra    GATE 2017 Mock    
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Question 63 Explanation: 
Given, A = { 1,2,3,4,…….∞ }
A: ( A,+ ) to be semi group, it has to satisfy Closure property & Associative property
Closure: Given, a + b = ab => 1 + 2 = 3
2 + 3 = 5
So what ever value we take for a,b their ab value belongs to A so it satisfies closure.
Associative: I n order to satisfy associative property it needs to hold
( a + b) + c = a + ( b + c )
Checking:
( a + b) + c = a + ( b + c )
ab + c = a + bc
abc = abc
So for all values for a,b,c associative property satisfies
Monoid:
a + e = e + a = a, ∀ a ∈ A
ae = a
e = 1
Identity element is 1, so A is monoid.
Group:
a + b = b + a = e
It doesn’t satisfy the property because for all values of a,b it is not equal to e. So it not a group.
There are 63 questions to complete.

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