Question 2: Consider the relation S on the set of real numbers R defined by aSb if and only if a^2+b^2≤1.Check whether S is reflexive, symmetric, and transitive

Let's analyze the relation \( S \) on the set of real numbers \( \mathbb{R} \) defined by \( aSb \) if and only if \( a^2 + b^2 \leq 1 \).

### 1. **Reflexivity**:
A relation \( S \) on a set \( \mathbb{R} \) is reflexive if for every element \( a \in \mathbb{R} \), the pair \( (a, a) \) is in the relation. In this context, this means \( aSa \) must hold for all \( a \in \mathbb{R} \).

For \( aSa \) to hold, we must have:
\[
a^2 + a^2 \leq 1 \implies 2a^2 \leq 1 \implies a^2 \leq \frac{1}{2}.
\]

This implies that \( |a| \leq \frac{1}{\sqrt{2}} \).

**Conclusion**: \( S \) is **reflexive** only for elements \( a \) such that \( |a| \leq \frac{1}{\sqrt{2}} \). Since it's not true for all \( a \in \mathbb{R} \), \( S \) is **not reflexive** on the entire set \( \mathbb{R} \).

### 2. **Symmetry**:
A relation \( S \) is symmetric if for all \( a, b \in \mathbb{R} \), whenever \( aSb \) holds, then \( bSa \) must also hold.

Given \( aSb \), we have:
\[
a^2 + b^2 \leq 1.
\]
Now, we need to check \( bSa \), which requires:
\[
b^2 + a^2 \leq 1.
\]

Since \( a^2 + b^2 \) is the same as \( b^2 + a^2 \), symmetry holds.

**Conclusion**: The relation \( S \) is **symmetric**.

### 3. **Transitivity**:
A relation \( S \) is transitive if for all \( a, b, c \in \mathbb{R} \), whenever \( aSb \) and \( bSc \) hold, then \( aSc \) must also hold.

Given \( aSb \) and \( bSc \), we have:
\[
a^2 + b^2 \leq 1 \quad \text{and} \quad b^2 + c^2 \leq 1.
\]
We want to determine if \( aSc \) holds, which requires:
\[
a^2 + c^2 \leq 1.
\]

However, knowing \( a^2 + b^2 \leq 1 \) and \( b^2 + c^2 \leq 1 \) does not necessarily imply that \( a^2 + c^2 \leq 1 \). For example, if \( a = 1, b = 0 \), and \( c = 1 \), we see that:
\[
1^2 + 0^2 = 1 \leq 1 \quad \text{and} \quad 0^2 + 1^2 = 1 \leq 1,
\]
but:
\[
1^2 + 1^2 = 2 \nleq 1.
\]
This violates the condition for transitivity.

**Conclusion**: The relation \( S \) is **not transitive**.

### **Summary**:
- The relation \( S \) is **not reflexive**.
- The relation \( S \) is **symmetric**.
- The relation \( S \) is **not transitive**.

Would you like to go into further details, or do you have any specific questions?

### Relative Questions:
1. What does it mean for a relation to be reflexive?
2. Can you provide an example of a reflexive relation on a set?
3. What is an example of a symmetric relation that is not transitive?
4. How do you prove whether a relation is transitive?
5. Can a relation be symmetric and transitive but not reflexive?
6. What are some real-world examples of symmetric relations?
7. How would the definition of \( S \) change if the inequality were strict (\( < \) instead of \( \leq \))?
8. How does the set where \( S \) is defined impact the reflexivity property?

**Tip**: Remember that relations are often easier to analyze by testing specific elements and using counterexamples for properties like transitivity.

Explore the analysis of relation S on the set of real numbers R defined by a^2 + b^2 ≤ 1. Check if the relation is reflexive, symmetric, and transitive with detailed explanations and examples. Suitable for advanced high school students and college-level learners.

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