Modulus • Topic 1 of 3

Absolute Value

The absolute value |x| is the distance of x from 0 on the number line, so it is never negative: |x| = x when x ≥ 0 and |x| = −x when x < 0. That "−x" trips up beginners — if x = −4, then −x = 4, which is exactly |−4|. Two equivalent forms are worth memorising: |x| = √(x²), and |x|² = x². The distance reading is the real CAT power-tool: |x − a| is simply "how far x is from a", so |x − 3| < 2 means "x is within 2 of 3", i.e. 1 < x < 5 — no algebra needed. Key properties: |ab| = |a||b|, |a/b| = |a|/|b|, and the triangle inequality |a + b| ≤ |a| + |b| (equality only when a and b have the same sign). Always remember |x| ≥ 0, so any equation forcing a modulus to equal a negative number has no solution.

✅ Solved examples

1. Evaluate |−7| + |3| − |−2|.
|−7| = 7, |3| = 3, |−2| = 2. So 7 + 3 − 2 = 8.
2. If x = −5, find the value of |x| − x.
|x| = |−5| = 5 and x = −5, so |x| − x = 5 − (−5) = 10.
3. Simplify |x − 4| for x = 1 and for x = 9.
At x = 1: |1 − 4| = |−3| = 3. At x = 9: |9 − 4| = |5| = 5. (Each is the distance from 4.)
4. For which real x is |x| = −x true?
|x| = −x exactly when −x ≥ 0, i.e. x ≤ 0. So it holds for all non-positive x.

✏️ Practice — try these, take hints as needed

1. Evaluate |−12| − |−5| + |0|.
Strip each modulus to its non-negative value.
|−12| = 12, |−5| = 5, |0| = 0.
12 − 5 + 0.
7
2. If a = −3, find |a| + a².
|a| = 3.
a² = 9.
Add them.
12
3. Read |x − 7| < 3 as a distance and give the range of x.
|x − 7| is the distance from 7.
Within 3 of 7.
7 − 3 < x < 7 + 3.
4 < x < 10
4. For x < 0, simplify |x| / x.
For x < 0, |x| = −x.
(−x)/x.
Cancel x.
−1
5. Find all x with |x| = 6.
Distance 6 from 0.
Two points.
+6 and −6.
x = 6 or x = −6

📝 Topic test — 8 questions

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