The probability that a relation R from {x, y} to {x, y} is both symmetric and transitive, is equal t

Q: Wait, why is the empty relation considered symmetric and transitive?

This is a classic point of confusion! In logic, we call this a 'vacuous truth.' A relation R is symmetric if for all a, b in S, (a, b) in R implies (b, a) in R. Since the empty relation contains no pairs, the condition '(a, b) in R' is always false. In formal logic, 'False implies anything' is a true statement. Therefore, the empty relation satisfies the definition of symmetry and transitivity by default.

Q: Why does including (x, y) and (y, x) force us to include (x, x) and (y, y)?

Great question! Let's look at the definition of transitivity: if (a, b) in R and (b, c) in R, then (a, c) must be in R. If we have (x, y) in R and (y, x) in R, then by transitivity, (x, x) must be in R. Similarly, if we have (y, x) in R and (x, y) in R, then (y, y) must be in R. So, if you choose to have any cross-connection, the self-loops are mandatory to keep the relation transitive.

Q: Is there a formula to find the number of equivalence relations?

Yes, there is! The number of equivalence relations on a set of n elements is given by the Bell number, B_n. For n = 2, B_2 = 2. However, this problem asks for relations that are symmetric and transitive, which are not necessarily reflexive. Therefore, we cannot simply use the Bell number. We must manually count the valid relations, which is why we found 5 instead of 2.

Question

The probability that a relation $R$ from $\{x, y\}$ to $\{x, y\}$ is both symmetric and transitive, is equal to:

Options:

(A) $\frac{5}{16}$

(B) $\frac{9}{16}$

(C) $\frac{11}{16}$

(D) $\frac{13}{16}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a mathematical garden, holding a set $S = \{x, y\}$. This set is small, yet it holds the key to understanding the profound structure of relations.

A relation $R$ on $S$ is simply a subset of the Cartesian product $S 	imes S$. To begin our journey, we must first map out our playground.

The Cartesian product $S 	imes S$ consists of all possible ordered pairs: $\{(x, x), (x, y), (y, x), (y, y)\}$. Since there are 4 elements in this product, the total number of possible relations is the number of subsets of a 4-element set, which is $2^4 = 16$.

This is our sample space. Our mission is to find the probability that a randomly chosen relation is both symmetric and transitive.

### The Rules of the Game

**Symmetry** is a mirror. It demands that if an arrow points from $a$ to $b$, a return arrow must exist from $b$ to $a$. Mathematically, $(a, b) \in R \implies (b, a) \in R$.

**Transitivity** is the bridge. It demands that if you can travel from $a$ to $b$ and then from $b$ to $c$, a direct path from $a$ to $c$ must exist. Mathematically, $(a, b) \in R \land (b, c) \in R \implies (a, c) \in R$.

These rules are strict, but they create a beautiful, rigid structure.

### The Systematic Search

Let us hunt for our favorable cases. We start with the most elusive one: the empty relation, $R_1 = \phi$. With no elements, it satisfies both symmetry and transitivity vacuously.

Next, we look at the self-loops. A relation containing only $(x, x)$, $R_2 = \{(x, x)\}$, is perfectly symmetric and transitive. Similarly, $R_3 = \{(y, y)\}$ works just as well.

What if we combine them? The identity relation, $R_4 = \{(x, x), (y, y)\}$, is also valid. We have found four cases so far.

Now, we approach the 'Cross-Connection Trap.' If we include $(x, y)$, symmetry forces us to include $(y, x)$.

But once we have both $(x, y)$ and $(y, x)$, transitivity demands that $(x, x)$ and $(y, y)$ must also be present. This forces us into the universal relation:

$$R_5 = \{(x, x), (y, y), (x, y), (y, x)\}$$

This is our fifth and final valid relation. Any other combination fails one of the two tests.

### The Final Revelation

We have meticulously identified 5 favorable relations out of 16 total possibilities. The probability is simply the ratio of these two numbers:

$$P = \frac{5}{16}$$

It is a simple fraction, but it represents a deep understanding of how symmetry and transitivity constrain the chaos of random relations. You have navigated the logic, avoided the traps, and arrived at the truth.

The beauty of mathematics lies in these small, perfect structures. The final probability is **$\frac{5}{16}$**.

Animated Solution for Mathematics - Sets and Relations: The probability that a relation $R$ from ${x, y}$ to ${x, y}$ is both symmetric and transitive, is equal to:

Select Answer:

Visualized Solution

Problem Setup

Total Number of Relations

Defining the Constraints

Case 1: The Empty Relation

Case 2: Singleton Self-loop on $x$

Case 3: Singleton Self-loop on $y$

Case 4: The Identity Relation

The Cross-Connection Trap

Case 5: The Universal Relation

Final Probability Calculation

The Sigma Insight: Types of Relations

Analyzing the Setup

The Rules of the Game

The Systematic Search

The Final Revelation

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Animated Solution for Mathematics - Sets and Relations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The probability that a relation R from {x,y} to {x,y} is both symmetric and transitive, is equal to: