Absolute value: distance, tolerance and cases

A part has a hypothetical nominal dimension of 50 mm. Its permitted deviation is at most 0.2 mm in either direction. Why are 49.9 mm and 50.1 mm equally acceptable? What matters is the size of the deviation, not its sign. Absolute value expresses that size.

Use real numbers and inequalities. This is a mathematical example, not a manufacturing standard; measurement uncertainty is not modeled.

Absolute value removes the direction of distance

For real $u$ , $|u|$ equals $u$ if $u\ge0$ , and $-u$ if $u<0$ . Thus $|3|=3$ , $|-3|=3$ and $|0|=0$ . In the second case $-u$ is positive: if $u=-3$ , then $-u=3$ . Absolute value is never negative, but it need not be positive.

$|x-a|$ is the distance between points $x$ and $a$ on the number line. Let $x$ be the numerical dimension in millimeters. The tolerance is:

|x-50|\le0.2\quad\Longleftrightarrow\quad49.8\le x\le50.2.

Both endpoints are included. A value of 49.7 fails: its distance, 0.3, exceeds 0.2. The ± notation describes the entire interval, not just its two endpoints.

Why a double inequality appears

Being at most $r\ge0$ away from $a$ means crossing neither boundary: $-r\le x-a\le r$ . Add $a$ to all three parts: $a-r\le x\le a+r$ . This uses “and.” Distance and inequalities, OpenStax §2.7.

Permitted dimensions

Distance from nominal

The same condition |x−50|≤0.2: the segment [49.8,50.2] and the part of the graph at or below y=0.2. Uniform scales; axes use different scales.

On the graph, each horizontal coordinate $x$ has vertical coordinate $y=|x-50|$ , the numerical deviation in mm. Left of 50, $y=50-x$ ; right of it, $y=x-50$ . The two lines meet at $(50,0)$ . Points at or below $y=0.2$ have exactly the permitted $x$ values.

An equation gives edges; an inequality gives a region

For $|2x-4|=6$ , the inner expression must equal 6 or −6. From $2x-4=6$ , get $x=5$ ; from $2x-4=-6$ , get $x=-1$ . Substitution confirms both. For $|2x-4|\le6$ , instead $-6\le2x-4\le6$ , hence the whole segment $-1\le x\le5$ . Equations and graphs, OpenStax §3.6.

The opposite requirement $|x-50|>0.2$ means $x<49.8$ or $x>50.2$ : two outer rays. Replacing “or” with “and” is impossible here: no number lies below the lower bound and above the upper bound simultaneously.

The expression inside determines the cases

In $|x-1|=2x+1$ , the right side must be nonnegative, so $x\ge-1/2$ . If $x\ge1$ , open the absolute value as $x-1$ : the equation gives $x=-2$ , contradicting this case. If $x<1$ , use $1-x=2x+1$ to obtain $x=0$ . It belongs to its case and satisfies the original equation, $1=1$ . Only 0 remains.

Check thresholds too: $|u|\le0$ means $u=0$ , whereas $|u|<0$ and $|u|=-2$ have no solutions. Distance suggests the correct cases before any manipulation: inside the bounds needs “and,” outside needs “or,” and negative distance is impossible.