Understanding Set S and the Equation \(x^2 + 1 = 0\)

The question asks us to identify the nature of the set S, defined as the set of all real numbers x that satisfy the equation \(x^2 + 1 = 0\). Let's break down the definition of set S.

The set S is given by:

\(S = \{x : x^2 + 1 = 0, \text{ x is real}\}\)

This means that an element 'x' belongs to the set S if and only if two conditions are met:

1. The number x satisfies the equation \(x^2 + 1 = 0\).
2. The number x is a real number.

Solving the Equation \(x^2 + 1 = 0\)

Let's first find the values of x that satisfy the equation \(x^2 + 1 = 0\).

The equation is:

\(x^2 + 1 = 0\)

To solve for \(x^2\), we subtract 1 from both sides:

\(x^2 = -1\)

Now, to find x, we take the square root of both sides:

\(x = \pm\sqrt{-1}\)

In mathematics, the square root of -1 is denoted by the imaginary unit 'i', where \(i^2 = -1\).

So, the solutions to the equation \(x^2 + 1 = 0\) are:

\(x = i \quad \text{and} \quad x = -i\)

Checking the Condition: x must be Real

Now we need to consider the second condition for elements of set S: x must be a real number.

• The number \(x = i\) is an imaginary number.
• The number \(x = -i\) is also an imaginary number.

Real numbers are numbers that can be found on the number line (like -3, 0, 1/2, \(\sqrt{2}\)). Imaginary numbers involve the unit 'i'. Numbers that are combinations of real and imaginary parts (like \(2 + 3i\)) are called complex numbers.

Since neither \(x=i\) nor \(x=-i\) are real numbers, there are no real values of x that satisfy the equation \(x^2 + 1 = 0\).

Identifying Set S: The Empty Set

The set S contains all real numbers x such that \(x^2 + 1 = 0\). We found that there are no such real numbers.

Therefore, the set S contains no elements.

A set that contains no elements is called an empty set. The empty set is often denoted by the symbol \(\emptyset\) or \{\}.

Based on our analysis, the set S is an empty set.

Evaluating the Options

Let's look at the given options:

1. \(\{-1\}\): This set contains the element -1. Is -1 in S? -1 is real, but \((-1)^2 + 1 = 1 + 1 = 2 \neq 0\). So, -1 is not in S.
2. \(\{0\}\): This set contains the element 0. Is 0 in S? 0 is real, but \((0)^2 + 1 = 0 + 1 = 1 \neq 0\). So, 0 is not in S.
3. \(\{1\}\): This set contains the element 1. Is 1 in S? 1 is real, but \((1)^2 + 1 = 1 + 1 = 2 \neq 0\). So, 1 is not in S.
4. an empty set: This indicates the set contains no elements. This matches our conclusion that S is an empty set.

Thus, the correct option is 'an empty set'.

Revision Table: Key Concepts Reviewed

Additional Information: Empty Sets and Number Systems

What is an Empty Set?

The empty set is a unique set with zero elements. It is a fundamental concept in set theory. For example, the set of all people who have walked on the sun is an empty set, because no person can survive on the sun.

Understanding Number Systems: Real vs. Complex Numbers

The numbers we commonly use are real numbers (\(\mathbb{R}\)). This includes natural numbers (1, 2, 3...), integers (...-1, 0, 1...), rational numbers (like 1/2, -3/4), and irrational numbers (like \(\sqrt{2}\), \(\pi\)).

Complex numbers (\(\mathbb{C}\)) are numbers of the form \(a + bi\), where 'a' and 'b' are real numbers, and 'i' is the imaginary unit (\(\sqrt{-1}\)).

• If \(b=0\), the complex number is \(a + 0i = a\), which is a real number. So, real numbers are a subset of complex numbers.
• If \(a=0\) and \(b \neq 0\), the complex number is \(0 + bi = bi\), which is a pure imaginary number.

The equation \(x^2 + 1 = 0\) has solutions in the system of complex numbers (\(x = i, x = -i\)), but it has no solutions in the system of real numbers (\(\mathbb{R}\)). Since the set S is restricted to real numbers, it ends up being empty.