Inequalities: Properties, Operations, and Solving Techniques

Inequalities

In mathematics, an inequality is a statement that compares two values, expressions, or quantities using one of the inequality symbols: "<<" (less than), ">>" (greater than), "\le" (less than or equal to), "\ge" (greater than or equal to) or "\neq" (not equal to).
Comparing numbers and expressions involves determining the relative sizes of different quantities. There are several methods for comparing numbers and expressions, including:

  • Comparison symbols: One of the simplest methods for comparing numbers is to use comparison symbols such as "less than" (<)(<), "greater than" (>)(>), "less than or equal to" ()(\le ) and "greater than or equal to" ()( \ge ). For example, to compare the numbers 3 and 5, we can write 3<53 < 5 to indicate that 3 is less than 5.
  • Number line: The number line is a visual representation of numbers on a line, with smaller numbers on the left and larger numbers on the right. To compare numbers on a number line, we can simply look at their positions relative to each other. For example, if we want to compare 3 and 5, we can see that 5 is to the right of 3 on the number line, so 5 is greater than 3.
  • Absolute value: The absolute value of a number is the distance of that number from zero on the number line. To compare two numbers with the same sign, we can compare their absolute values. For example, to compare -3 and -5, we can compare the absolute values of these numbers, which are 3 and 5, respectively. Since 5 is greater than 3, we can say that The absolute value of -5 is greater than The absolute value of -3.
  • Algebraic manipulation: We can use algebraic manipulation to compare expressions by simplifying them and comparing the resulting expressions. For example, to compare the expressions 2x+32x+3 and 3x13x-1 , we can simplify them by combining like terms and get 2x+3<3x12x+3 < 3x-1 . Then, we can isolate the variable on one side of the inequality and get x>4x> 4 .
  • Common denominator: When comparing fractions, we can find a common denominator and then compare the numerators. For example, to compare 14\frac{1}{4} and 25\frac{2}{5} , we can find a common denominator of 20 and get 520\frac{5}{20} and 820\frac{8}{20} . Then, we can see that 820\frac{8}{20} is greater than 520\frac{5}{20} , so 25\frac{2}{5} is greater than 14\frac{1}{4} .

These are just a few of the methods used to compare numbers and expressions in mathematics. The choice of method depends on the specific problem and the tools available.

Properties of inequalities.

Inequalities have several properties that govern how they behave, including the following:

  • Transitivity: If a<ba < b and b<cb < c, then a<ca < c . This property means that if one value is less than another value, and the second value is less than a third value, then the first value is also less than the third value.
  • Reflexivity: aaa \le a and aaa \ge a for any value of aa . This property means that any value is equal to itself, and it is both less than or equal to and greater than or equal to itself.
  • Symmetry: If a<ba < b , then b>ab> a . This property means that if one value is less than another value, then the second value is greater than the first value.
  • Addition property: If a<ba < b and cc is any number, then a+c<b+ca+c < b+c . This property means that if you add the same number to both sides of an inequality, the inequality remains true.
  • Subtraction property: If a<ba < b and cc is any number, then ac<bca-c < b-c . This property means that if you subtract the same number from both sides of an inequality, the inequality remains true.
  • Multiplication property: If a<ba < b and cc is a positive number, then ac<bcac < bc . This property means that if you multiply both sides of an inequality by a positive number, the inequality remains true.
  • Division property: If a<ba < b and cc is a positive number, then ac<bc\frac{a}{c} < \frac{b}{c} . This property means that if you divide both sides of an inequality by a positive number, the inequality remains true.
  • Inverse property: If a<ba < b , then b<a-b < -a . This property means that if you negate both sides of an inequality, the inequality reverses direction.
  • Transpose property: If a<ba < b and c<dc < d , then a+c<b+da+c < b+d . This property means that if you add two inequalities together, the result is also an inequality.

These properties are fundamental to working with inequalities in mathematics. They allow us to manipulate inequalities and arrive at new inequalities that are still true, which helps us to solve problems and prove mathematical statements.

Addition and multiplication of inequalities

Addition and multiplication of inequalities are operations used in algebra to manipulate and solve equations and inequalities.

Addition of Inequalities: If we have two inequalities a<ba < b and c<dc < d , we can add them together to get a+c<b+da+c < b+d . This property is sometimes referred to as the addition property of inequalities.
For example, let's say we want to solve the inequality
2x5>72x-5 > 7 . We can add 5 to both sides of the inequality to get 2x>122x > 12 . Then, we can divide both sides by 2 to get x>6x > 6 . This tells us that any value of xx greater than 6 will satisfy the inequality.

Multiplication of Inequalities: If we have two inequalities a<ba < b and c>0c> 0 , we can multiply them together to get ac<bcac < bc . This property is sometimes referred to as the multiplication property of inequalities.
However, if cc is less than 0, we must reverse the direction of the inequality, since multiplying by a negative number changes the order of the inequality.
For example, if we have the inequality 3<5-3 < 5 and we multiply both sides by -2 , we get 6>106> -10 .
For example, let's say we want to solve the inequality 2x5<72x-5 < 7. We can add 5 to both sides to get 2x<122x < 12. Then, we can divide both sides by 2 to get x<6x < 6 . This tells us that any value of xx less than 6 will satisfy the inequality.
Similarly, if we have the inequality 3x>12-3x > 12 , we can divide both sides by -3, but since we're dividing by a negative number, we need to flip the direction of the inequality. This gives us x<4x < -4 .

These operations are essential tools in algebra and can help us to solve a wide range of problems involving inequalities. However, it is important to remember to always check the signs of the numbers we're multiplying or adding, and to be careful when flipping the direction of an inequality.

Numerical intervals.

In mathematics, a number interval is a range of numbers between two specified values. Intervals are often used to describe solutions to equations and inequalities, and they are represented using brackets and parentheses.

There are several types of number intervals, each with its own notation and meaning:

  • Closed interval: A closed interval includes both endpoints and is denoted by square brackets. For example, the interval [a,b][a,b] includes all values of xx between aa and bb , including aa and bb themselves. So, if a=1a=1 and b=5b=5 , then [1,5][1,5] includes 1, 2, 3, 4 and 5.
  • Open interval: An open interval excludes both endpoints and is denoted by parentheses. For example, the interval (a,b)(a,b) includes all values of xx between aa and bb, but not including aa and bb themselves. So, if a=1a=1 and b=5b=5, then (1,5)(1,5) includes 2, 3 and 4, but not 1 or 5
  • Half-open interval: A half-open interval includes one endpoint and excludes the other, and is denoted by a combination of brackets and parentheses. For example, the interval [a,b)[a,b) includes all values of xx between aa and bb, including aa but not including bb. So, if a=1a=1 and b=5b=5 , then [1,5)[1,5) includes 1, 2, 3 and 4, but not 5.
  • Half-closed interval: A half-closed interval excludes one endpoint and includes the other, and is denoted by a combination of parentheses and brackets. For example, the interval (a,b](a,b] includes all values of xx between aa and bb, including bb but not including aa. So, if a=1a=1 and b=5b=5, then (1,5](1,5] includes 2, 3, 4 and 5, but not 1.
  • Infinite interval: An infinite interval has one or both endpoints at infinity, and is denoted by the symbols \infty or -\infty. For example, the interval (a,)(a, \infty ) includes all values of xx greater than a, while the interval (,b)( -\infty , b ) includes all values of xx less than bb.

Number intervals are useful for describing the solution sets to equations and inequalities, and they are commonly used in calculus and real analysis. Understanding the notation and meaning of number intervals is essential for interpreting mathematical expressions and solving problems involving inequalities and equations.

Solving linear inequalities in one variable.

In mathematics, a linear inequality in one variable is an inequality that can be expressed in the form ax+b<cax+b < c , ax+b>cax+b> c, ax+bcax+b \le c or ax+bcax+b\ge c , where aa, bb and cc are constants and xx is a variable.
To solve a linear inequality in one variable, we need to isolate the variable on one side of the inequality symbol, and then determine the range of values that satisfy the inequality. Here are the steps for solving a linear inequality in one variable:

  • When you pass an inequality from one side to the other with the opposite sign, you obtain an equivalent inequality.
  • When you multiply or divide both sides of an inequality by the same positive number, you obtain an equivalent inequality.
  • When you multiply or divide both sides of an inequality by the same negative number, you obtain an equivalent inequality by changing the sign of the inequality.

NLet's look at some examples to see how this works in practice:

Example 1: Solve the inequality 2x+3<92x+3 < 9
Solution: We can start by subtracting 3 from both sides of the inequality to get 2x<62x < 6. Then, we can divide both sides by 2 to get x<3x < 3. The solution set is the open interval (,3)( -\infty , 3).

Example 2: Solve the inequality 5x+213-5x+2\ge -13
Solution: We can start by subtracting 2 from both sides of the inequality to get 5x15-5x \ge -15. Then, we can divide both sides by -5, but since we're dividing by a negative number, we need to flip the direction of the inequality symbol to get x3x \le 3. The solution set is the closed interval (,3](-\infty,3].

Example 3: Solve the inequality 3x4>5x+23x-4 > 5x+2
Solution: We can start by subtracting 3x3x from both sides of the inequality to get 4>2x+2-4 > 2x+2. Then, we can subtract 2 from both sides to get 6>2x-6 > 2x. Finally, we can divide both sides by 2 to get x<3x < -3. The solution set is the open interval (,3)(-\infty,-3).

It's important to check our solutions by plugging in values from the solution set and making sure they satisfy the original inequality. Solving linear inequalities in one variable is an important skill in algebra, and it has applications in many areas of mathematics and science.

Solving double inequalities

Double inequalities are a type of inequality that involves two inequality symbols and a variable in between them. A double inequality expresses a range of values for the variable that satisfy the inequality. The most common type of double inequality is the one that involves the symbols "<<" and ">>" .
The general form of a double inequality is: a<x<ba < x < b
where aa and bb are real numbers, and xx is the variable that we are interested in. This inequality tells us that xx must be greater than aa and less than bb. In other words, xx must fall between the two values aa and bb.
To solve a double inequality, we need to find the range of values for xx that satisfy the inequality. To do this, we need to solve each of the two inequalities separately and then combine the solutions.

For example: let's say we have the following double inequality: 3<2x+1<7-3 < 2x+1 < 7 .
To solve this inequality, we need to isolate x in the middle of the inequality by solving each of the two inequalities separately.
First, we'll solve the left inequality: 3<2x+1-3 < 2x+1 Subtracting 1 from both sides, we get: 4<2x-4 < 2x
Dividing both sides by 2, we get: 2<x-2 < x
Next, we'll solve the right inequality: 2x+1<72x+1 < 7
Subtracting 1 from both sides, we get: 2x<62x < 6
Dividing both sides by 2, we get: x<3x < 3
Now we have the two solutions: 2<x<3-2 < x < 3
This means that x must be greater than -2 and less than 3 to satisfy the original double inequality.
Another type of double inequality involves the symbols "\le" and "\ge". This type of double inequality expresses a range of values for the variable that includes the endpoints. The general form of a double inequality with "\le" and "\ge" symbols is: axba \le x \le b , where aa and bb are real numbers, and xx is the variable that we are interested in.
To solve this type of double inequality, we follow a similar process as with the "<<" and ">>" symbols. We solve each of the two inequalities separately and then combine the solutions.

For example , let's say we have the following double inequality: 23x15-2 \le 3x-1 \le 5 .
To solve this inequality, we first solve the left inequality: 23x1-2 \le 3x-1
Adding 1 to both sides, we get: 13x-1 \le 3x
Dividing both sides by 3, we get: 13x-\frac{1}{3} \le x
Next, we solve the right inequality: 3x153x-1 \le 5
Adding 1 to both sides, we get: 3x63x \le 6
Dividing both sides by 3, we get: x2x \le 2
Now we have the two solutions: 13x2-\frac{1}{3} \le x \le 2
This means that xx must be greater than or equal to 13-\frac{1}{3} and less than or equal to 2 to satisfy the original double inequality.

In summary, solving double inequalities involves finding the range of values for the variable that satisfies the inequality. To solve a double inequality, we need to isolate the variable by solving each of the two inequalities separately and then combine the solutions.

Simple inequalities with a variable inside the modulus sign. Absolute value inequalities.

Inequalities with a variable inside the modulus notation are called absolute value inequalities. The absolute value of a number represents the distance of that number from zero on the number line. The absolute value of a variable xx is written as x|x| and is always non-negative.
The general form of a simple absolute value inequality with a variable xx is: ax+b<c|ax+b| < c , where aa, bb and cc are constants.

To solve a simple absolute value inequality, we need to isolate the variable xx by considering the two possible cases:


Case 1: ax+bax+b is positive or zero, so ax+b=ax+b|ax+b|=ax+b.
In this case, we can solve the inequality as follows:
ax+b<cax<cbx<cbaax+b < c \rightarrow ax < c-b \rightarrow x < \frac{c-b}{a}

Case 2: ax+bax+b is negative, so ax+b=(ax+b)|ax+b| = -(ax+b).
In this case, we can solve the inequality as follows:
(ax+b)<caxb<c\small -(ax+b) < c \rightarrow -ax-b < c \rightarrow ax>(c+b)x>c+ba\small \rightarrow ax> -(c+b) \rightarrow x > -\frac{c+b}{a}

To summarize, to solve a simple absolute value inequality ax+b<c|ax+b| < c, we need to consider two cases: ax+bax+b is positive or zero and ax+bax+b is negative. Then, we can isolate xx in each case to find the solution.

Let's work through an example to see how this works in practice: 2x+1<5|2x+1| < 5 .

Case 1: 2x+102x+1 \ge 0
2x+1<52x+1 < 5
2x<42x < 4
x<2x < 2

Case 2: 2x+1<02x+1 < 0
2x1<5-2x-1 < 5
2x<6-2x < 6
x>3x > -3

Therefore, the solution to the absolute value inequality 2x+1<5|2x+1| < 5 is 3<x<2-3 < x < 2. (3;2)(-3;2).