Which of the following is the empty set?

Options

A. `{x : x in R, x^(2) - 1=0}`
B. `{x : x in N,x^(2) - 4=0}`
C. `{x: x in R,x^(2)-x - 1=0}`
D. `{x : x in N, x^(2) + 4x + 3 = 0}`

Answer

To determine which of the given options is the empty set, we will analyze each option one by one. An empty set is defined as a set that contains no elements.

### Step-by-Step Solution:

**Option 1:**
- The set is defined by the equation \(x^2 - 1 = 0\).
- Solving this, we get:
\[ x^2 = 1 \implies x = \pm 1 \]
- Therefore, the set is \(\{1, -1\}\).
- Since this set contains two elements, it is **not** an empty set.

**Option 2:**
- The set is defined by the equation \(x^2 - 4 = 0\).
- Solving this, we get:
\[ x^2 = 4 \implies x = \pm 2 \]
- However, the condition states that \(x\) must belong to the set of natural numbers.
- The only valid solution is \(x = 2\) (since \(-2\) is not a natural number).
- Thus, the set is \(\{2\}\).
- Since this set contains one element, it is **not** an empty set.

**Option 3:**
- The set is defined by the equation \(x^2 - x - 1 = 0\).
- We will use the quadratic formula to solve this:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]
- This gives us two values:
\[ x = \frac{1 + \sqrt{5}}{2} \quad \text{and} \quad x = \frac{1 - \sqrt{5}}{2} \]
- Both of these values are real numbers, hence the set is not empty.
- Therefore, this set is **not** an empty set.

**Option 4:**
- The set is defined by the equation \(x^2 + 4x + 3 = 0\).
- We can factor this as:
\[ (x + 1)(x + 3) = 0 \]
- This gives us the solutions:
\[ x = -1 \quad \text{and} \quad x = -3 \]
- The condition states that \(x\) must belong to the set of natural numbers.
- Since both \(-1\) and \(-3\) are not natural numbers, there are no valid solutions.
- Thus, the set is empty, represented as \(\emptyset\).

### Conclusion:
The empty set is **Option 4**.