Interval notation

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Order and inequations

The usual order of numbers is determined by their size. The signs , , and are used to show the order

of numbers or algebraic expressions. In this section we will properly define these signs and the rules for their use.

The rules for the use of the order signs can all be proven from these definitions. The following rules are more correctly called theorems. This means that they can be proven from definitions and basic facts (axioms). All of the following apply to less than () in the same way as they are stated for greater than ().

Linear inequations can be solved in a similar way to linear equations. The difference is that multiplication or division by a negative number reverses the order sign.

Solutions may be graphed on a number line or shown using interval notation.

Inequality operators

The inequality operator (greater than) is defined as:

ab⇔ (ab) is a positive number. The operator (less than) is defined as:

abba

For convenience, we define (greater than or equal) and (less than or equal) as:

abab or a=b abab or a=b

It also follows from these definitions that:

ab⇔ (ab) is a negative number.

!

Order theorems

Trichotomy: For any two numbers x and y, exactly one of xy, x=y or xy is true.

Transitivity: If xyANDyz, then xz. Operations: For all numbers w, x, y, z:

xyx+zy+zxyxzyz

xyANDz 0 ⇒xzyzxyANDz 0 ⇒xzyz

xyANDz 0 ⇒ ■ xyANDz 0 ⇒

xyANDwzx+wy+z

x z

-- y

z

-- x

z

-- y

z

--!

Detailed Proof

Graphical notation

■ A closed circle ( ) means the boundary is included.

■ An open circle ( ) means the boundary is not included.

■ A straight line on or above the number line shows the section included.

Interval notation

■ Numbers in brackets are used to indicate an interval (part of the number line).

■ Square brackets [ ] indicate the boundary is included.

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Show the following using both graphs and interval notation.

a −2 x 3 b 4 x 9

c x 5 d x−3

Solution

a

b

c

d

−4 −3 −2 −1 0 1 2 3 4 or [−2, 3] as interval notation

2 3 4 5 6 7 8 9 10 or [4, 9) as interval notation

or (5, …] or (5, ∞] as interval notation

2 3 4 5 6 7 8 9 10

or […, −3] or [−∞, −3] as interval notation

−7 −6 −5 −6 −3 −2 −1 0 1

Example 1

Show the solutions of:

a y= 2x+ 2 where −3 x 3 b 12x+ 3y− 24 = 0 where −5 x 5

Solution

a Write the equation. y = 2x+ 2

Find the value of y when x=−3. y = 2 ×−3 + 2 =−4

Find the value of y when x= 3. y = 2 × 3 + 2 = 8

State the solution. −4 y 8

Now graph the solution and write the interval.

b Write the equation. 12x+ 3y− 24 = 0

Rearrange. 3y =−12x+ 24

Divide by 3. y =−4x+ 8

Find the value of y when x=−5. y =−4 ×−5 + 8 = 28

Find the value of y when x= 5. y =−4 × 5 + 8 =−12

State the solution. −12 y 28

Now graph the solution and write the interval.

or [−4, 8]

−6 −4 −2 0 2 4 6 8 10

16 20 24 28

or (−12, 28]

−16 −12 −8 −4 0 4 8 12

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Solve and show the solutions of:

a 5x+ 11 2x− 7 b 7 − 4yy+ 4

c p− 3 4p+ 12 3p+ 10 d

Solution

a Write the inequation. 5x+ 11 2x− 7

Collect variables (LHS) and constants (RHS). 5x− 2x−7 − 11

Simplify. 3x−18

Divide both sides by 3. x−6

Now graph the solution and write the interval.

b Write the inequation. 7 − 4yy+ 4

Collect variables and constants. −4yy 4 − 7

Simplify. −5y−3

Divide both sides by −5. Remember, reverse the

sign when dividing by a negative. y

Now graph the solution and write the interval.

c Write the inequation.

This is solved as two separate problems.

p− 3 4p+ 12 3p+ 10

Write the first inequation. p− 3 4p+ 12

Collect variables and constants. p− 4p 12 + 3

Simplify. −3p 15

Divide by −3 and reverse the sign. p−5

Write the second inequation. 4p+ 12 3p+ 10

Collect variables and constants. 4p− 3p 10 − 12

Simplify. p−2

The answers can now be put together.

Now graph the solution and write the interval.

−5p−2

d Write the inequation. −

Multiply by the lowest common

denominator, 10. 10 − 10 10

Cancel where possible. 2(7d− 8) − (2d+ 7) 5(5d− 2)

Expand the brackets. 14d− 16 − 2d− 7 25d− 10

Simplify. 12d− 23 25d− 10

Collect variables and constants. 12d− 25d −10 + 23

Simplify. −13d 13

Divide by −13 and reverse the sign. d −1

7d –8 5

--- 2d+7 10

--- 5d –2 2

---or [−∞, −6]

−9 −8 −7 −6 −5 −4 −3 −2 −1

3 5

---or [−∞, )3 5

---−4 −3 −2 −1 0 3 1 2 3 4

5

---or (−5, −2]

−7 −6 −5 −4 −3 −2 −1 0 1

7d– 8 5

--- 2d+ 7 10

--- 5d–2 2

---7d– 8 5

--- 2d+ 7 10

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Exercise

Order and inequations

1 Solve and show the following on a number line.

a 6x− 5 2x+ 31 b 2r+ 7 5r− 23

c v− 6 3v+ 12 d 5e− 3 e+ 9

e 3m+ 10 11m+ 34 f 5 −ii− 7

g 3q+ 5 5 − 3q h 6k+ 8 7k− 6

i 2b− 9 7b− 12 j 4w− 6 15 − 3w k 19 − 3a−16 − 8a l 12f+ 6 3f− 27

2 Solve and show the following on a number line.

a y= 3x+ 1 where −4 x 4 b y=−2x+ 5 where 0 x 8

c 2y= 6x+ 10 where −9 x 0 d y− 3x= 15 where −3 x 3

e 8x+ 2y− 12 = 0 where −5 x 4 f 9x+ 3y+ 21 = 0 where −7 x 3

3 Solve and show the following as an interval and on a number line.

a t− 7 2t− 4 6 b 12 − 2u 5 − 3u 9 −u c 3n+ 4 4n+ 9 1 d −19 3g− 7 g+ 1

e −2x− 1 7 −x 1 − 2x f 5z− 11 3z+ 5 4z+ 8

g 13h+ 4 9h− 8 15h+ 16 h 7 − 5j 9 − 6j 3

4 Solve the following.

ab + 0

c + y− 4 d − +

e − − f − −

g + 1 hk

5 Prove each of the following theorems for real numbers w, x, y, z.

a xyANDyzxz b xyxzyz

c xyx+zy+z d xyxzyz e xyANDz 0 ⇒xzyz f xyANDz 0 ⇒xzyz

g xyANDz 0 ⇒xzyz h xyANDz 0 ⇒

i xyANDz 0 ⇒ j xyANDz 0 ⇒

k xyANDz 0 ⇒

l wxANDyzw+yx+z (Hint: Use transitivity after adding y to the first relation and x to the second relation.)

m xy 0 ⇒ 0 (Hint: Use an operation with xy.)

n For all values of x, y 0: xyx2y2 (You must prove both ways.) 3x –1

4

--- 2x+3 3

--- 4x– 9 12

--- 2w–1

7

--- 2w+9 3

---2y –5 3

--- y– 4 2

--- 3p–8

5

--- 2p+1 4

--- 3p–5 10 --- 1 4 ---t 6

--- 3t–2 8

--- 4t+11 12

--- t–2 4

--- 3v+5

8 --- v

3

--- v+3 3

--- v+ 7 4

---2(m+5) 5

--- 3m+1 4

--- 2 5

--- 3k+1

8 --- 1

6

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---Exercise

Answers

1 a x 9 b r 10

c v−9 d e 3

e m−3 f i 6

g q 0 h k 14

i b j w 3

k a−7 l f−3

2 a −11 y 13 b −11 y 5

c −22 y 5 d 6 y 24

e −10 y 26 f −16 y 14

3 a −3 t5 [−3, 5] b −7 u−2 (−7, −2)

c −5 n−2 [−5, −2] d −4 g4 (−4, 4)

e −8 x−6 (−8, −6] f −3 z8 [−3, 8)

g −4 h−3 (−4, −3] h 1 j2 [1, 2)

4 a x−2 b w−3 c y−2 d p−8

e t−4 f v 33 g m 1 h k−17

6 7 8 9 10 12 13 14 15 7 8 9 10 11 12 13 14 15

−12 −11 −10 −9 −8 −7 −6 −5 −4 0 1 2 3 4 5 6 7 8

−6 −5 −4 −3 −2 −1 −2 −3 −4 3 4 5 6 7 8 9 10 11

−3 −2 −1 0 1 2 3 4 5 11 12 13 14 15 16 17 18 19

3 5

---−5 −2 −1 0 1 2

3 5

---0 1 2 3 4 5

3 4 5 6 7 8

2 3

---−6 −5 −4 −3 −2 −1

2 3

---−3

−10 −9 −8 −7 −6 −5 −6 −7 −8 −2 −3 −4

−12 −8 −4 0 4 8 12 16

13

−11

−12−10 −8 −6 −4 −2 0 2

5

−11

4 6

20

−30−20 −10 0 10 20 30 40

5

−22

4 8 12 16 20 24 28

6

50 60 32 36 40

−10 −5 0 5 10 15 20 25

26

30 −20−16−12 −8 −4 0 4

14

8 12 16

−8 −6 −4 −2 0 2 4 6 8 −8 −7 −6 −5 −4 −3 −2 −1 0

−8 −7 −6 −5 −4 −3 −2 −1 0 −8 −6 −4 −2 0 2 4 6 8

−10 −9 −8 −7 −6 −5 −4 −3 −2 −8 −6 −4 −2 0 2 4 6 8

−7 −6 −5 −4 −3 −2 −1 0 1 −4 −3 −2 −1 0 1 2 3 4

Figure

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References

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