How this instrument works
|x − a| < b asks a single question with many answers at once: which numbers x sit closer to a than b units away? Absolute value strips away direction and leaves only distance, so |x − a| < b unfolds by definition into the double inequality −b < x − a < b, and adding a across every part slides it into a − b < x < a + b — the interval this sheet reports directly as lower and upper.
Change that strict less-than to an equals sign and the picture changes entirely: |x − a| = b keeps only the two boundary numbers, a − b and a + b, discarding every point strictly between them, while this inequality keeps the entire stretch of numbers in between and discards the two edges themselves. Same bars, same a, same b — an interval's interior versus its endpoints, and swapping one symbol changes the shape of the answer from a filled stretch of line into a pair of isolated points.
The vertical-bar notation is younger than the algebra behind it: Karl Weierstrass wrote |x| for absolute value in an 1841 essay on power series, though the essay itself stayed unpublished until his collected works appeared decades later. One edge case worth knowing here: set b to zero or below and the interval this sheet reports is empty, in the sense that no real x can sit strictly closer to a than zero units — a distance of exactly zero is never less than itself, let alone less than a negative number.
- Enter the center point in the 'a, in |x − a| < b' field — the value the whole interval is built around.
- Enter the distance bound in the 'b, in |x − a| < b' field; it must be positive, since b ≤ 0 leaves no interval at all.
- Read 'Lower bound (a − b)' for where the interval opens.
- Read 'Upper bound (a + b)' for where it closes — every x strictly between the two satisfies the inequality.
Worked example — a safe-passage corridor five kilometres wide
A lighthouse marks the 3-kilometre point along a surveyed coastal line, and the charted safe-passage corridor for small boats extends 5 kilometres to either side of it. A boat's position x, measured in kilometres along that same line, sits inside the corridor exactly when |x − 3| < 5 — within 5 km of marker 3 in either direction.
This instrument takes a = 3 and b = 5 and returns lower = a − b = 3 − 5 = −2 and upper = a + b = 3 + 5 = 8, so the corridor runs from the −2 km marker to the 8 km marker. A boat sitting at the 6 km marker is inside it, since −2 < 6 < 8; one sitting at the 9 km marker is outside, since 9 misses the 8 km limit by a full kilometre.
Questions
What does |x − a| < b actually mean?
It means x sits closer to a than b units, in either direction — a distance statement, not a sign statement. Expanding the bars by definition gives −b < x − a < b, and adding a throughout produces the interval a − b < x < a + b, which is exactly what this sheet's lower and upper bounds report.
How is this different from solving |x − a| = b?
The equation |x − a| = b keeps only the two boundary numbers, a − b and a + b, and rejects everything between them. This inequality keeps everything between those same two numbers and rejects the boundary points themselves — same formula, opposite halves of the same picture.
What changes if the inequality is |x − a| > b instead?
The solution splits from one connected interval into two separate rays: x < a − b or x > a + b. Carrying over the earlier 'and' structure out of habit is the single most common error with this family of problems, since 'less than' joins a range while 'greater than' breaks it into two.
What happens if b is zero or negative?
No x satisfies the inequality. A distance of exactly zero is never strictly less than zero, and a distance can never be less than a negative number in the first place, so the interval this sheet would report collapses to nothing whenever b is zero or smaller.
Why is the interval open rather than closed at its ends?
Because the inequality is strict — a plain < rather than ≤. Swap in ≤ and the two endpoints a − b and a + b join the solution set, closing the interval; with strict <, x may approach either edge as closely as it likes but never actually reach it.
Does the center a have to be a positive number?
No — a can be any real number, positive, negative, or zero; it only marks a location on the number line, not a size. Try a = −4 and b = 2, for instance, and the interval runs from −6 to −2, sitting entirely on the negative side, while b alone still sets its width.