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Chapter 0: Problem 43

In Exercises 39-48, evaluate the expression. $$ |-1|-|-2| $$

Short Answer

Expert verified
The result of the expression |-1|-|-2| is -1.

Step by step solution

01

Define Absolute Value

The absolute value of a number is its distance from zero on the number line. It is always positive or zero. The absolute value of -1, written as |-1|, is 1. Likewise, the absolute value of -2, written as |-2| is 2.
02

Perform the Operation

We are asked to subtract |-2| from |-1|. Since |-1| equals 1 and |-2| equals 2, this is equivalent to subtracting 2 from 1. So, the expression \(-1-|-2|\) becomes \(1-2\).
03

Final Calculation

The subtraction operation \(1-2\) results in -1.

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Key Concepts

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

Number Line
The number line is a fundamental tool in mathematics that helps us understand numbers and their relationships. Imagine a straight horizontal line where each point corresponds to a number. Zero is typically at the center, with positive numbers extending to the right and negative numbers stretching to the left. This line allows us to visually grasp the idea of distance between numbers and their position relative to zero.

When dealing with absolute values, the number line is handy because it emphasizes distance. The absolute value of any number is simply how far it is from zero, regardless of direction. Therefore, both -1 and 1 have the same absolute value because they are both one unit away from zero on the number line.
  • Useful for visualizing addition and subtraction.
  • Shows the concept of distance without direction.
  • Helps in understanding concepts like absolute value clearly.
Positive Numbers
Positive numbers are the set of numbers that are greater than zero. On the number line, they are found to the right of zero. These are values like 1, 2, 3, and so on. Positive numbers represent quantities or amounts and are used in various real-world scenarios, such as measuring height or computing total earnings.

In the context of absolute value, the result is always a positive number or zero. Even if you start with a negative, absolute value strips away the sign, turning the number into its positive counterpart. This is why when dealing with absolute values like \(|-1|\) or \(|-2|\), the results are positive: 1 and 2, respectively.
  • Represent values greater than zero.
  • Always to the right on a number line.
  • In absolute values, negatives are treated as positives.
Subtraction Operation
Subtraction is a basic arithmetic operation used to determine the difference between numbers or to remove quantity from a set. In subtraction, you take one number away from another. Recognized commonly through the minus (-) sign, it contrasts with addition, which combines values.

In expressions involving absolute values, like \(1 - 2\) derived from \(|-1| - |-2|\), subtraction tells us how many units one number is away from another. Here, it shows that the result is -1 because you end up one unit to the left of zero on the number line.
  • Finds the difference between numbers.
  • Uses the minus sign to express operation.
  • Movement left on the number line indicates a negative result.

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Most popular questions from this chapter

In Exercises 55-68, determine whether the function has an inverse function. If it does, find the inverse function. $$ f(x)=\sqrt{x-2} $$

In Exercises 55-68, determine whether the function has an inverse function. If it does, find the inverse function. $$ f(x)= \begin{cases}-x, & x \leq 0 \\ x^{2}-3 x, & x>0\end{cases} $$

In Exercises 33-38, use a graphing utility to graph the function, and use the Horizontal Line Test to determine whether the function is one-to-one and so has an inverse function. $$ f(x)=10 $$

Find the average rate of change of the function from \(x_{1}\) to \(x_{2}\). $$ \begin{array}{cc} \text { Function } & x \text {-Values } \\ f(x)=-\sqrt{x+1}+3 &\quad x_{1}=3, x_{2}=8 \end{array} $$

In Exercises 27 and 28, use the table of values for \(y=f(x)\) to complete a table for \(y=f^{-1}(x)\). $$ \begin{array}{|l|r|r|r|r|r|r|} \hline x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline f(x) & -2 & 0 & 2 & 4 & 6 & 8 \\ \hline \end{array} $$
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