A number a is greater than a number b if a − b is positive. In symbols this is written as a > b.
Thus | 2 > 1 | because | 2 − 1 = 1 is positive | |
3 > −1 | because | 3 − (−1) = 4 is positive | ||
BUT | −1 > 2 is false | because | −1 − 2 = −3 is negative |
Example 1: Prove or disprove the following inequalities:
Solution: As a decimal, 1 ⁄ 4 = 0.25 and so 0.4 − 1 ⁄ 4 = 0.4 − 0.25 = 0.15, which is positive. Thus 0.4 > 1 ⁄ 4 is true.
Solution: Here (0.7)^{2} = 0.7 × 0.7 = 0.49. As a fraction 1 ⁄ 2 is 0.5. In this case, (0.7)^{2} − 1 ⁄ 2 = 0.49 − 0.5 = −0.01, which is negative. This means that the inequality (0.7)^{2} > 1 ⁄ 2 is false.
For this latter example we would write (0.7)^{2} < 1 ⁄ 2, or in words, (0.7)^{2} is less than 1 ⁄ 2. In general we say: A number a is less than a number b if a − b is negative. In symbols this is written as a < b.
If a < b then b > a and vice versa |
Example 2: In each of the following pairs of numbers, use one of the symbols > or < to give the correct ordering of the numbers for the order in which they appear:
Solution: Taking a = −1 and b = 2 the difference a − b, becomes a − b = (−1) − 2 = −3, which is negative. The correct inequality is: −1 < 2
Solution: In decimal form 1 ⁄ 4 = 0.25 and 1 ⁄ 5 = 0.2. Since 0.25 − 0.2 = 0.05, and this is positive, the correct inequality is: 1 ⁄ 4 > 1 ⁄ 5
In addition to these two inequalities there are two further symbols, ≥ and ≤. The first of these is read as 'greater than or equal to' and the second as 'less than or equal to'.
Exercise 1: For each of the following pairs of numbers use one of the symbols >, <, ≥ or ≤ to give the correct relationship for the order in which they appear:
− 1 2 = − 1 × 3 2 × 3 − 3 6 | and | − 1 3 = − 1 × 2 3 × 2 − 2 6 |
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Quiz 1: Which one of the following inequalities is correct?
Solution: The solution to this is obtained from:
3^{2} = 9 and 2^{3} = 8, and 9 > 8
The section following this one will deal with the solution of inequalities. As with the solution of equations, there are certain rules that may be used. In the case of inequalities these are:
Rule 1 | An equal quantity may be added to (or subtracted from) both sides of an inequality without changing the inequality. |
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Rule 2 | An equal positive quantity may multiply (or divide) both sides of an inequality without changing the inequality. |
Rule 3 | If both sides of an inequality are multiplied (or divided) by a negative quantity then the inequality is reversed. |
N.B. It is very important to be careful with the last rule!
Before looking at the solution of inequalities it is useful to see why the above rules hold. This is done in the following example.
Example 3: Given the (true) inequality 4 > −1, verify each of the rules:
Solution:
Adding 3 to both sides gives 7 > 2, which is true
Solution:
Subtracting 3 from both sides gives 4 − 3 > −1 − 3 or 1 > −4, which is true.
Solution:
Multiplying both sides by 3 gives 12 > −3, which is true.
Solution:
Multiplying the inequality 4 > −1 by −3 gives, according to Rule 3, 4 × −3 < −1 × −3, or −12 < 3, which is trueExample 4: Solve the following inequalities:
Solution:
x − 3 | > | 5 | add 3 to both sides (Rule 1): | |
x − 3 + 3 | > | 5 + 3 | ||
x | > | 8 |
Solution:
2x − 1 > 7 | > | 5 | add 1 to both sides (Rule 1) | |
2x | > | 8 | divide both sides by 2 (Rule 2) | |
x | > | 4 |
Solution:
3 − 2x | > | −5 | subtract 3 from both sides (Rule 1) | |
−2x | > | −8 | divide both sides by −2 (Rule 3) | |
x | < | 4 | note that the inequality is now reversed |
Here are some exercises on solving inequalities:
Exercise 2: Solve each of the following inequalities using the rules given in Section 2
Solution:
3x − 4 + 4 | < | 5 + 4 |
3x | < | 9 |
x | < | 3 |
Solution:
x + 1 − 1 | < | 0 − 1 |
x | < | −1 |
Solution:
2x − 6 + 6 | ≥ | 10 + 6 |
2x | ≥ | 16 |
x | ≥ | 8 |
Solution:
2x − x | ≥ | x − 3 − x |
x | ≥ | −3 |
Solution:
3x + 1 − 1 | ≥ | 2x + 5 − 1 |
3x | ≥ | 2x + 4 |
3x − 2x | ≥ | 2x + 4 − 2x |
x | ≥ | 4 |
Solution:
3x − 3 | ≥ | 2 − 2x |
3x − 3 + 3 | ≥ | 2 − 2x + 3 |
3x | ≥ | 5 + 2x |
3x− 2x | ≥ | 5 + 2x − 2x |
x | ≥ | 5 |
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Quiz 2: Which of the following is the solution to the inequality: 15 − x > 7 + x ?
15 - x | > | 7 + x |
15 - x + x | > | 7 + x + x |
15 | > | 7 + 2x |
15 − 7 | > | 7 + 2x - 7 |
8 | > | 2x |
4 | > | x |
or equivalently: x | < | 4 |
The first step in this solution was adding x to both sides so that (the positive) 2x appears on the right. This meant that the subsequent division was by the positive number 2. Division by positive numbers is always preferable as it generally leads to fewer mistakes.
Some inequalities contain more information and need further development.
Example 4: Solve the inequality: x − 10 < 2x − 2 < x
Solution:
The method is the same as before but now there are two inequalities to solve, i.e. x − 10 < 2x − 2 and 2x − 2 < x. The first of these is solved below in the left-hand column, the second in the right-hand column:
x − 10 | < | 2x − 2 | 2x − 2 | < | x | |
x − 10 + 2 | < | 2x − 2 + 2 | 2x − 2 + 2 | < | x + 2 | |
x − 8 | < | 2x | 2x | < | x + 2 | |
x − 8 − x | < | 2x − x | 2x − x | < | x + 2 − x | |
−8 | < | x | x | < | 2 |
Both of these must hold, so the solution is −8 < x < 2, i.e. x must be larger than −8 and smaller than 2.
Exercise 3: Find the solution to each of the following inequalities:
Solution: Here there are two inequalities to be solved, −3 ≤ 3x and 3x ≤ 18. The first of these is:
−3 | ≤ | 3x | divide both sides by 3 | |
-1 | ≤ | x |
The second is:
3x | ≤ | 18 | divide both sides by 3 | |
x | ≤ | 6 |
In both of the above inequalities the divisor is 3, which is positive, so the division does not reverse the inequalities.
The solution to the inequality is thus: -1 ≤ x ≤ 6
Solution: There are two inequalities to be solved here, 10 ≤ 2x and 2x ≤ x + 9.
The first of these is:
10 | ≤ | 2x | divide both sides by 2 | |
5 | ≤ | x |
The second is:
2x | ≤ | x + 9 | subtract x from both sides | |
x | ≤ | 9 |
In both of the above inequalities the divisor is 3, which is positive, so the division does not reverse the inequalities.
The solution to the inequality is thus: 5 ≤ x ≤ 9
x | < | 3x - 1 | add 1 | ||
x + 1 | < | 3x | subtract x | ||
1 | < | 2x | divide by 2 | ||
1 / 2 | < | x | |||
and: | |||||
3x - 1 | < | 2x + 7 | add 1 | ||
3x | < | 2x + 8 | subtract 2x | ||
x | < | 8 |
The solution to the inequality is: 1 / 2 < x < 8
Solution: The two inequalities in this case are: 2x − 7 < 8 and 8 < 3x − 11. The solutions to each are:
2x − 7 | < | 8 | add 7 | ||
2x | < | 15 | divide by 2 | ||
x | < | 15 / 2 | |||
and: | |||||
8 | < | 3x - 11 | add 11 | ||
19 | < | 3x | divide by 3 | ||
19 / 3 | < | x |
The solution to this is: 19 / 3 < x < 15 / 2
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Choose the solutions from the options given