Discover Top Posts Tagged with #real line

Real Numbers and the Real Line

Calculus depends on properties of the real number system.

Real numbers are numbers that can be expressed as decimals.

5 = 5.000... -3/4 = -0.75000... 1/3 = 0.333... √2 = 1.414... π = 3.141...

In each example above, the three dots "..." indicate that the sequence of decimal digits goes on forever.

For the first three examples above, the pattern of the digits are obvious, where the subsequent digits are easily known. However, for √2 and π, there is no obvious pattern of their decimal digits.

Real Line The real numbers can be represented geometrically as points on a number line called the real line.

The symbol ℝ is used to denote either the real number system or the real number line.

Properties of Real Numbers The properties of real numbers fall into three categories: algebraic properties, order properties, and completeness.

The algebraic properties assert that the real numbers can be added, subtracted, multiplied, and divided (except zero) to produce another real number, and so the rules of arithmetic are valid.

Order Properties The order properties of the real numbers refer to the order the real numbers appears on the number line.

If x lies to the left of y on the real line, then x is less than y, or y is greater than x, written as x < y, y > x, respectively.

The inequality x ≤ y means either x = y or x < y.

The following are order properties of the real numbers:

The symbol => means “implies” while <=> (seen later) means “equivalent to.”

For Rules 1-4 and 6 (for 0 < a), they also hold if < and > are replaced with ≤ and ≥, respectively.

Note that, for Rules 3 and 4, the rules for multiplying or dividing an inequality by a positive number c preserves the inequality. If the number c is negative, the inequality is reversed.

Completeness Property If A is any set of real numbers having at least one number in the set, and if there exists a real number y with the property x ≤ y for every x in A, then there exists a smallest number y with this same property.

Therefore, there are no holes on the real line, where every point corresponds to a real number.

The study of infinite sequences will use the property of completeness.

Subsets of the Real Numbers The set of real numbers ℝ has some special subsets.

1. The set of natural numbers ℕ = {1, 2, 3, 4, ...}. 2. The set of integers ℤ = {..., -2, -1, 0, 1, 2, ...} 3. The set of rational numbers ℚ, which is the set of numbers that can be expressed as a fraction m/n, where m, n ∈ ℤ and n ≠ 0.

Rational Numbers Rational Numbers are real numbers with decimal expansion that does either of the following:

1. Terminate: ending with an infinite string of zeros, such as 3/4 = 0.75000..., or 2. Repeat: ending with a string of digits that repeat infinitely, such as 23/11 = 2.090909... = 2.̄0̄9̄ . The bar over the digits indicate the pattern of repeating digits.

Real numbers that are not rational are called irrational numbers.

The set of all rational numbers possesses all algebraic and order properties of the real numbers, but it does not possess the completeness property. For example, √2 is irrational, so there is a "hole" on the rational line where √2 should be.

Because the real line has no such "holes," it is the appropriate setting for the study of calculus.

Intervals A subset of the real line is called an interval if it contains at least two numbers and all real numbers between any two of its elements.

For example, the set of real numbers x such that x > 6 is an interval, but the set of real numbers such that y ≠ 0 is not an interval, because it consists of two intervals.

If a and b are real numbers and a < b:

1. The open interval from a to b, denoted as (a, b), consists of all real numbers x satisfying a < x < b.

2. The closed interval from a to b, denoted as [a, b], consists of all real numbers x satisfying a ≤ x ≤ b.

3. The half-open interval from a to b, denoted as [a, b), consists of all real numbers x satisfying a ≤ x < b.

4. The half-open interval from a to b, denoted as (a, b], consists of all real numbers x satisfying a < x ≤ b.

Note that, hollow dots are used to indicate endpoints that are not included in the interval, and the solid dots are used to indicate endpoints of the interval that are included. The endpoints are also called boundary points.

The above type of intervals are called finite intervals, such that each interval has a finite length of b – a.

Intervals with infinite length are called infinite intervals.

The following are the infinite intervals (a, ∞) and (-∞, a]:

The whole real line ℝ is an infinite interval denoted as (-∞, ∞).

Infinity ∞ does not denote a real number, and so it is never allowed to belong to an interval.

Union and Intersection of Intervals The symbol ⋃ is used to denote the union of intervals.

A real number is in the union of intervals if it is in at least one of the intervals.

For example, [1, 3) ⋃ [2, 4] = [1, 4]. Even though 3 is not included in the first interval, it is included in the second interval, and so the union of these two intervals is simply from 1 to 4, inclusive.

The symbol ⋂ is used to denote the intersection of intervals.

A real number is in the intersection of intervals if it is in every one of those intervals.

For example, [1, 3) ⋂ [2, 4] = [2, 3). Even though 4 is included in the second interval, and 1 is included in the first interval, they are not included in both intervals.

Whenever "and" is mentioned in conditions for intervals, it will be one interval. Whenever "or" is mentioned in conditions for intervals, it will be a union of intervals.

The Absolute Value The absolute value, or magnitude, of a number x, denoted as |x|, is defined as the following:

The vertical lines in the symbol |x| are called absolute value bars.

For example, |3| = 3, |0| = 0, and |-5| = 5.

Note that, √a always denotes the non-negative square root of a, and so the alternative definition for |x| is |x| = √x². It is important to remember that √a² = |a| and not just a, unless it is known that a ≥ 0.

Geometrically, |x| represents the non-negative distance from x to 0 on the real line.

In general, |x – y| represents the non-negative distance between x and y on the real line, since this distance is the same as the distance between x – y and 0.

The following demonstrates |x – y| = distance from x to y:

Properties of Absolute Value The absolute value function has the following properties:

Equations and Inequalities Involving Absolute Values The equation |x| = D, where D > 0, has two solutions: x = -D and x = D, which are two points on the real line that lie at distance D from the origin 0.

Equations and inequalities involving absolute values can be solved algebraically by using cases according to the definition of absolute value. Another way to solve them is geometrically interpreting absolute values as distances.

For example, the inequality |x – a| < D means the distance from x to a is less than D, and so x must lie between a – D and a + D, or a must lie between x – D and x + D.

If D is a positive number, then:

1. |x| = D <=> either x = -D or x = D |x – a| = D <=> either x = a – D or x = a + D

2. |x| < D <=> -D < x < D |x – a| < D <=> a – D < x < a + D

3. |x| ≤ D <=> -D ≤ x ≤ D |x – a| ≤ D <=> a – D ≤ x ≤ a + D

4. |x| > D <=> either x < -D or x > D |x – a| > D <=> either x < a – D or x > a + D

PDF reference: 27/1

Objectives

1. Writing repeating decimals. 2. Converting fractions into repeating decimals. 3. Converting repeating decimals into fractions. 4. Solving linear inequalities and graphing the solution set. 5. Solving quadratic inequalities. 6. Solving absolute value equations and inequalities. 7. Using the Triangle Inequality to prove other inequalities.

Show that each of the following numbers is a rational number by expressing it as a quotient of two integers:

a) 1.323232... = 1.3̄2̄ b) 0.3405405405... = 0.34̄0̄5̄

Let x = 1.323232...

Then the non-decimal digit is subtracted from both sides.

x – 1 = 0.323232...

Then 100 is multiplied on both sides of the original equation to get one repeating pattern on the non-decimal side.

100x = 132.323232... = 132 + 0.323232... = 132 + x – 1

Using algebraic techniques forms the decimal into a fraction and hence prove the number is rational.

100x = 132 + x – 1 99x = 131 x = 131/99

Let x = 0.3405405405...

Since the non-repeating digit is in the decimal side, a multiple of 10 must be multiplied to make it into a non-decimal digit.

10x = 3.405405405...

Then the non-decimal digit can be subtracted from both sides.

10x – 3 = 0.405405405...

Then 10000 is multiplied on both sides of the original equation to get one repeating pattern on the non-decimal side.

10000x = 3405.405405405... = 3405 + 0.405405405... = 3405 + 10x - 3

Using algebraic techniques forms the decimal into a fraction and hence prove the number is rational.

10000x = 3405 + 10x – 3 9990x = 3402 x = 3402/9990 = 63/185

Express the following rational numbers as a repeating decimal. Use a bar to indicate the repeating digits:

a) 2/9 b) 1/11

Using long division, it is found that 2/9 = 0.222... = 0.2̄ .

Using long division, it is found that 1/11 = 0.090909... = 0.0̄9̄ .

Express the following repeating decimal as a quotient of integers in lowest terms:

a) 0.1̄2̄ b) 3.27̄

Let x = 0.121212...

Since there are no non-repeating digits, many steps are skipped and now 100 is multiplied on both sides to get one repeating pattern on the non-decimal side.

100x = 12.121212... = 12 + 0.121212 = 12 + x

Using algebraic techniques forms the decimal into a fraction and then simplified into its lowest terms.

100x = 12 + x 99x = 12 x = 12/99 = 4/33

Let x = 3.2777...

Since there is a non-repeating digit on the decimal side, 10 is multiplied on both sides.

10x = 32.777..

The non-repeating digits are subtracted from both sides.

10x – 32 = 0.777...

Now the original equation is multiplied by 100 to get one repeating pattern on the non-decimal side.

100x = 327.777... = 327 + 0.777... = 327 + 10x – 32

Using algebraic techniques forms the decimal into a fraction and then simplified into its lowest terms.

100x = 327 + 10x – 32 90x = 295 x = 295/90 = 59/18

Solve the following inequalities and express the solution sets in terms of intervals. Graph their intervals.

a) 2x – 1 > x + 3 b) -x/3 ≥ 2x - 1 c) 2/(x – 1) ≥ 5

2x – 1 > x + 3 x – 1 > 3 x > 4

Solution: (4, ∞)

-x/3 ≥ 2x – 1 x ≤ -6x + 3 7x ≤ 3 x ≤ 3/7

Solution: (-∞, 3/7]

This inequality has x in the denominator, and so it is possible for x to be undefined at a certain point on the real line.

The inequality will be written such that all terms are on one side and then simplified into one fraction.

2/(x – 1) ≥ 5 2/(x – 1) – 5 ≥ 0 (2 – 5x + 5)/(x – 1) ≥ 0 (7 – 5x)/(x – 1) ≥ 0

The numerator is examined to determine when it equals 0.

7 – 5x = 0 x = 7/5

The denominator is examined to determine when it equals 0.

x – 1 = 0 x = 1

Since the fraction is undefined when x = 1, the interval will have a round bracket on 1 and a square bracket on 7/5.

Solution: (1, 7/5]

Solve the following systems of inequalities:

a) 3 ≤ 2x + 1 ≤ 5 b) 3x – 1 < 5x + 3 ≤ 2x + 15

First, the left inequality is solved:

3 ≤ 2x + 1 2 ≤ 2x x ≥ 1

Then the right inequality is solved:

2x + 1 ≤ 5 2x ≤ 4 x ≤ 2

Therefore, since x ≥ 1 and x ≤ 2, the interval is [1, 2].

First, the left inequality is solved:

3x – 1 < 5x + 3 -4 < 2x x > -2

Then the right inequality is solved:

5x + 3 ≤ 2x + 15 3x ≤ 12 x ≤ 4

Since x > -2 and x ≤ 4, the interval is (-2, 4].

Solve the following quadratic inequalities:

a) x² – 5x + 6 < 0 b) 2x² + 1 > 4x

x² – 5x + 6 < 0 (x – 2)(x – 3) < 0 x = 2, 3

Drawing a number line with labeled solutions, points between -∞ and 2 are positive, points between 2 and 3 are negative, and points between 3 and ∞ are positive.

Therefore, (2, 3).

2x² + 1 > 4x 2x² – 4x + 1 > 0

The quadratic formula is used to determine the solutions.

x = 1 ± √2/2

(x - 1 - √2/2)(x - 1 + √2/2) > 0

Drawing a number line with labeled solutions, points between -∞ and 1 - √2/2 are positive, points between 1 - √2/2 and 1 + √2/2 are negative, and points between 1 + √2/2 and ∞ are positive.

Therefore, the solution is the union of the intervals (-∞, 1 - √2/2) ⋃ (1 + √2/2, ∞).

Express the set of all real numbers x satisfying the following conditions as an interval or a union of intervals.

a) x ≥ 0 and x ≤ 5 b) x < 2 and x ≥ -3 c) x > -5 or x < -6 d) x ≤ -1 e) x > -2 f) x < 4 or x ≥ 2

Since "and" is used, the interval must satisfy when x ≥ 0 and when x ≤ 5.

[0, 5]

Since "and" is used, the interval must satisfy when x < 2 and when x ≥ -3.

[-3, 2)

Since "or" is used, the interval must satisfy when x > -5 or x < -6.

(-∞, -6) ⋃ (-5, ∞)

x ≤ -1 means that the interval contains all real numbers such that x is less than or equal to -1.

(-∞, -1]

x > -2 means that the interval contains all real numbers such that x is greater than -2.

(-2, ∞)

Since "or" is used, the interval must satisfy when x < 4 or x ≥ 2, which is just the real line.

(-∞, 4) ⋃ [2, ∞) = (-∞, ∞)

Solve the following inequality and graph the solution set.

3/(x – 1) < -2/x

Bring all terms to one side.

3/(x – 1) + 2/x < 0 (3x + 2x – 2)/x(x – 1) = (5x – 2)/x(x – 1) < 0

Examine the numerator and determine when it is equal to zero.

5x – 2 = 0 x = 2/5

Examine the denominator and determine when it is equal to zero.

x(x – 1) = 0 x = 0, 1

The denominator is undefined when either x = 0 or x = 1 or both.

Testing points between 0, 2/5, and 1 to determine whether it is negative or positive.

Points to the left of zero are negative, points between 0 and 2/5 are positive, points between 2/5 and 1 are negative, and points to the right of one are positive.

Since it is less than zero and not less than or equal to zero, all three endpoints use a round bracket.

(-∞, 0) ⋃ (2/5, 1)

Solve the following inequalities, giving the solution set as an interval or union of intervals.

a) -2x > 4 b) 3x + 5 ≤ 8 c) 5x – 3 ≤ 7 - 3x d) (6 – x)/4 ≥ (3x – 4)/2 e) 3(2 – x) < 2(3 + x) f) x² < 9 g) 1/(2 – x) < 3 h) (x + 1)/x ≥ 2 i) x² – 2x ≤ 0 j) 6x² – 5x ≤ -1 k) x³ > 4x l) x² – x ≤ 2 m) x/2 ≥ 1 + 4/x n) 3/(x – 1) < 2/(x + 1)

-2x > 4 <=> x < -2 <=> (-∞, -2)

3x + 5 ≤ 8 <=> 3x ≤ 3 <=> x ≤ 1 <=> (-∞, 1]

5x – 3 ≤ 7 – 3x <=> 8x – 3 ≤ 7 <=> 8x ≤ 10 <=> x ≤ 5/4 <=> (-∞, 5/4]

(6 – x)/4 ≥ (3x – 4)/2 <=> 12 – 2x ≥ 12x – 16 <=> -14x ≥ -28 <=> x ≤ 2 <=> (-∞, 2]

3(2 – x) < 2(3 + x) <=> 6 – 3x < 6 + 2x <=> -5x < 0 <=> x > 0 <=> (0, ∞)

x² < 9 <=> x² – 9 < 0 <=> (x – 3)(x + 3) < 0 => x = -3, 3

Drawing a number line with labeled solutions, points between -∞ and -3 are positive, points between -3 and 3 are negative, and points between 3 and ∞ are positive.

Therefore, (-3,3).

1/(2 – x) < 3 <=> 1/(2 – x) – 3 < 0 <=> (1 – 6 + 3x)/(2 – x) <=> (-5 + 3x)/(2 – x) => x = 5/3, 2

Drawing a number line with labeled solutions, points between -∞ and 5/3 are negative, points between 5/3 and 2 are positive, and points between 2 and ∞ are negative.

Since the fraction is undefined when x = 2, 2 is not allowed to be included in the interval.

Therefore, (-∞, 5/3) ⋃ (2, ∞).

(x + 1)/x ≥ 2 <=> (x + 1)/x – 2 ≥ 0 <=> (x + 1 – 2x)/x ≥ 0 <=> (-x + 1)/x ≥ 0 => x = 0, 1 Drawing a number line with labeled solutions, points between -∞ and 0 are negative, points between 0 and 1 are positive, and points between 1 and ∞ are negative.

Since the fraction is undefined when x = 0, 0 is not allowed to be included in the interval.

Therefore, (0, 1].

x² – 2x ≤ 0 <=> x(x – 2) ≤ 0 => x = 0, 2

Drawing a number line with labeled solutions, points between -∞ and 0 are positive, points between 0 and 2 are negative, and points between 2 and ∞ are positive.

Therefore, [0, 2].

6x² – 5x ≤ -1 <=> 6x² – 5x + 1 ≤ 0 <=> 6x² – 3x – 2x + 1 ≤ 0 <=> 3x(2x – 1) – (2x – 1) ≤ 0 <=> (3x – 1)(2x – 1) ≤ 0 => x = 1/3, 1/2

Drawing a number line with labeled solutions, points between -∞ and 1/3 are positive, points between 1/3 and 1/2 are negative, and points between 1/2 and ∞ are positive.

Therefore, [1/3, 1/2].

x³ > 4x <=> x³ – 4x > 0 <=> x(x² – 4) > 0 <=> x(x – 2)(x + 2) > 0 => x = -2, 0, 2

Drawing a number line with labeled solutions, points between -∞ and -2 are negative, points between -2 and 0 are positive, point between 0 and 2 are negative, and points between 2 and ∞ are positive.

Therefore, (-2, 0) ⋃ (2, ∞).

x² – x ≤ 2 <=> x² – x – 2 ≤ 0 <=> (x – 2)(x + 1) ≤ 0 => x = -1, 2

Drawing a number line with labeled solutions, points between -∞ and -1 are positive, points between -1 and 2 are negative, and points between 2 and ∞ are positive.

Therefore, [-1, 2].

x/2 ≥ 1 + 4/x <=> 0 ≥ 1 – x/2 + 4/x <=> 0 ≥ 0 ≥ (x – x²/2 + 4)/x <=> 0 ≥ (-2/-2)[(x – x²/2 + 4)/x] <=> 0 ≥ (x² – 2x – 8)/(-2x) <=> 0 ≥ (x – 4)(x + 2)/(-2x) <=> (x – 4)(x + 2)/(-2x) ≤ 0 => x = -2, 0, 4

Drawing a number line with labeled solutions, points between -∞ and -2 are positive, points between -2 and 0 are negative, point between 0 and 4 are positive, and points between 4 and ∞ are negative.

Since the fraction is undefined when x = 0, 0 is not allowed to be included in the interval.

Therefore, [-2, 0) ⋃ [4, ∞).

3/(x – 1) < 2/(x + 1) <=> 3/(x – 1) – 2/(x + 1) < 0 <=> (3x + 3 – 2x + 2)/(x - 1)(x + 1) < 0 <=> (x + 5)/(x – 1)(x + 1) < 0 => x = -5, -1, 1

Drawing a number line with labeled solutions, points between -∞ and -5 are negative, points between -5 and -1 are positive, point between -1 and 1 are negative, and points between 1 and ∞ are positive.

Since the fraction is undefined when x = -1 or 1, -1 and 1 are not allowed to be included in the interval.

Therefore, (-∞, -5) ⋃ (-1, 1).

Solve the following:

a) |2x + 5| = 3 b) |3x – 2| ≤ 1

Case 1: x > 0 |2x + 5| = 3 <=> 2x + 5 = 3 <=> x = -1

Case 2: x < 0 |2x + 5| = 3 <=> -(2x + 5) = 3 <=> 2x + 5 = -3 <=> x = -4

Since the question deals with "less than or equal to," the equivalent form is used.

|3x – 2| ≤ 1 <=> -1 ≤ 3x – 2 ≤ 1

The left inequality is solved first.

-1 ≤ 3x – 2 1 ≤ 3x x ≥ 1/3

The right inequality is solved last.

3x – 2 ≤ 1 3x ≤ 3 x ≤ 1

Therefore, [1/3,1].

Solve the equation |x + 1| = |x – 3|.

Case 1: x < 0 |x + 1| = |x – 3| <=> -(x + 1) = x – 3 <=> -x – 1 = x – 3 <=> -2x = -2 <=> x = 1

Case 2: x > 0 |x + 1| = |x – 3| <=> x + 1 = x – 3 <=> no solution

What values of x satisfy the following inequality?

Since the inequality uses "less than," the following equivalent form is used.

-3 < 5 – 2/x < 3 <=> -8 < -2/x < -2 <=> 4 > 1/x > 1

Solve the left inequality first.

4 > 1/x <=> 4x > 1 <=> x > 1/4

Solve the right inequality last.

1/x > 1 <=> 1 > x <=> x < 1

Therefore, the values of x that satisfy the above inequality is in the interval (1/4, 1).

Solve the following equations:

a) |x – 3| = 7 b) |2t + 5| = 4 c) |1 – t| = 1 d) |s/2 – 1| = 1

|x – 3| = 7

Case 1: x < 0 -(x – 3) = 7 <=> x – 3 = -7 <=> x = -4

Case 2: x > 0 x – 3 = 7 <=> x = 10

Therefore, x = -4, 10.

|2t + 5| = 4

Case 1: x < 0 -(2t + 5) = 4 <=> 2t + 5 = -4 <=> 2t = -9 <=> t = -9/2

Case 2: x > 0 2t + 5 = 4 <=> t = -1/2

Therefore, t = -1/2, -9/2.

|1 – t| = 1

Case 1: x < 0 -(1 – t) = 1 <=> 1 – t = -1 <=> t = 2

Case 2: x > 0 1 – t = 1 <=> t = 0

Therefore, t = 0, 2.

|s/2 – 1| = 1

Case 1: x < 0 -(s/2 – 1) = 1 <=> s/2 – 1 = -1 <=> s = 0

Case 2: x > 0 s/2 – 1 = 1 <=> s/2 = 2 <=> s = 4

Therefore, s = 0, 4.

Write the interval defined by the following inequalities:

a) |x| < 2 b) |x| ≤ 2 c) |t + 2| < 1 d) |3x – 7| < 2 e) |x/2 – 1| ≤ 1 f) |2 – x/2| < 1/2

|x| < 2 <=> -2 < x < 2 <=> (-2, 2)

|x| ≤ 2 <=> -2 ≤ x ≤ 2 <=> [-2, 2]

|t + 2| < 1 <=> -1 < t + 2 < 1 <=> -3 < t < -1 <=> (-3, -1)

|3x – 7| < 2 <=> -2 < 3x – 7 < 2 <=> 5 < 3x < 9 <=> 5/3 < x < 3 <=> (5/3, 3)

|x/2 – 1| ≤ 1 <=> -1 ≤ x/2 – 1 ≤ 1 <=> 0 ≤ x/2 ≤ 2 <=> 0 ≤ x ≤ 4 <=> [0, 4]

|2 – x/2| < 1/2 <=> -0.5 < 2 – 0.5x < 0.5 <=> -2.5 < -0.5x < 2.5 <=> 5 > x > 3 <=> (3, 5)

Solve the equation |x – 1| = 1 – x.

Case 1: x < 0 -(x – 1) = 1 – x <=> -x + 1 = 1 – x <=> no solution

Case 2: x > 0 x – 1 = 1 – x <=> 2x = 2 <=> x = 1

Additionally, the equation holds if |x – 1| = -(1 – x), since this is when they are always equal to each other. For this to be true, x – 1 < 0, or x < 1.

Therefore, x < 1.

Show that the following inequality holds for all real numbers a and b:

|a – b| ≥ ||a| - |b||

Begin with the Triangle Inequality.

|x + y| ≤ |x| + |y| <=> |x| + |y| ≥ |x + y| <=> |x| ≥ |x + y| - |y|

Let x = a – b and y = b.

|x| ≥ |x + y| - |y| <=> |a – b| ≥ |a – b + b| - |b| = |a| - |b|

Similarly, |a – b| = |b – a| ≥ |b| - |a|.

Then ||a| - |b|| is equal to either |a| - |b| or |b| - |a|, and since |a – b| ≥ |b| - |a| and |a – b| ≥ |a| - |b|:

|a – b| ≥ ||a| - |b||

#math #calculus #preliminaries #single variable #real line #real numbers #absolute value #inequalities #decimals #repeating decimals #rational numbers #study blog #study notes