Modulus or Absolute Value of a Real Number
Definition
The modulus, absolute value, or magnitude of a real number \( x \) is denoted by \( x \) and is defined as the distance of \( x \) from the origin on the real number line. This distance is always nonnegative.
Consider the numbers 5, 5, and 0:  For \( x = 5 \), the modulus is \( 5 = 5 \). This is the distance from 0 to 5 on the number line, which is 5 units.  For \( x = 5 \), the modulus is \( 5 = 5 \). Although 5 is 5 units left of the origin, distance is measured positively, thus it is also 5 units.  For \( x = 0 \), the modulus is \( 0 = 0 \). The point 0 itself is at the origin, so its distance from the origin is zero.
The modulus \( x \) of a number \( x \) is defined by the following piecewise expression:
In this piecewise definition:  If \( x \) is positive or zero (\( x \geq 0 \)), then \( x = x \). Here, the output is the same as the input because the distance from the origin to a positive number or zero is the number itself.  If \( x \) is negative (\( x < 0 \)), then \( x = x \). This calculation effectively removes the negative sign, turning \( x \) into a positive value. This is because the distance, being a measure of magnitude without direction, must always be positive or zero.
Graph
To draw the graph of \( y = x \):
Begin by recognizing that when the input \( x \) is not negative, the output \( y \) is equal to \( x \). Place marks on the graph where the input and the output match: (0,0), (1,1), (2,2), (3,3), and so on. There are infinite inputs and correspoding outputs, which results in infinte points. these infinite points form a continuous path that extends indefinitely to the right, starting from the origin and moving diagonally upward in a straight direction.
Conversely, for inputs \( x \) that are less than zero, the output \( y \) becomes the opposite of \( x \). Mark the points (1,1), (2,2), (3,3), continuing this pattern. Connecting these points forms a continuous path that extends indefinitely to the left, starting from the origin and moving diagonally upward in a straight direction but mirrored compared to the previous path.
Merging these two paths at the origin, the complete graph forms a V shape with the bottom point at the origin, rising to the right and left. This V represents all pairs of inputs and their corresponding outputs for \( y = x \), where the distance from the origin reflects the magnitude of the input without regard to whether it was originally a negative or a positive number.
Properties of the Absolute Value Function

Nonnegativity
 The absolute value of any real number \( x \), denoted \( x \), is always nonnegative. This aligns with the geometric interpretation of absolute value as the distance of \( x \) from the origin on the real number line, and distance is never negative.
 The expression \( x = 0 \) if and only if \( x = 0 \). This means the only scenario where the distance from the origin is zero is when the point is at the origin itself.
 If \( x > 0 \), it implies that \( x \) is in the set of all real numbers except zero, \( \mathbb{R} \setminus \{0\} \). This is because a nonzero distance from the origin indicates the point is not at the origin.

Square Root Relationship:
 The equation \( \sqrt{x^2} = x \) holds for any real number \( x \). This can be understood through the squaring and square root operations:
 Squaring any real number \( x \) results in a nonnegative number \( x^2 \), as the square of both positive and negative numbers is positive (e.g., \( (2)^2 = 4 \) and \( (2)^2 = 4 \)).
 Taking the square root of a nonnegative number yields a nonnegative result. Thus, \( \sqrt{x^2} \) effectively removes the sign of \( x \), yielding the nonnegative magnitude of \( x \), which is exactly what \( x \) represents.
 The equation \( \sqrt{x^2} = x \) holds for any real number \( x \). This can be understood through the squaring and square root operations:

Identity of Inverse: \( x = x \)
The absolute value of the additive inverse (negative) of \( x \) is equal to the absolute value of \( x \).

Multiplicative Property: \( xy = xy \)
The absolute value of the product of \( x \) and \( y \) is the product of their respective absolute values.

Quotient Property: \( \left \frac{x}{y} \right = \frac{x}{y} \) for \( y \neq 0 \)
The absolute value of the quotient of \( x \) and \( y \) is the quotient of their respective absolute values, provided \( y \) is nonzero.

Power Property: \( x^n = x^n \) for any integer \( n \)
The absolute value of \( x \) raised to the power of \( n \) is equal to the absolute value of \( x \) raised to the power of \( n \).

Product of Absolute Values: \( a_1 a_2 \ldots a_n = a_1 a_2 \ldots a_n \)
The absolute value of the product of several numbers \( a_1, a_2, \ldots, a_n \) is the product of the absolute values of each number.

\[ x^2 = x^2 \]
Proof: The absolute value of \( x \), denoted \( x \), is defined by:
\[ x = \begin{cases} x & \text{if } x \geq 0 \\ x & \text{if } x < 0 \end{cases} \]Squaring \( x \) gives us:
\[ x^2 = \begin{cases} x^2 & \text{if } x \geq 0 \\ (x)^2 & \text{if } x < 0 \end{cases} \]Since \( (x)^2 = x^2 \) for all real numbers \( x \), whether \( x \) is positive or negative, we have:
\[ x^2 = x^2 \quad \text{for all } x \in \mathbb{R} \]
Solving Equations

If \( x = a \) and \( a > 0 \), then \( x = \pm a \). This indicates two possible real solutions for \( x \) since both \( a \) and \( a \) have the same absolute value when \( a \) is positive.

If \( x = 0 \), then \( x = 0 \). The only number with an absolute value of zero is zero itself.

If \( x = a \) and \( a < 0 \), then no real solution for \( x \) exists. Absolute values are always nonnegative; thus, they cannot equal a negative number.
Example
Find all solutions of the equation \( (x  1)(x + 2)(x  3) = 0 \).
Solution: Set each factor to zero and solve for \( x \):
\( x  1 = 0 \) or \( x + 2 = 0 \) or \( x  3 = 0 \)
From \( x  1 = 0 \): \( x = 1 \) \( x = \pm 1 \)
From \( x + 2 = 0 \): No solution since \( x = 2 \) has no real solutions (as per property 3).
From \( x  3 = 0 \): \( x = 3 \) \( x = \pm 3 \)
The set of solutions is \( x \in \{ 3, 1, 1, 3 \} \).
Example
Solve:
\[ 2x  3 = 1 \]Solution:
\[\begin{align*} &2x  3 = 1 \\ \implies \quad &2x  3 = \pm 1 \\ \implies \quad &2x = 4 \quad \text{or} \quad 2x = 2 \\ \implies \quad &x = 2 \quad \text{or} \quad x = 1 \\ \implies \quad &x = \pm 2 \quad \text{or} \quad x = \pm 1 \end{align*}\]Hence, \( x \in \{2, 1, 1, 2\} \).
Example
Solve:
\[ x^2  4x + 3 = 0 \]Sol.: Let \( x = t \) for convenience, then the equation becomes
\[ t^2  4t + 3 = 0 \]Factor this quadratic equation to get:
\[ (t  3)(t  1) = 0 \]Thus, \( t = 1 \) or \( t = 3 \)
Since \( t = x \):
\( x = 1 \) gives \( x = \pm 1 \) \( x = 3 \) gives \( x = \pm 3 \)
Hence, the solution set is \( x \in \{3, 1, 1, 3\} \).

If \( x = y \), then \( x = \pm y \). This indicates that if two numbers have the same absolute value, they are either equal or opposites of each other.
Example
Solve:
\[ x + 1 = 2  3x \]Sol.: Using the property, we have two cases:
\[ x + 1 = 2  3x \quad \text{or} \quad x + 1 = (2  3x) \]For the first case:
\[ x + 1 = 2  3x \]\[ 4x = 1 \]\[ x = \frac{1}{4} \]For the second case:
\[ x + 1 = 2 + 3x \]\[ 2x = 3 \]\[ x = \frac{3}{2} \]Thus, the solutions are \( x = \frac{1}{4} \) or \( x = \frac{3}{2} \).

Triangle's Inequality
The triangle inequality is a fundamental property of real numbers that states:
For any two real numbers \( x \) and \( y \), the absolute value of their sum is less than or equal to the sum of their absolute values. Mathematically, this is represented as:
\[ x + y \leq x + y \]This inequality holds for all \( x, y \in \mathbb{R} \) and can be understood with following examples:
 For \( x = 5 \) and \( y = 3 \), the direct sum \( 5 + 3 \) equals 8, which is exactly the sum of their absolute values \( 5 + 3 \).
 When both \( x \) and \( y \) are negative, such as \( x = 5 \) and \( y = 3 \), the absolute value of their sum \( 5 + (3) \) is also 8, matching \( 5 + 3 \).
 If \( x \) and \( y \) have opposite signs, for instance \( x = 5 \) and \( y = 3 \), the absolute value of their sum \( 5 + (3) \) is 2, which is less than the sum of their absolute values.
 When one of the numbers is zero, such as \( x = 5 \) and \( y = 0 \), the absolute value of their sum \( 5 + 0 \) equals the absolute value of the nonzero number.
The strict inequality \( x + y < x + y \) is observed specifically when \( x \) and \( y \) are of opposite signs because the effect of one number's negation partially cancels out the magnitude of the other when summed.
Conversely, the equality \( x + y = x + y \) occurs in two cases:
 When both numbers have the same sign, their magnitudes directly add together without any cancellation.
 When one of the numbers is zero, it does not contribute to the magnitude of the sum, effectively leaving the absolute value of the other number.
Misconcpt
For any two real numbers \(x\) and \(y\), the statement \( x+y > x + y \) is ALWAYS false. It contradicts the triangle inequality which correctly states that \( x + y \leq x + y \).
Summary
The inequality \( x + y < x + y \) holds if and only if \( x \) and \( y \) have opposite signs, which is mathematically represented as \( xy < 0 \). This is because the product of a positive and a negative number is negative, reflecting the opposing signs of \( x \) and \( y \).
The equality \( x + y = x + y \) is true if and only if \( x \) and \( y \) are either both nonnegative or both nonpositive, or one of them is zero. This condition is described mathematically as \( xy \geq 0 \).
It is not possible for \( x + y \) to be greater than \( x + y \); such a case does not exist for any real numbers \( x \) and \( y \) because it would contravene the triangle inequality principle.
Example
Example 1:
Show that the inequality \( x + 1 > x + 1 \) has no solution.
Solution: Consider the triangle inequality, which states that for any real numbers \( x \) and \( y \), the inequality \( x + y \leq x + y \) always holds. Since \( 1 \) is positive, the inequality \( x + 1 > x + 1 \) contradicts the triangle inequality, as it suggests that the absolute value of \( x + 1 \) could be greater than the sum of the absolute value of \( x \) and \( 1 \), which is never true. Hence, there is no real number \( x \) that satisfies this inequality.
Example 2:
Prove that the inequality \( x + 1 < x + 1 \) holds if and only if \( x < 0 \).
Solution:
The inequality \( x + y \leq x + y \) becomes strict (i.e., \( x + y < x + y \)) only when \( x \) and \( y \) have opposite signs. Thus, \( x + 1 \leq x + 1 \), which can be rewritten as \( x + 1 \leq x + 1 \), can be true only when \(x\times 1<0\). Thus, \(x<0\)

Second Triangle's Inequality
For all \( x, y \in \mathbb{R} \),
\[  x  y  \leq x  y \]
The strict inequality \( x  y < x  y \) occurs when \( x \) and \( y \) have opposite signs, which is mathematically stated as \( xy < 0 \). When \( x \) and \( y \) are of opposite signs, the difference \( x  y \) does not fully capture the distance between \( x \) and \( y \) on the number line, since it neglects the fact that they are in different directions from zero.

The equality \( x  y = x  y \) occurs when \( x \) and \( y \) have the same sign or at least one is zero, which is represented as \( xy \geq 0 \). When both \( x \) and \( y \) are nonnegative or nonpositive, \( x  y \) precisely measures the distance between them. If one of them is zero, the absolute values and the actual difference are equivalent.
Example
Find the solutions of the inequality: \( 2x+1 < x + x+1 \).
Solution:
Rewrite the left side of the inequality to emphasize the sum of two terms:
\[ 2x+1 = x + (x+1) \]By the triangle inequality, we know that:
\[ a + b \leq a + b \]for any real numbers \( a \) and \( b \). The strict inequality \( a + b < a + b \) holds if and only if \( a \) and \( b \) have opposite signs.
Apply this to our expression:
\[ x + (x+1) < x + x+1 \]According to the triangle inequality, this strict inequality holds if and only if \( x \) and \( x+1 \) have opposite signs, i.e., when:
\[ x(x+1) < 0 \]The wavy curve method is used to find the range of \( x \) for which this inequality is true. This method considers the change in sign of the expression across its roots. The roots of \( x(x+1) \) are \( x = 0 \) and \( x = 1 \). Plot these on a number line and test the intervals:
 For \( x > 0 \), \( x(x+1) > 0 \) (positive).
 Between \( x = 1 \) and \( x = 0 \), \( x(x+1) < 0 \) (negative).
 For \( x < 1 \), \( x(x+1) > 0 \) (positive).
Therefore, the inequality \( 2x+1 < x + x+1 \) is satisfied when \( x \) is in the interval \( (1,0) \).

Example
Solve the equation \( x + 4 + x = 3 \).
Solution:
Our first goal is to eliminate the modulus symbol so that we can do the usual algebra on it and isolate the variable \(x\) to find the solutions. This we do by first considering two separate cases corresponding to the two scenarios of the absolute value.
Case I:
When \( x + 4 \geq 0 \), that is \(x\geq4\), the expression inside the absolute value is nonnegative. Hence, \( x + 4 = x + 4 \). Substituting this into our equation, we have:
Since \( x = \frac{1}{2} \) satisfies \( x + 4 \geq 0 \), it is a valid solution.
Case II:
When \( x + 4 < 0 \), that is, \(x<4\) the expression inside the absolute value is negative, and \( x + 4 = (x + 4) \). The equation then becomes:
This statement leads to a contradiction, indicating no solutions exist for this case.
Considering both cases, the solution to the equation \( x + 4 + x = 3 \) is \( x = \frac{1}{2} \).
Example
Solve the equation \( 2x  2 = 3x \).
Solution:
To address the equation involving the absolute value, we assess it in a piecewise manner, taking into account the two conditions that govern the behavior of \( x  2 \) based on the value of \( x \).
Case I:
Consider when \( x  2 \geq 0 \), which implies \( x \geq 2 \). The absolute value function \( x  2 \) can be replaced directly with \( x  2 \) under this condition. The equation then becomes:
However, substituting \( x = 4 \) back into the original equation gives \( 2 \cdot 2 \neq 3 \cdot 4 \), which is not true. Therefore, \( x = 4 \) is not a solution.
Case II:
Now consider when \( x  2 < 0 \), which means \( x < 2 \). In this case, \( x  2 \) becomes \( (x  2) \). Substituting this into our original equation yields:
Checking this solution in the original equation, we find that it satisfies the equation when \( x = \frac{4}{5} \), which falls within the range of \( x < 2 \).
Therefore, the only solution to the equation \( 2x  2 = 3x \) is \( x = \frac{4}{5} \).
Handling two modulus in an expression
Solve:
Solution:
Solution:
When we have two modulus symbols, as in this example, we want to isolate \( x \) but cannot do so directly due to the modulus symbols. To solve this, we first need to remove the modulus symbols because until we do that, we cannot perform usual algebraic manipulations.
Step 1: Analyze each expression inside the modulus and exactly determine when it is positive or negative. Combine this information to remove the modulus sign.
In this example, we observe:
 \( x1 \) is positive when \( x > 1 \), negative when \( x < 1 \), and zero when \( x = 1 \).
 \( x3 \) in the second modulus is positive when \( x > 3 \), negative when \( x < 3 \), and zero when \( x = 3 \).
Step 2: Express these conditions by drawing a real line as shown in the picture. Then make three cases based on \( x \):

For \( x < 1 \):
 \( x1 < 0 \) (hence \( x1 = (x1) = 1x \))
 \( x3 < 0 \) (hence \( x3 = (x3) = 3x \))

For \( 1 \leq x < 3 \):
 \( x1 \geq 0 \) (hence \( x1 = x1 \))
 \( x3 < 0 \) (hence \( x3 = (x3) = 3x \))

For \( x \geq 3 \):
 \( x1 \geq 0 \) (hence \( x1 = x1 \))
 \( x3 \geq 0 \) (hence \( x3 = x3 \))
Case Analysis:
Case I: \( x \geq 3 \)
\(x = \frac{13}{3} \) satifies the case as, clearly, \(x = \frac{13}{3} > 3\). Thus it is a correct solution.
Case II: \( 1 \leq x < 3 \)
(Note: \( x = 7 \) does not satisfy \( 1 \leq x < 3 \), so it is not a solution in this case.)
Case III: \( x < 1 \)
\(x=1\) is also a correct solution as it also satisfies the case.
Final Solutions:
\(\blacksquare\)
How to decide cases?
Deciding cases when dealing with absolute value equations involves identifying the points where the expressions within the absolute value symbols change from positive to negative or vice versa. These points are called "critical points," and they occur where the expression inside the absolute value equals zero. The cases are then built around these critical points to ensure all possible scenarios are considered for the variable \( x \). Here's a detailed stepbystep on how to decide the cases:

Identify Critical Points
For the equation \( 2x1 + x3 = 8 \), the critical points come from the expressions within the absolute values:
 \( x1 = 0 \) leads to \( x = 1 \)
 \( x3 = 0 \) leads to \( x = 3 \)
These points are where the expressions change their nature from positive to negative or vice versa.

Divide the Number Line into Intervals
The critical points divide the number line into different intervals. In this case, we have:
 \( x < 1 \)
 \( 1 \leq x < 3 \)
 \( x \geq 3 \)

Decide the Sign of Each Expression in Each Interval
For each interval, determine the sign of the expressions inside the absolute values:

For \( x < 1 \):
 \( x1 < 0 \) (hence \( x1 = (x1) = 1x \))
 \( x3 < 0 \) (hence \( x3 = (x3) = 3x \))

For \( 1 \leq x < 3 \):
 \( x1 \geq 0 \) (hence \( x1 = x1 \))
 \( x3 < 0 \) (hence \( x3 = (x3) = 3x \))

For \( x \geq 3 \):
 \( x1 \geq 0 \) (hence \( x1 = x1 \))
 \( x3 \geq 0 \) (hence \( x3 = x3 \))


Solve for Each Case
For each interval:
 If \( x < 1 \):
 Solve \( 2(1x) + (3x) = 8 \)
 If \( 1 \leq x < 3 \):
 Solve \( 2(x1) + (3x) = 8 \)
 If \( x \geq 3 \):
 Solve \( 2(x1) + (x3) = 8 \)

Check All Solutions
Ensure that any solutions found fit the correspoding cases they were derived from and confirm whether the solution satisfies the original equation. If an \( x \) value found in an interval does not logically fit within that interval upon reevaluation, discard it.
Solving Inequations
The following inequations help you solve inequations:

For \(a>0\),
\[ x < a \implies a < x < a \]Proof:
To validate this, consider the nature of absolute values and break it down into two cases based on the value of \( x \):
Case 1: \( x \geq 0 \)
 For nonnegative values of \( x \), the absolute value \( x \) is simply \( x \) itself. Hence, the condition \( x < a \) translates to:
\[ x < a \] This means that for \( x \geq 0 \), the values of \( x \) must be less than \( a \). Thus, \(x\in[0,a)\)
Case 2: \( x < 0 \)
 For negative values of \( x \), the absolute value \( x \) is \( x \) (since absolute value measures distance from zero on a number line, which is always positive). Therefore, the condition \( x < a \) translates to:
\[ x < a \] Multiplying both sides of the inequality by 1 (and flipping the inequality because of the multiplication by a negative number) yields:
\[ x > a \] Thus, for \( x < 0 \), the values of \( x \) must be greater than \( a \). Thus, \(x\in(a,0)\)
Combining the Solutions:
 Combining both solutions \(x\in[0,a)\) and \(x\in(a,0)\), we get
\[ a < x < a \]When \(a<0\)
If \( a \) is negative in the inequality \( x < a \), there are no solutions. This is because no real number \( x \) can satisfy the condition where its absolute value is less than a negative number. For example: \( x < 3 \implies\) no real solutions.

For \(a > 0\), the inequality \(x > a\) implies \(x > a\) or \(x < a\).
Proof:
To prove this, consider the definition of absolute value, which states \(x = x\) if \(x \geq 0\) and \(x = x\) if \(x < 0\):
 If \(x \geq 0\), then \(x = x\), and \(x > a\) simplifies to \(x > a\).
 If \(x < 0\), then \(x = x\), and \(x > a\) simplifies to \(x > a\). Multiplying both sides by \(1\) (and reversing the inequality) gives \(x < a\).
Combining both cases, \(x\) must either be greater than \(a\) or less than \(a\), which completes the proof.
Example
Solve the inequality \( 2x  1 > 2 \).
Solution:
Given the inequality, we split it into two cases based on the definition of absolute value:

For the positive part of the inequality:
\[ 2x  1 > 2 \]\[\implies 2x > 2 + 1 \]\[\implies 2x > 3 \]\[\implies x > \frac{3}{2} \] 
For the negative part of the inequality:
\[ 2x  1 < 2 \]\[\implies 2x < 2 + 1 \]\[\implies 2x < 1 \]\[\implies x < \frac{1}{2} \]
Combining the Solutions:
The two parts of the solution state that \( x \) can be either greater than \( \frac{3}{2} \) or less than \( \frac{1}{2} \). Hence, in interval notation:
This provides the complete set of values for \( x \) that satisfy the original inequality.