Relation Practice Questions

A relation $S$ is given to be reflexive and circular.
To prove $S$ is an equivalence relation, we need to show it is also symmetric and transitive.
Given $xSy$, and $S$ is reflexive, so $ySy$ is true. By the circular property ($xSy$ and $ySy$ imply $ySx$), $S$ is symmetric.
Given $xSy$ and $ySz$. By the circular property, this implies $zSx$.
Since $S$ is symmetric (as proved above), $zSx$ implies $xSz$. Thus, $S$ is transitive.
As $S$ is reflexive, symmetric, and transitive, it is an equivalence relation.

Let the given set be $A = \{1, 2, 3\}$, so the cardinality is $|A| = 3$.
The total number of binary relations on $A$ is $2^{|A| \times |A|} = 2^{3 \times 3} = 2^9$.
For a relation to be reflexive, all pairs $(x, x)$ for $x \in A$ must be present. There are $|A|$ such pairs, which are $(1,1), (2,2), (3,3)$.
The remaining $3^2 - 3 = 6$ pairs can either be present or absent in a reflexive relation. Thus, the number of reflexive relations is $2^{|A|^2 - |A|} = 2^{3^2 - 3} = 2^6$.
The probability that a randomly chosen relation is reflexive is the ratio of the number of reflexive relations to the total number of relations:
$P(\text{reflexive}) = \frac{2^6}{2^9} = \frac{1}{2^3} = \frac{1}{8} = 0.125$

The probability is 0.125.

[CORRECT_OPTION: 0.125]

For relation R1 (conjugacy, $a=g^{-1}bg$): It is reflexive ($a=e^{-1}ae$), symmetric (if $a=g^{-1}bg$, then $b=(g^{-1})^{-1}a(g^{-1})$), and transitive (if $a=g_1^{-1}bg_1$ and $b=g_2^{-1}cg_2$, then $a=(g_2g_1)^{-1}c(g_2g_1)$). Thus, R1 is an equivalence relation.
For relation R2 ($a=b^{-1}$): It is not reflexive, because $a=a^{-1}$ is not true for all elements in an arbitrary group. Since it is not reflexive, it cannot be an equivalence relation.

A relation R is reflexive if for every element x, (x)R(x). For the given relation on $\mathbb{N}\times \mathbb{N}$, we check if (a,b)R(a,b). This condition is true if $a \leq a$ or $b \leq b$. Since $a \leq a$ is always true, the relation is reflexive. Thus, proposition P is true.

A relation is transitive if (a,b)R(c,d) and (c,d)R(e,f) implies (a,b)R(e,f). Consider the counterexample: (2,4)R(3,2) is true because $2 \leq 3$. Also, (3,2)R(1,3) is true because $2 \leq 3$. However, (2,4)R(1,3) is false because neither $2 \leq 1$ nor $4 \leq 3$ is true. Therefore, the relation is not transitive, and proposition Q is false.

The relation R on ordered pairs of positive integers is defined as $((p,q),(r,s)) \in R$ if and only if $p-s = q-r$. This can be rearranged to $p+r = q+s$.

Reflexivity: We check if $((p,q),(p,q)) \in R$ for any pair $(p,q)$. The condition becomes $p-q = q-p$, which simplifies to $2p = 2q$, or $p=q$. Since this does not hold for all pairs (e.g., $(1,2)$), the relation is not reflexive.

Symmetry: We check if $((p,q),(r,s)) \in R$ implies $((r,s),(p,q)) \in R$.

• Given: $p-s = q-r$.
• To check: $r-q = s-p$.
  Multiplying the given equation by -1 gives $-(p-s) = -(q-r)$, which is $s-p = r-q$. This is the condition we need to check. Therefore, the relation is symmetric.

The relation is not reflexive but is symmetric.

The relation R is defined on positive integers such that aRb if a ≠ b and gcd(a, b) > 1.

1. Reflexivity: R is not reflexive. For any integer 'a', aRa is not possible because the definition requires the elements to be distinct (a ≠ a is false).
2. Symmetry: R is symmetric. If a and b have a common divisor greater than 1, then b and a also have that same common divisor.
3. Transitivity: R is not transitive. For example, 3R15 (common divisor 3) and 15R5 (common divisor 5), but 3 and 5 are not related since their only common divisor is 1.

A relation $R$ is symmetric if for every $(a,b) \in R$, we have $(b,a) \in R$.
Given $(x,y) \in R$, but $(y,x) \notin R$. Thus, $R$ is NOT symmetric.

A relation $R$ is antisymmetric if for every $(a,b) \in R$ and $(b,a) \in R$, it implies $a=b$.
Given $(x,z) \in R$ and $(z,x) \in R$. However, $x \neq z$. Thus, $R$ is NOT antisymmetric.

Since $R$ is neither symmetric nor antisymmetric, option D is true.