Euclidean Distance in Rn

From Coordinates to Distance

A point in \(\mathbb{R}^n\) is an ordered list of \(n\) real coordinates. To measure the separation of two such points, subtract their coordinates and combine the resulting differences into one nonnegative number. The familiar formula in the plane extends directly to every finite dimension. Its triangle inequality is not a separate geometric assumption: it follows algebraically from the Cauchy–Schwarz inequality.

Throughout, \(n\in\mathbb{N}\), and \(\mathbb{R}^n\) is equipped with the Euclidean distance defined below. For \(x=(x_1,\dots,x_n)\) and \(y=(y_1,\dots,y_n)\), their dot product is the real number \(\langle x,y\rangle=\sum_{i=1}^n x_i y_i\). The usual norm is defined from this dot product.

Each coordinate difference is squared, so it contributes a nonnegative amount; taking the nonnegative square root then gives a nonnegative distance. In \(\mathbb{R}^2\), this is the familiar square-root-of-the-sum-of-squares formula. The same rule applies without change in \(\mathbb{R}^3\) and in every higher finite dimension. To show that this formula is a metric, the main step is proving its triangle inequality.

The Cauchy–Schwarz Inequality

Proof. First suppose \(y=0\). Then \(\langle x,y\rangle=0\) and \(\|y\|=0\), so both sides of the claimed inequality are zero. Now suppose \(y\ne 0\), so \(\|y\|^2>0\). Set \(t=\langle x,y\rangle/\|y\|^2\). A squared norm is nonnegative, and expanding the following one using the dot product gives

Multiplication by the positive number \(\|y\|^2\) yields \(\langle x,y\rangle^2\leq\|x\|^2\|y\|^2\). Both sides are nonnegative, so taking square roots gives \(|\langle x,y\rangle|\leq\|x\|\|y\|\). This also proves the stated result in the case \(y\ne0\); the case \(y=0\) was checked separately. \(\square\)

The absolute value matters: the dot product can be negative, while the norms cannot. In particular, the lemma implies the one-sided bound \(\langle x,y\rangle\leq\|x\|\|y\|\), which is the form used to prove the triangle inequality.

The Triangle Inequality and the Metric Axioms

Proof. A metric must be nonnegative, vanish exactly when its two inputs agree, be symmetric, and satisfy the triangle inequality. By definition, \(d_2(x,y)\) is a square root of a sum of squares, so \(d_2(x,y)\geq0\). If \(d_2(x,y)=0\), then \(\sum_{i=1}^n(x_i-y_i)^2=0\). Every summand is nonnegative, so each is zero. Thus \(x_i=y_i\) for every coordinate \(i\), and \(x=y\). Conversely, if \(x=y\), every difference is zero and \(d_2(x,y)=0\). Also, \((x_i-y_i)^2=(y_i-x_i)^2\) for each \(i\), so \(d_2(x,y)=d_2(y,x)\).

It remains to prove the triangle inequality. The identity \(x-z=(x-y)+(y-z)\) gives, by expansion of the dot product and then Cauchy–Schwarz,

Both distances, and their sum, are nonnegative. Therefore, taking nonnegative square roots preserves the inequality and gives \(d_2(x,z)\leq d_2(x,y)+d_2(y,z)\). This verifies every metric axiom. \(\square\)

The proof uses no assumption that the points are distinct. If some or all of \(x,y,z\) coincide, the same identities and inequalities still apply; for example, if \(x=y\), then \(d_2(x,z)=d_2(y,z)\) and the right side is \(0+d_2(y,z)\). Thus the degenerate cases are included.

A Common Calculation Error

Common pitfall. For the points \((1,2)\) and \((4,6)\), one might calculate the coordinate differences as \(3\) and \(4\), then incorrectly add their magnitudes and report the Euclidean distance as \(3+4=7\). That calculation does not follow the Euclidean formula. The correct computation is

The incorrect value \(7\) comes from replacing the square root of a sum by a sum of square roots: \(\sqrt{9+16}=5\), whereas \(\sqrt9+\sqrt{16}=3+4=7\). These expressions are not equal. Checking the correct answer by squaring gives \(5^2=25=9+16\), as required by the defining formula. The order of operations is essential: square the coordinate differences, add the squares, and only then take the nonnegative square root.

Why the Formula Works in Every Finite Dimension

The coordinate formula is useful because it turns distance into a finite calculation, while the proof of the metric property does not depend on any special feature of the plane. In the triangle-inequality proof, the only dimension-dependent operation was the dot product \(\sum_{i=1}^n x_i y_i\). Cauchy–Schwarz applies to that sum for every positive integer \(n\), so the same argument covers all finite-dimensional Euclidean spaces.

It is also important to distinguish the norm of a vector from the distance between two points. The norm \(\|x\|\) measures the length associated to a vector \(x\); the distance \(d_2(x,y)\) applies that norm to the difference \(x-y\). The subtraction makes the distance depend on the separation of the points, not on their individual coordinates. For example, translating both points by the same vector does not change their difference: \((x+a)-(y+a)=x-y\). This observation follows directly from the formula and requires no extra metric assumption.

The metric axioms establish that this formula is a valid distance. In particular, the triangle inequality says that the direct distance between two points cannot exceed the sum of the distances along a route through a third point. The examples show how to verify that claim numerically; the Cauchy–Schwarz argument explains why it holds for every triple of points in every \(\mathbb{R}^n\).