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INAIO Guide Stage 1 1Numerical Foundations

1.3 Absolute values and inequalities

Measure distance without a sign, compare numbers, and solve for every value that satisfies a limit.

Absolute values and inequalities turn descriptions such as “within two units” or “at least five” into mathematics. Absolute value measures the size of a difference, regardless of its sign. An inequality describes a range of acceptable values instead of one exact answer. You will use both when comparing measurements, checking errors, or deciding whether a constraint is met.

This is foundational material. If you can compare negative numbers and solve a simple equation, you can start here. For a first pass, use Khan Academy's lessons on negative numbers and absolute value. The OpenStax section on inequalities adds longer examples and practice.

A basic idea: distance on a number line

The absolute value ∣x∣|x| is the distance from xx to zero. So ∣−3∣=3|-3|=3 and ∣3∣=3|3|=3: the two numbers lie on opposite sides of zero but equally far away. More generally, ∣x−a∣|x-a| is the distance from xx to aa.

Two number lines. The first shows minus 3 and 3 each three units from zero. The second shows the closed interval from minus 1 to 5, centred at 2: every point in the interval is within three units of 2.Two number lines. The first shows minus 3 and 3 each three units from zero. The second shows the closed interval from minus 1 to 5, centred at 2: every point in the interval is within three units of 2.
Absolute value turns a signed difference into distance. The lower line shows every value within 3 units of 2.

The order on the line also tells you how to compare signed numbers: a number farther right is larger. Thus −2>−5-2>-5, even though 55 has the greater absolute value.

A quick tour of rules

Absolute values

The definition is

∣x∣={xif x≥0,−xif x<0.|x|=\begin{cases} x & \text{if }x\ge 0,\\ -x & \text{if }x<0. \end{cases}

Absolute value is never negative, and changing a sign does not change a distance: ∣x∣≥0|x|\ge0 and ∣−x∣=∣x∣|-x|=|x|. The distance between two numbers is ∣a−b∣=∣b−a∣|a-b|=|b-a|. Combining two changes cannot create a greater distance than adding their separate distances, so ∣a+b∣≤∣a∣+∣b∣|a+b|\le|a|+|b|.

For r≥0r\ge0, two useful translations are

∣x−a∣≤r⟺a−r≤x≤a+r,|x-a|\le r\quad\Longleftrightarrow\quad a-r\le x\le a+r,

∣x−a∣>r⟺x<a−r or x>a+r.|x-a|>r\quad\Longleftrightarrow\quad x<a-r\ \text{or}\ x>a+r.

“Within” gives one interval around aa; “farther than” gives two regions outside it. A strict sign, << or >>, excludes the boundary. A sign with equality, ≤\le or ≥\ge, includes it.

Inequalities and numerical comparisons

You can add or subtract the same number on both sides without changing the comparison. Multiplication depends on the sign:

Starting comparison Operation Result
a<ba<b Add cc a+c<b+ca+c<b+c
a<ba<b Multiply by c>0c>0 ac<bcac<bc
a<ba<b Multiply by c<0c<0 ac>bcac>bc

Multiplying by a negative reverses the order: 2<52<5, but −2>−5-2>-5. The same rule applies when you divide by a negative. If the sign of a multiplier is unknown, find it or split into cases before multiplying.

Application 1: a measurement with tolerance

A sensor reports 47.647.6 units. Its true reading xx can be at most 1.21.2 units away from the reported value. What readings are possible?

  1. Write the distance. The difference from 47.647.6 is ∣x−47.6∣|x-47.6|. “At most” includes the limit, so ∣x−47.6∣≤1.2|x-47.6|\le1.2.
  2. Turn distance into a range. Subtract and add the tolerance: 47.6−1.2≤x≤47.6+1.247.6-1.2\le x\le47.6+1.2.
  3. Simplify and check. The range is 46.4≤x≤48.846.4\le x\le48.8. Both endpoints are exactly 1.21.2 units away; 4949 is 1.41.4 away and does not qualify.

The sign of an error tells you whether the sensor read high or low. Its absolute value tells you how large the error was.

Application 2: an energy limit

A robot begins with 1212 energy units. Each move uses 33 units, and it must keep at least 22 units for its final action. What is the greatest whole number of moves it can make?

  1. Write the remaining energy. After mm moves it has 12−3m12-3m units, so the requirement is 12−3m≥212-3m\ge2.
  2. Solve the inequality. Subtract 1212 to get −3m≥−10-3m\ge-10. Divide by −3-3 and reverse the sign: m≤10/3m\le10/3.
  3. Use the whole-number constraint. The robot can make at most 33 moves. After 33 moves it has 33 units left; after 44 it has 00, which fails the requirement.

This question needs both the inequality rule and a comparison between a fractional bound and possible whole-number moves.

Resources

SourceTitleWhy read it
Khan AcademyNegative numbers and absolute valueStart here for number lines and absolute value practice.
OpenStaxLinear and absolute value inequalitiesContinue with inequality rules, intervals, and worked problems.

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