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Chapter 1: Problem 92

Finding Inverses Find (a) the additive inverse and (b) the multiplicative inverse of \(-8\).

Short Answer

Expert verified
The additive inverse of \(-8\) is \(8\) and the multiplicative inverse of \(-8\) is \(-\frac{1}{8}\).

Step by step solution

01

Find the additive inverse

The additive inverse of a number is what we get when we change the sign of the number. So the additive inverse of \(-8\) is just \(8\), because when we add \(-8\) and \(8\) together, we get \(0\).
02

Find the multiplicative inverse

The multiplicative inverse of a number \(x\) (x鈮�0) is \(\frac{1}{x}\). So the multiplicative inverse of \(-8\) is \(\frac{1}{-8}\), which simplifies to \(-\frac{1}{8}\). This is because when we multiply \(-8\) and \(-\frac{1}{8}\) together, we get \(1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Additive Inverse
Understanding the additive inverse is foundational for solving many types of algebraic problems. The additive inverse of any number is another number that, when added to the original, results in zero鈥攅ssentially negating its value. For example, for the given number \( -8 \), its additive inverse is \( 8 \) because \( -8 + 8 = 0 \). This concept is crucial when we're trying to isolate variables in equations and particularly helpful in operations that involve subtraction. By adding the opposite, or the additive inverse, we effectively remove the number from one side of an equation, simplifying the problem.
Multiplicative Inverse
The multiplicative inverse, also known as the reciprocal, is a value which, when multiplied with the original number, results in one. In simpler terms, for any non-zero number \( x \), its multiplicative inverse is \( \frac{1}{x} \). Applying this to our example, the multiplicative inverse of \( -8 \) is \( \frac{1}{-8} \), simplifying further to \( -\frac{1}{8} \). Multiplicative inverses enable us to solve for variables in equations where they are multiplied by a coefficient. By multiplying by the reciprocal, we can 'cancel out' the coefficient and isolate the variable.
Algebraic Operations
Algebraic operations include addition, subtraction, multiplication, and division鈥攖he cornerstone processes in algebra. They can be easily remembered with the acronym 'PEMDAS', standing for Parentheses, Exponents, Multiplication and Division (from left to right), and Addition and Subtraction (from left to right). When approaching algebraic expressions or equations, understanding how to manipulate these operations using inverse operations is fundamental. It's how we simplify expressions, solve for unknowns, and manipulate equations to reveal underlying relationships between the variables involved.
Elementary Algebra
Elementary algebra is often the first step into the world of abstract mathematics for many students. It's where we first encounter symbols and letters used to represent numbers and where we learn to perform operations on these symbols. The concepts of additive and multiplicative inverses play a pivotal role in this field as they are the basic tools for solving linear equations, simplifying expressions, and understanding function properties. Mastery of elementary algebra sets the stage for more advanced study in mathematics, so grasping these foundational concepts is crucial for continued success and confidence in math.

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Most popular questions from this chapter

\(15+3 \cdot 4\)

Matching Match each expression in the first column with its value in the second column. \(\begin{array}{cc}\text { Expression } & \text { Value } \\ (6+2) \cdot(5+3) & 19 \\ (6+2) \cdot 5+3 & 22 \\ 6+2 \cdot 5+3 & 64 \\ 6+2 \cdot(5+3) & 43\end{array}\)

\([127-(13+4)] \div 10\)

\(\frac{16}{2} \cdot 2 \neq 4\)

\((19-4) \div(7-2)\)
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