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Coordinates in One Dimension Help

By — McGraw-Hill Professional
Updated on Aug 31, 2011

Introduction to Coordinates in One Dimension

We envision the real numbers as laid out on a line, and we locate real numbers from left to right on this line. If a < b are real numbers then a will lie to the left of b on this line. See Fig. 1.1.

 

Basics 1.2 Coordinates in One Dimension

Fig. 1.1

Examples

Example 1

On a real number line, plot the numbers −4, −1, 2, 6. Also plot the sets S = { x Basics 1.2 Coordinates in One Dimension Basics 1.2 Coordinates in One Dimension : − 8 ≤ x < −5} and T = {t Basics 1.2 Coordinates in One Dimension Basics 1.2 Coordinates in One Dimension : 7 < t ≤ 9}. Label the plots.

Solution 1

Figure 1.2 exhibits the indicated points and the two sets. These sets are called half-open intervals because each set includes one endpoint and not the other.

 

Basics 1.2 Coordinates in One Dimension

Fig. 1.2

Math Note: The notation S = { x Basics 1.2 Coordinates in One Dimension Basics 1.2 Coordinates in One Dimension : − 8 ≤ x < −5} is called set builder notation . It says that S is the set of all numbers x such that x is greater than or equal to −8 and less than 5. We will use set builder notation throughout the book.

If an interval contains both its endpoints, then it is called a closed interval . If an interval omits both its endpoints, then it is called an open interval . See Fig. 1.3.

 

Basics 1.2 Coordinates in One Dimension

Fig. 1.3

Examples

Example 2

Find the set of points that satisfy x − 2 < 4 and exhibit it on a number line.

Solution 2

We solve the inequality to obtain x < 6. The set of points satisfying this inequality is exhibited in Fig. 1.4 .

 

Basics 1.2 Coordinates in One Dimension

Fig. 1.4

Example 3

Find the set of points that satisfy the condition

Basics 1.2 Coordinates in One Dimension

and exhibit it on a number line.

Solution 3

In case x + 3 ≥ 0 then ∣ x + 3∣ = x + 3 and we may write condition (*) as

x + 3 ≤ 2

or

x ≤ −1.

Combining x + 3 ≥ 0 and x ≤ −1 gives −3 ≤ x ≤ −1.

On the other hand, if x + 3 < 0 then ∣ x + 3∣ = −( x + 3). We may then write condition (*) as

−( x + 3) ≤ 2

or

−5 ≤ x .

Combining x + 3 < 0 and −5 ≤ x gives −5 ≤ x < −3.

We have found that our inequality ∣ x + 3∣ ≤ 2 is true precisely when either −3 ≤ x ≤ −1 or −5 ≤ x < −3. Putting these together yields −5 ≤ x ≤ −1. We display this set in Fig. 1.5 .

 

Basics 1.2 Coordinates in One Dimension

Fig. 1.5

You Try It: Solve the inequality ∣ x −4∣ > 1. Exhibit your answer on a number line.

You Try It: On a real number line, sketch the set { x : x 2 − 1 < 3}.

Practice problems for this concept can be found at: Calculus Basics Practice Test.

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