If `S={x: x^(2)+1=0, x` is real }, then S is:

Options

A. `(-1)`
B. `(0)`
C. `(1)`
D. An empty set

Answer

To solve the problem, we need to analyze the set \( S = \{ x : x^2 + 1 = 0, x \text{ is real} \} \).

### Step-by-Step Solution:

1. **Understanding the Equation**: We start with the equation given in the set definition:
\[ x^2 + 1 = 0 \]

2. **Rearranging the Equation**: We can rearrange this equation to isolate \( x^2 \):
\[ x^2 = -1 \]

3. **Analyzing the Square of a Real Number**: We know that the square of any real number is always non-negative. This means:
\[ x^2 \geq 0 \quad \text{for all real } x \]

4. **Conclusion from the Analysis**: Since \( x^2 = -1 \) implies that \( x^2 \) is negative, which contradicts the property of real numbers, we conclude that there are no real solutions to the equation \( x^2 + 1 = 0 \).

5. **Defining the Set**: Therefore, the set \( S \) contains no elements:
\[ S = \emptyset \]

6. **Final Answer**: Thus, the answer to the question is that \( S \) is an empty set.

### Answer:
The set \( S \) is an empty set.

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