Five lose myth on the subject of mathematics

INEQUALITY
A. Definition
1. Inequality Notation
Assume if a and b are real number.
a. a is called less than b, is written a < b if and only if a - b is negative. For example, 7 < 12 because 7 – 12 = -5 and -5 is negative.
b. A is called more than b, is written a > b if and only if a – b is positive. For example, 5 > 2 because 5 – 2 = 3, and 3 is positive.
c. A is called less than or equals by b, is written a ≤ b if and only if b if and only if a < b or a = b. In another word, a ≤ is negation from a > b. for example, 4 ≤ 7 is true because 4 > 7 is false.
d. A is called more than or equals by b, is written a ≥ b if only if a > b or a = b. in another word a ≥ b is negation from a < b. for example, 7 ≥ 3 is true because 7 < 3 is false.
2. Inequality Definition
Above description, were given notation from … a ≤ b and a ≥ b. inequality is an open sentence that related by notation “<”, “>“, “≤“, or “≥”. The general forms of inequality are
u (x) < v (x) u (x) ≤ v (x)
u (x) > v (x) u (x) ≥ v (x)
Example 4.1
An inequality form’s example
a. X – 2 < 5 d. l5 – 3xl < 13
b. x² - 1 ≥ 2 e.
c. (x – 5)(x + 1)²(x + 3) > 0
3. Interval
Before we learn about solution of inequality form, do we learn about solution of inequality form, do you still remember about an interval? For example R is a real number set. We can explain a set that is collection of the real number x… For example, {x l x < 4, x Є R}, {x l x ≥ 1, x Є R}, {x l 2 < x ≤ 5, x Є R}. the sets can be explain as bellow.
There are 8 kind of interval forms possibility that often we find if solve an inequality generally.
Table 4.1
Set Notation Number Line Kind of Interval
1. {x l x < a, x Є R}
2. {x l x > a, x Є R}
3. {x l x ≤ a, x Є R}
4. {x l x ≥ a, x Є R}
5. {x l a < x < b, x Є R}
6. {x l a < x ≤ b, x Є R}
7. {x l a ≤ x < b, x Є R}
8. {x l a < x < b, x Є R} Opened interval at a
Opened interval at a
Closed interval at b
Closed interval at b
Opened interval at a and b
Opened interval at a and closed interval at b
Closed interval at a and opened interval at b
Closed interval at a and b
Interval in 1-4 is named infinite interval, whereas interval in 5-8 is named finite interval.
4. Inequality Characteristic
Interval definition explain presence a solve from the inequality. Determine of the solve or the solution is based of theorems bellow.
- Theorem 4.1
Interval notation an inequality is not change if there is added or reduction a number (statement) at both of parts of an equation that same.
- Proof
Let c is a number (statement) which will added in a < b inequality. Because of a < b, so a – b = n, n < 0. Thus
a – b + c – c = n, c – c = 0
⇔ (a + c) – (b + c)=n < 0
⇔ a + c < b + c
Example 4.2
Determine the solution of each inequality bellow!
a. X – 2 < 5
b. X – 4 > -6
c. X + 9 < 11
Answer :
a. X – 2 < 5 b. x – 4 > -6 c. x + 9 < 11
⇔ x – 2 + 2 < 5 + 2 ⇔ x – 4 – 4 > -6 + 4 ⇔ x + 9 – 9 < 11 – 9
⇔ x < 7 ⇔ x > 2 ⇔ x < 2

 Note :
Let x – 2 < 5
The left part of inequality above is x – 2, whereas the right part of inequality above is 5.

- Theorem 4. 2
Notation an inequality is not change if make a positive number (statement) multiplication or division in both of parts of an inequality.
- Proof
Let a < b and k > 0. Because of a < b, so a – b = n, n < 0
Thus
k(a - b) = kn
⇔ ka – kb = kn < 0
⇔ ka < kb
Example 4.3
Determine the solution of each inequality bellow!
a. 2x > 14
b. 3x < 18
c. 2x + 3 < 9
Answer
a. 2x > 14 b. 3x < 18 c. 2x + 3 < 9
⇔ ½ . 2x > ½ . 4 ⇔ ⇔ 2x + 3 – 3 < 9 - 3
⇔ x > 7 ⇔ x < 6 ⇔ 2x < 6
⇔
⇔ x < 3
- Theorem 4.3
Notation of inequality is change if make a negative number (statement) multiplication or division in both of parts of an inequality.
- Proof
For example a < b and k < 0
Because of a < b, so a – b = n, n < 0
Thus
k(a - b) = kn
⇔ ka – kb = kn > 0
⇔ ka > kb
Example 4.4
a. -5x < 10
b. -4x > -20
c. -3x + 5 < x – 1
Answer
a. -5x < 10 b. -4x > -20 c. -3x + 5 < x - 1
⇔ ⇔ ⇔ -3x + 5 – 5 < x – 1 – 5
⇔ x > -2 ⇔ x < 5 ⇔ -3x < x – 6
⇔ -3x – x < x –x – 6
⇔ -4x <-6
⇔
⇔ x > 3/2
 Playing a number
Is 7 < 4 true ?
1. Let a = 3 and b = 4, thus : a < b
2. Multiplying by a : a² < ab
3. Reducing by b² : a² - b² < ab - b²
4. Make a factor : (a +b)(a - b) < b (a - b)
5. Over by (a – b) : a + b < b
6. Replacing a with 3 and b with 4 : 7 < 4
Where are the false of argument above?

B. Linear Inequality
1. Definition
Linear inequality is inequality that contain variable with its highest power is 1 (one). Example 4.2 – 4.4 showed same linear inequality forms with variable x.
2. Solve the Linear Inequality
In this topic, our term is only for one variable linear inequality. The solution linear inequality is divided become two, they are :
a. Absolute linear inequality
This inequality is true to all of its variables. For example, x + 2 < x + 5
b. Conditional Linear inequality
This inequality is true to certain value of its variable. For example, x – 2 > 8 is true for x > 10
Solution of this linear inequality can showed by a set or a number line (interval) like was showed in example 4.2 – 4.4.
Example 4.5
Determine the solution set in number line bellow of inequalities bellow!
a. 4x + 2 < 10
b. 7x – 5 ≥ x – 17
c. 3x + 5 < 5x + 7
Answer
a. 4x + 2 < 10 c. 3x + 5 < 5x – 7
⇔ 4x + 2 – 2 < 10 – 2 ⇔ 3x + 5 – 3x - 7 < 5x + 7 – 3x - 7
⇔ 4x < 8 ⇔ -2 < 2x
⇔ ⇔ -1 < x
⇔ x < 2 SS = {x l x > -1, x Є R}
SS = {x l x < 2, x Є R}
b. 7x – 5 ≥ x – 17
⇔ 7x – 5 + 5 ≥ x – 17 + 5
⇔ 7x ≥ x – 12
⇔ 7x – x ≥ x – x – 12
⇔ 6x ≥ -12
⇔
⇔x ≥ -2
SS = {x l x ≥ -2, x Є R}
Exercise 4.1
In exercise 1-8, observant
1. x < 6 ⇔ x + 2 < 8 5. a > 3 ⇔ 5a > 15
2. y > 8 ⇔ y – 5 > -2 6. P < -4 ⇔ 3p > 3p > -12
3. z – 1 ≤ 4 ⇔ x ≤ 5 7. n ≥ 2 ⇔ -3n ≥ -6
4. m + 2 ≥ 1 ⇔ m ≥ -1 8. m > 7 ⇔ -m > -7
In exercise 9-20, determine inequality set to all of real number component by set notation!
9. 5 ≤ k 13. 8 > q + 6 17.
10. a < -2 14. 5n < 20 18. -3 < m + 2 < 4
11. p + 1 < 3 15. –w ≥ 20 19. 2x – 1 ≤ 3x + 5 ≤ 2x + 6
12. t – 2 > 1 16. 20. x – 1 ≤ 2x + 1 ≤ 3 + x
In exercise 21-30, determine the set solution to all of real number component in a number line.
21. x < 1 25. 10 – 2z ≥ 6 28. 2x – 1 < x
22. 2y + 1 < 3 26. 5p – 6 > 4 29. 4(y - 2) ≤ 0
23. 2 + 5k ≤ 12 27. 2m + 1 < 0 30. 3 < 2x – 7 ≤ 5
24. 1 – p > 2 + 7p