Absolute Value as Distance

Distance Between Real Numbers

The definition of absolute value tells us that \(|x|\) is nonnegative, but it also gives a way to interpret what that number measures: the distance from \(x\) to zero. To measure the distance between two arbitrary real numbers, we use their difference. The order of subtraction may change the sign of the difference, but absolute value removes that sign without changing the distance.

The notation \(d(x,y)\) names a function of two real numbers. The difference \(x-y\) records how far apart the numbers are with a sign: it is positive when \(x\) is to the right of \(y\), and negative when \(x\) is to the left of \(y\). Taking absolute value gives the nonnegative distance regardless of which number comes first.

For example, the distance from \(8\) to \(3\) is \(|8-3|=5\). Reversing the order gives \(|3-8|=|-5|=5\), the same distance. This interpretation matches the familiar number line: the distance is the length of the interval between the two numbers, not the signed difference between them.

Basic Properties of Distance

Distance inherits several important properties from absolute value. The Basic Properties of Absolute Value theorem and the Triangle Inequality established earlier in “Absolute Value” supply the facts we need. In the statements below, these facts are expressed in terms of the distance function \(d\).

Proof. By definition, \(d(x,y)=|x-y|\). The Basic Properties of Absolute Value theorem gives \(|x-y|\geq0\), and says that this absolute value is zero if and only if \(x-y=0\). By the definition of subtraction, \(x-y=0\) if and only if \(x=y\). This proves the first property.

For symmetry, the Basic Properties of Absolute Value theorem gives \(|-u|=|u|\) for every real \(u\). Since \(y-x=-(x-y)\), it follows that

For the final property, write \(x-z=(x-y)+(y-z)\). The Triangle Inequality, applied to the two terms \(x-y\) and \(y-z\), gives

All three properties follow. \(\square\)

The last property is the triangle inequality for distances: going from \(x\) to \(z\) directly can be no longer than going from \(x\) to \(y\) and then from \(y\) to \(z\). It includes cases where \(y\) lies outside the interval with endpoints \(x\) and \(z\); the inequality still holds. These distance properties also explain why the absolute value of a difference, rather than the difference alone, is the appropriate quantity for measuring separation.

Distance from a Fixed Point Describes an Interval

When one number is fixed, a distance bound describes all points within a specified amount of that number. If \(a\) is the fixed point and \(r\geq0\) is the allowed distance, then the points at distance at most \(r\) from \(a\) form the closed interval from \(a-r\) to \(a+r\). This is the absolute-value bound criterion from “Absolute Value,” applied to the difference \(x-a\).

Proof. By definition, \(d(x,a)=|x-a|\). Apply the Absolute-Value Bound Criterion with \(x-a\) in place of \(x\). Since \(r\geq0\), it gives

Adding \(a\) to all parts of the double inequality preserves order, by translation of order. Therefore

Combining the equivalences proves the result. \(\square\)

The condition \(r\geq0\) matters. A distance is nonnegative, so no point can be within a negative distance of \(a\). The interval description also includes its endpoints: \(a-r\) and \(a+r\) are exactly distance \(r\) from \(a\).

There is a parallel description for a strict distance bound. If \(r>0\), then \(d(x,a)<r\) means that \(x\) lies strictly between \(a-r\) and \(a+r\). To see this, put \(u=x-a\). When \(u\geq0\), the definition gives \(|u|=u\), so \(|u|<r\) means \(u<r\); the other endpoint inequality \(-r<u\) follows from \(u\geq0\) and \(r>0\). When \(u<0\), the definition gives \(|u|=-u\), so \(|u|<r\) means \(-u<r\), or \(u>-r\); also \(u<r\), since \(u<0<r\). In both cases, \(-r<u<r\), which is equivalent to \(a-r<x<a+r\) after adding \(a\). Conversely, if \(-r<u<r\), then either \(u\geq0\), giving \(|u|=u<r\), or \(u<0\), giving \(|u|=-u<r\). Thus the strict bound has precisely the stated interpretation.

Points at Exactly a Given Distance

An equation specifying an exact distance gives at most two points: one on either side of the center. For radius zero, the two descriptions coincide at the center. The nonnegative condition on the radius again follows from the fact that distance cannot be negative.

Proof. By definition, the equation is \(|x-a|=r\). If \(x-a\geq0\), then \(|x-a|=x-a\), so the equation implies \(x-a=r\), hence \(x=a+r\). If \(x-a<0\), then \(|x-a|=-(x-a)\), so the equation implies \(x-a=-r\), hence \(x=a-r\). This proves that at least one of the two stated values must hold.

For the converse, if \(x=a+r\), then \(x-a=r\), and \(|x-a|=|r|=r\) because \(r\geq0\). If \(x=a-r\), then \(x-a=-r\), and \(|x-a|=|-r|=r\), using the definition of absolute value and \(r\geq0\). Thus either listed value gives distance \(r\), completing the proof. \(\square\)

Using the Distance Viewpoint Carefully

Writing an absolute value as a distance is useful whenever the expression inside it is a difference. The expression \(|x-a|\) measures how far \(x\) is from the fixed point \(a\); it is not generally the distance of \(x\) from zero. For example, \(|x-4|\) is the distance from \(4\), while \(|x|\) is the distance from \(0\).

A common mistake is to treat \(x-a\) as the distance itself. The difference can be negative: if \(x<a\), then \(x-a<0\). The distance is \(|x-a|\), which is always nonnegative. Another mistake is to report only one solution to an exact-distance equation with positive radius. There are two points, \(a-r\) and \(a+r\), and both should be checked. For an inequality with “at most,” include the endpoints; for a strict inequality, exclude them.

These translations turn absolute-value statements into geometric descriptions on the real line: an upper bound on distance gives an interval, while an exact positive distance gives two points. They preserve the original meaning while making the location of the solutions explicit.