Problem 7 - Algebra 2

Question

Identify the additive inverse and multiplicative inverse for each number. $$ -8 $$

Step-by-Step Solution

Answer

Additive inverse: 8; Multiplicative inverse: \(-\frac{1}{8}\).

Step 1: Identify the Additive Inverse

The additive inverse of a number is what you need to add to the original number to get zero. For the number \(-8\), add \(8\). Hence, the additive inverse of \(-8\) is \(8\) because \(-8 + 8 = 0\).

Step 2: Identify the Multiplicative Inverse

The multiplicative inverse of a number is what you multiply the original number by to get one. For the number \(-8\), multiply by \(-\frac{1}{8}\). Thus, the multiplicative inverse of \(-8\) is \(-\frac{1}{8}\) because \(-8 imes -\frac{1}{8} = 1\).

Key Concepts

Additive Inverse

In algebra, the additive inverse of a number is what you add to that number to achieve zero. It's like finding the balancing act. Let's understand this with an example based on the number \(-8\).
\(-8\) might seem far from zero, but when you add \(8\) to it, you reach zero. That's why \(8\) is the additive inverse of \(-8\).
This concept can be extended to any number, positive or negative. Here's a quick reminder:

• The additive inverse of any number \(x\) is \(-x\).
• When you add a number to its additive inverse, the sum is always zero.

By understanding and applying this rule, you can easily find the additive inverse for any problem involving algebraic operations.

Multiplicative Inverse

The multiplicative inverse, also known as the reciprocal, is what you multiply by a number to get one. This is a vital concept in algebra because it allows us to "undo" multiplication operations.
Let's take \(-8\) as an example. To find its multiplicative inverse, you need a number that when multiplied by \(-8\), results in one. This number is \(-\frac{1}{8}\).
Here's the reasoning:

• The product of \(-8\) and \(-\frac{1}{8}\) is \(1\).
• Therefore, \(-\frac{1}{8}\) is the multiplicative inverse of \(-8\).

Remember:

• The multiplicative inverse of a non-zero number \(x\) is \(\frac{1}{x}\).
• Inverting negatives is a little different; if you start with \(-x\), its inverse is \(-\frac{1}{x}\) as shown in our example.

This understanding is critical when dealing with fractions and solving equations, as it helps isolate variables and find solutions efficiently.

Algebraic Operations

In algebra, operations include addition, subtraction, multiplication, and division. Each of these operations interacts with numbers and inverses in unique ways. Grasping these interactions is crucial for performing algebraic manipulations correctly.
Let's examine how these operations work with addditive and multiplicative inverses:

• Addition and Additive Inverse: Adding a number to its additive inverse results in zero. For example, \(-8 + 8 = 0\).
• Multiplication and Multiplicative Inverse: Multiplying a number by its multiplicative inverse yields one. As seen with \(-8\times -\frac{1}{8} = 1\).

When solving algebraic equations, inverses are particularly helpful:

• Use additive inverses to move terms from one side of an equation to the other by adding or subtracting them.
• Employ multiplicative inverses to isolate variables by dividing both sides of an equation.

Understanding these relationships simplifies solving equations and empowers you to navigate complex algebraic expressions with confidence.