## Sets-And Relation

Question 1 |

Let R be the set of all binary relations on the set {1,2,3}. Suppose a relation is chosen from R at random. The probability that the chosen relation is reflexive (round off to 3 decimal places) is _____.

0.125 |

The number of reflexive relations is 2^(n^2-n).

The total number of relations on a set with n elements is 2^ (n^2).

The probability of choosing the reflexive relation out of set of relations is

= 2^(n^2-n) /2^ (n^2)

= 2^( n^2-n- n^2)

= 2^(-n)

Given n=3, the probability will be 2

^{-n}= ⅛ = 0.125

Question 2 |

Let (S, ≤) 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?

P(x) = True for all x ∈ S such that x ≠ b
| |

P(x) = False for all x ∈ S such that x ≠ a and x ≠ c | |

P(x) = False for all x ∈ S such that b ≤ x and x ≠ c | |

P(x) = False for all x ∈ S such that a ≤ x and b ≤ x |

a or b the minimal element in set.

P(a) = True for all x ∈ S such that a ≤ x and b ≤ x.

Option D is False.

Question 3 |

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?

X = Y | |

X ⊂ Y | |

Y ⊂ X | |

None of these |

B = {1, 3, 4, 5}

C = {2, 4, 5, 6}

X = (A - B) - C

X = {2, 6} - {2, 4, 5, 6}

= ∅

Y = (A - C) - (B - C)

= {1, 3} - { 1, 3}

= ∅

**X = Y**

X = (A - B) - C

= (1, 5) - (5, 7, 4, 3)

= (1)

Y = (A - C) - (B - C)

= (1, 4) - (2, 4)

= (1)

**X = Y**

Question 4 |

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:

3 and 13 | |

2 and 11 | |

4 and 13 | |

8 and 14 |

Inverse of 4 = m; Inverse of 7 = n

(4×m)%15=1; (7*n)%15=1

Option A: m=3 n=13

12%15≠1 (✖️) 91%15=1 (✔️)

Option B: m=2 n=11

8%15≠1 (✖️) 11%15≠1 (✖️)

Option C: m=4 n=13

16%15=1(✔️) 91%15=1 (✔️)

Option D: m=8 n=14

120%15≠1(✖️) 98%15≠1(✖️)

Question 5 |

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?

It is not closed | |

2 does not have an inverse | |

3 does not have an inverse | |

8 does not have an inverse |

Option A:

It is not closed under multiplication. After multiplication modulo (10) we get ‘0’. The ‘0’ is not present in the set.

(2*5)%10 ⇒ 10%10 = 0

Option B:

2 does not have an inverse such as

(2*x)%10 ≠ 1

Option C:

3 have an inverse such that

(3*7)%10 = 1

Option D:

8 does not have an inverse such that

(8*x)%10 ≠ 1

Question 6 |

Consider the set S = {a,b,c,d}. Consider the following 4 partitions π_{1}, π_{2}, π_{3}, π_{4} on Let p be the partial order on the set of partitions S' = {π_{1}, π_{2}, π_{3}, π_{4}} defined as follows: π_{i} p π_{j} if and only if π_{i} refines π_{j}. The poset diagram for (S', p) is:

_{4}= refines every partition. So it has to be bottom of poset diagram.

And, neither π

_{2}refines π

_{3}, nor π

_{3}refines π

_{2}.

Here, only π

_{1}refined by every set, so it has to be at the top.

Finally, option C satisfies all the property.