Inequalities path · I03

Before this lesson: Inequality rules and signs, Arithmetic and exact calculation

Your goal: Translate absolute values into distance or justified cases.

Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.

Absolute-value inequalities: the key idea

For a real x, |x| is its distance from zero: x when x≥0 and −x when x<0. Thus |x|≤a with a≥0 means −a≤x≤a. The triangle inequality |x+y|≤|x|+|y| says the direct displacement is no longer than two successive displacements. The reverse triangle inequality is ||x|−|y||≤|x−y|. To solve an absolute-value equation, split at the points where the expressions inside the bars change sign. Never remove the bars using one sign on an entire domain where that sign changes.

A worked example

Solve |x−2|≤3.

The expression is the distance from x to 2. The allowed points lie within three units of 2. Algebraically −3≤x−2≤3, so −1≤x≤5.

Your turn: change one thing

Find the minimum of |x−1|+|x−5| over real x.

Try this on paper before opening the explanation.

Compare your reasoning

The sum is at least |(x−1)−(x−5)|=4. For every x between 1 and 5 the sum equals (x−1)+(5−x)=4, so the minimum is 4, attained on the whole interval.

Pause and check

A trap to avoid: Removing modulus signs without checking signs.

Practise and adjust the level

Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.

Interactive practice loads here. You can also use the complete question set below.

Write a complete argument

Prove the reverse triangle inequality ||x|−|y||≤|x−y|.

Planning hint

List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.

Read the full solution after your attempt

Write x=(x−y)+y and use the triangle inequality to get |x|−|y|≤|x−y|. Interchanging x,y gives |y|−|x|≤|x−y|. These two inequalities together bound the absolute value of |x|−|y| by |x−y|.

My proof notebook

Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.

All nine practice questions

Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.

1. Evaluate |−7|.

Foundation

  1. 0
  2. 14
  3. 7
  4. -7
Hint

Absolute value is a nonnegative distance.

Answer and reasoning

7. The distance from −7 to zero is 7.

2. For real x, when does |x|=−x?

Foundation

  1. For every x
  2. Only when x=1
  3. When x≤0
  4. When x>0
Hint

Check the sign in the definition.

Answer and reasoning

When x≤0. For nonpositive x, −x is nonnegative and equals |x|.

3. Solve |x−3|<2.

Core

  1. 1≤x≤5
  2. −2<x<2
  3. 1<x<5
  4. x<1 or x>5
Hint

Use a double inequality.

Answer and reasoning

1<x<5. −2<x−3<2 gives 1<x<5.

4. Find the minimum of |x−2|+|x−8|.

Core

  1. 4
  2. 10
  3. 6
  4. 0
Hint

Use the distance between the fixed endpoints.

Answer and reasoning

6. The triangle inequality gives at least 8−2=6, attained for 2≤x≤8.

5. When is |a+b|=|a|+|b| for real a,b?

Stretch

  1. When ab≥0
  2. Only when a=b
  3. When ab<0
  4. Never
Hint

The displacements must not oppose each other unless one is zero.

Answer and reasoning

When ab≥0. Equality holds for matching signs, including a zero factor, equivalently ab≥0.

6. Which bound is always valid?

Stretch

  1. |a+b|>|a|+|b|
  2. ||a|−|b||≤|a−b|
  3. ||a|−|b||≥|a−b|
  4. |a−b|=|a|−|b|
Hint

Apply the triangle inequality in both directions.

Answer and reasoning

||a|−|b||≤|a−b|. Bounding |a|−|b| and |b|−|a| by |a−b| gives the reverse triangle inequality.

7. What is |−7|?

Foundation

  1. 7
  2. −7
  3. 0
  4. 49
Hint

Absolute value is distance from zero.

Answer and reasoning

7. Distances are nonnegative.

8. Solve |x−4|≤1.

Core

  1. −1≤x≤1
  2. x=4 only
  3. 3≤x≤5
  4. x≤3 or x≥5
Hint

Convert to a double inequality.

Answer and reasoning

3≤x≤5. −1≤x−4≤1 gives 3≤x≤5.

9. What is the minimum of |x−1|+|x−5|?

Stretch

  1. 2
  2. 6
  3. 4
  4. 0
Hint

Use the distance between endpoints.

Answer and reasoning

4. The sum is at least |5−1|=4, attained throughout 1≤x≤5.

Choose your next step

Continue to Sum of squares. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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