1
The second case is a negation, not a negative
−x means the additive inverse of x. If x = −7, then −x = 7. Reading −x as "a negative number" makes the definition look as though it sometimes returns a negative absolute value, which it never does.
Lesson · Foundations of Algebra · The number system
Absolute Value
The absolute value of a number is its distance from zero, which is never negative. Getting that from a formula takes two cases, and the second case — where the answer is −x — is the one that has to be read carefully.
|x| = x when x ≥ 0
−x when x < 0
|x − y| = the distance from x to y
|−7| = 7, |0| = 0, −|−7| = −7
The definition, and what it measures
Core ideas
What this lesson establishes
One symbol, one definition, and a single reading error that accounts for most of the trouble it causes.
2
Distance is the meaning; the cases are the machinery
|x| is how far x is from 0, and |x − y| is how far x is from y. Each related use — the equations and inequalities in Linear Equations and Inequalities, the error bounds in any applied subject — reads as a distance.
3
An absolute value equation has two solutions, one, or none
|x| = 3 has two, |x| = 0 has one, |x| = −1 has none. Which case you are in is decided by the number on the right before any algebra happens, and "no solution" is a complete answer.
The material
Distance, written as a formula
The two-case definition, what it guarantees, and how an equation in one absolute value is split.
How far is −7 from 0? Seven. Distance does not carry a sign. So there has to be an operation that removes the sign of a number without pretending the number was positive all along, and it needs a definition that works for a letter whose sign you do not know.
Definition · Absolute value
For a real number x, |x| = x when x ≥ 0, and |x| = −x when x < 0.
The boundary case x = 0 is assigned to the first line by choice; both lines give 0 there, so the choice is harmless.
Run it three times. |5|: 5 ≥ 0, so the first line applies and |5| = 5. |−5|: −5 < 0, so the second applies and |−5| = −(−5) = 5. |0|: the first line, giving 0.
The middle one is the whole lesson. The second line does not say the answer is negative; it says the answer is the inverse of a negative, which is positive. When x is a letter you cannot see the sign of, that is exactly why both cases have to be carried.
Theorem · What absolute value guarantees
|x| ≥ 0 for every real x, and |x| = 0 only when x = 0.
If x ≥ 0 then |x| = x ≥ 0. If x < 0 then −x > 0, so |x| = −x > 0. Every real falls into one of those two cases, so |x| ≥ 0 always.
For the second part: the first case gives 0 only at x = 0, and the second case gives a value strictly greater than 0, so no other x produces 0.
□
Distance between two numbers
Definition · Distance
The distance between reals a and b is |a − b|.
|a − b| = |b − a|, because b − a = −(a − b) and a number and its inverse have the same absolute value. So it does not matter which end you subtract from.
|3 − 10| = |−7| = 7
|10 − 3| = |7| = 7
|−2 − 5| = |−7| = 7 −2 and 5 are 7 apart
Solving an equation with one absolute value
Read |x − a| = c as "x is at distance c from a". On a line there are two such points when c is positive, one when c is zero, and none when c is negative.
|x| = 3 → x = 3 or x = −3 two solutions
|x| = 0 → x = 0 one solution
|x| = −1 → no solution |x| is never negative
The two solutions are not optional and neither is the check. The formal route splits the equation into the two cases of the definition, and that split is exhaustive — every real x satisfies either x − a ≥ 0 or x − a < 0 — which is what tells you there are no further solutions hiding.
Absolute value passes through products and quotients but not through sums: |ab| = |a| · |b| is true for all reals, and |a + b| = |a| + |b| is not. At a = 3 and b = −5, the left side is |−2| = 2 and the right side is 8. What is always true is |a + b| ≤ |a| + |b|.
Example · Reading the centre off the bars
|x − 4| = 3 asks for the points 3 away from 4, namely 1 and 7. |x + 4| = 3 is |x − (−4)| = 3, so the centre is −4 and the points are −7 and −1. The sign inside flips the centre, not the distance.
Interactive
Absolute value as distance
Both cases of the definition, and the one that applies
Method
Working with an absolute value
Decide the sign of what is inside the bars. Everything else follows from that.
1
Look at the sign of the contents, not of the whole expression
The definition branches on whether the quantity inside the bars is negative. |3 − 10| is decided by 3 − 10 being −7, so the second line applies and the result is 7.
2
Apply the case that fires
If the contents are ≥ 0, drop the bars. If they are < 0, drop the bars and negate the whole contents — every term of them, which is the distributive law from “Properties of the Real Numbers”.
3
For an equation, check the right-hand side first, then split
If it is negative, there are no solutions and you are finished. If it is zero, there is one. If it is positive, write the two equations: contents equal to it, and contents equal to its negative.
4
Substitute every candidate back
Put each answer into the original equation, bars and all. The split itself guarantees you have found all of them; the substitution guarantees each one is real rather than an arithmetic slip.
Bars and distance
Score 0 / 0Choose the answer you can justify.
Solving |x − 4| = 3, two ways
The same equation read as a distance and read through the definition. The second way is the one that generalises.
by distance
|x − 4| is the distance from x to 4
3 to the left: 4 − 3 = 1
3 to the right: 4 + 3 = 7
by the definition
case x − 4 ≥ 0: x − 4 = 3 → x = 7
case x − 4 < 0: −(x − 4) = 3
x − 4 = −3 → x = 1
check
x = 7: |7 − 4| = |3| = 3
x = 1: |1 − 4| = |−3| = 3
solutions x = 1 and x = 7
The two checks confirm that 1 and 7 satisfy the equation. They do not, on their own, say there is nothing else. That comes from the case split: every real number makes x − 4 either non-negative or negative, so the two cases cover all of them and each yielded one value.
Compare with |x − 4| = −3. The distance reading kills it immediately — no point is at distance −3 from anything — and the definition agrees, since |x − 4| ≥ 0 for every x. Writing "no solution" is the finished answer.
Fade the case table now. With a = −6 and b = 1, decide which line of the definition fires and find the distance before opening the matching lab preset: |−6| = −(−6) = 6 and |−6 − 1| = 7. A negative first answer reads −x as a negative number; a distance of 5 subtracts the magnitudes instead of measuring between the two points.
Common mistakes
1
Reading the second line of the definition as producing a negative|x| = −x when x < 0 gives a positive answer, because −x is the inverse of a negative number. At x = −7 it reads |−7| = −(−7) = 7.
2
Keeping only the positive solution|x| = 3 has solutions 3 and −3. Dropping the second is the same omission as writing √ where ± is needed in Quadratics and Complex Numbers, and it loses half the answer every time.
3
Splitting the bars over a sum|a + b| is not |a| + |b|. At a = 3 and b = −5 those are 2 and 8. Products are different: |ab| = |a| · |b| really does hold, which is why the false version looks plausible.
Check your understanding
✓
Finish when both cases of the definition are automatic and "no solution" is an answer you will write.
Evaluate |−8|, −|−8| and |−3 − 5| from the definition; then mark the points described by |x + 3| = 4 using distance alone and reject |2x − 1| = −4 on sight. Linear Equations and Inequalities supplies the general equation method; it assumes the meaning of the bars is automatic.
Note: The two-case shape returns in Lines, Functions and Graphs as a piecewise function — |x| is the standard first example of one — and again in “Roots and Radicals” here, where √(x²) is |x| rather than x. The reason is the same both times: a formula that has to work for inputs of either sign needs a branch.