Absolute value

LVL: FREE

MODULE: Number Sense and Basic Intuition

[EXEC: MICRO_CORE]

✖️ 1. Geometric definition of absolute value and the non-negativity property

📏 Geometric Definition & Non-Negativity

  • Absolute value x|x| measures the distance from zero on the number line.
  • Distance is always non-negative, so x0|x| \geq 0 for all real numbers.
  • If xx is positive or zero, then x=x|x| = x.
  • If xx is negative, then x=x|x| = -x (which makes it positive).
  • The absolute value of zero is zero: 0=0|0| = 0.

Example: 5=5|5| = 5 and 5=5|-5| = 5 because both are 5 units from zero.

💡 Think of absolute value as erasing the sign to get pure distance.

[EXEC: DEEP_COMPUTE]

1. Geometric definition of absolute value and the non-negativity property

Geometric Definition of Absolute Value

The absolute value of a real number xx, denoted x|x|, is the distance from xx to zero on the number line. Distance is always measured as a non-negative quantity, so x0|x| \geq 0 for all real numbers xx.

Intuitively, absolute value strips away the sign of a number, leaving only its magnitude.

Core Rules:

  • x=x|x| = x if x0x \geq 0
  • x=x|x| = -x if x<0x < 0
  • x0|x| \geq 0 for all xRx \in \mathbb{R} (non-negativity)
  • x=0|x| = 0 if and only if x=0x = 0

This non-negativity property ensures that absolute value always produces a result on the non-negative portion of the number line.

Example: 5=5|5| = 5 because 5 is 5 units from zero; 5=5|-5| = 5 because -5 is also 5 units from zero.

TASK_1[0 / 3]
LVL_2
EXEC: ALGORITHM

Solve the equation x4=6|x - 4| = 6. Enter the positive solution.

DEEP_COMPUTE
ULTRA
[EXEC: MICRO_CORE]

✖️ 2. Evaluating absolute value of negative numbers and the distance between two numbers

🔢 Evaluating Negatives & Distance Between Numbers

  • For any negative number xx, flip the sign: 7=7|-7| = 7.
  • The distance between two numbers aa and bb is ab|a - b|.
  • Order does not matter: ab=ba|a - b| = |b - a|.
  • Subtracting inside absolute value gives the gap on the number line.
  • Always compute the inside first, then take absolute value.

Example: Distance between 3 and 8 is 38=5=5|3 - 8| = |-5| = 5 units.

💡 ab|a - b| is the ruler measurement between aa and bb.

[EXEC: DEEP_COMPUTE]

2. Evaluating absolute value of negative numbers and the distance between two numbers

Distance Between Two Numbers

For any negative number x<0x < 0, the absolute value x|x| equals x-x, which is positive. The expression ab|a - b| represents the distance between aa and bb on the number line, regardless of their order.

This distance interpretation is symmetric: ab=ba|a - b| = |b - a| because distance does not depend on direction.

Core Rules:

  • If x<0x < 0, then x=x>0|x| = -x > 0
  • ab|a - b| measures the gap between aa and bb
  • ab=ba|a - b| = |b - a| (symmetry)
  • Distance is always non-negative

This formulation unifies the concept of separation between any two points.

Example: 7=7|-7| = 7; the distance between 3 and 8 is 38=5=5|3 - 8| = |-5| = 5, which equals 83=5|8 - 3| = 5.

TASK_1[0 / 3]
LVL_3
EXEC: ALGORITHM

Solve the equation: x5=8|x - 5| = 8. Find the smallest solution.

DEEP_COMPUTE
ULTRA
[EXEC: MICRO_CORE]

✖️ 3. Understanding the piecewise nature of the absolute value function

🔀 Piecewise Nature of Absolute Value

  • The function f(x)=xf(x) = |x| has two different rules depending on the sign of xx.
  • For x0x \geq 0: x=x|x| = x (identity line with slope 1).
  • For x<0x < 0: x=x|x| = -x (reflected line with slope -1).
  • The graph forms a V-shape with the vertex at the origin.
  • The function is not differentiable at x=0x = 0 because of the sharp corner.

Example: 3=3|3| = 3 uses the first rule, 3=3|-3| = 3 uses the second rule.

💡 Absolute value is a two-faced function that switches behavior at zero.

[EXEC: DEEP_COMPUTE]

3. Understanding the piecewise nature of the absolute value function

Piecewise Nature of Absolute Value

The absolute value function f(x)=xf(x) = |x| is defined piecewise because its formula changes depending on whether the input is non-negative or negative. This creates a V-shaped graph with a corner at the origin.

The function is continuous everywhere but not differentiable at x=0x = 0 due to the sharp turn.

Core Rules:

  • x=x|x| = x when x0x \geq 0 (right branch, slope = 1)
  • x=x|x| = -x when x<0x < 0 (left branch, slope = -1)
  • The graph has a vertex at (0,0)(0, 0)
  • Not differentiable at x=0x = 0 (sharp corner)

This piecewise structure is fundamental to solving equations and inequalities involving absolute value.

Example: For x=3x = -3, use the second piece: 3=(3)=3|-3| = -(-3) = 3. For x=4x = 4, use the first piece: 4=4|4| = 4.

TASK_1[0 / 3]
LVL_2
RSN: CONSTRAINTS

How many solutions does the equation 2x7=4|2x - 7| = -4 have?

DEEP_COMPUTE
ULTRA
[EXEC: MICRO_CORE]

✖️ 4. Solving basic absolute value equations and simple inequalities

⚖️ Solving Equations & Inequalities

  • For x=k|x| = k where k>0k > 0, there are two solutions: x=kx = k or x=kx = -k.
  • If k=0k = 0, then x=0x = 0 is the only solution.
  • If k<0k < 0, there is no solution (absolute value cannot be negative).
  • For x<k|x| < k, the solution is k<x<k-k < x < k (numbers within kk units of zero).
  • For x>k|x| > k, the solution is x<kx < -k or x>kx > k (numbers farther than kk units from zero).

Example: x=4|x| = 4 gives x=4x = 4 or x=4x = -4.

💡 Absolute value equations split into two cases, one positive and one negative.

[EXEC: DEEP_COMPUTE]

4. Solving basic absolute value equations and simple inequalities

Solving Absolute Value Equations and Inequalities

An equation x=k|x| = k where k>0k > 0 has exactly two solutions: x=kx = k or x=kx = -k, because both values are distance kk from zero. If k<0k < 0, there is no solution since absolute value cannot be negative.

For inequalities, x<k|x| < k means k<x<k-k < x < k (values within distance kk of zero), while x>k|x| > k means x<kx < -k or x>kx > k (values beyond distance kk).

Core Rules:

  • x=k|x| = k (where k>0k > 0) gives x=kx = k or x=kx = -k
  • x<k|x| < k gives k<x<k-k < x < k
  • x>k|x| > k gives x<kx < -k or x>kx > k
  • If k0k \leq 0, then x=k|x| = k has no solution

Example: x=7|x| = 7 yields x=7x = 7 or x=7x = -7. The inequality x<3|x| < 3 gives 3<x<3-3 < x < 3.

TASK_1[0 / 3]
LVL_2
EXEC: ALGORITHMRSN: DEBUG

A student solves the equation x5=2x1|x - 5| = 2x - 1 and finds two candidate solutions: x=4x = -4 and x=2x = 2.

Which of these is an extraneous solution?

DEEP_COMPUTE
ULTRA
[EXEC: MICRO_CORE]

✖️ 5. Applications: Calculating total magnitude of forces or net change volatility in economics

🌍 Applications in Forces & Economics

  • In physics, F|F| represents the magnitude of a force regardless of direction.
  • Net change volatility uses change|\text{change}| to measure total movement ignoring sign.
  • Stock price fluctuation over time is measured as PfinalPinitial|P_{\text{final}} - P_{\text{initial}}|.
  • Temperature differences use absolute value: T1T2|T_1 - T_2| gives the gap in degrees.
  • Absolute value captures size without caring about positive or negative.

Example: If a stock moves from 50 dollars to 42 dollars, volatility is 4250=8|42 - 50| = 8 dollars.

💡 Absolute value measures how much, not which way.

[EXEC: DEEP_COMPUTE]

5. Applications: Calculating total magnitude of forces or net change volatility in economics

Applications of Absolute Value

Absolute value quantifies magnitude without regard to direction, making it essential in physics for computing total force magnitudes and in economics for measuring volatility. In finance, ΔP|\Delta P| captures price change size regardless of whether the market rose or fell.

This abstraction allows analysts to assess variability, risk, or total effect independent of sign.

Core Rules:

  • Use F|F| to find force magnitude (ignoring direction)
  • Sum Δxi\sum |\Delta x_i| to compute total volatility or cumulative change
  • Absolute deviations xμ|x - \mu| measure spread from a mean
  • Always non-negative, suitable for distance-based metrics

Absolute value transforms signed quantities into comparable magnitudes.

Example: If stock prices change by -5 dollars, +3 dollars, -2 dollars over three days, total volatility is 5+3+2=5+3+2=10|-5| + |3| + |-2| = 5 + 3 + 2 = 10 dollars.

TASK_1[0 / 3]
LVL_2
EXEC: ALGORITHM

A machine part has a target length of 120120 mm and a tolerance of 0.50.5 mm. What is the maximum acceptable length for this part?

DEEP_COMPUTE
ULTRA

AWAITING_CONFIRMATION

CONFIRM KNOWLEDGE ACQUISITION TO UPDATE SYSTEM ANALYTICS.