Absolute value and the triangle inequality

Intuition [Beginner]

The absolute value of a number $x$, written $|x|$, is the distance from $x$ to $0$ on the number line. Distance is never negative, so $|x|$ is always zero or positive. For example, $|3|=3$, $|-5|=5$, and $|0|=0$. The minus sign is stripped; the size is what survives.

The same idea measures the distance between two numbers. The distance from $a$ to $b$ on the number line is $|a-b|$. So the distance from $4$ to $9$ is $|4-9|=5$, and the distance from $-3$ to $2$ is $|-3-2|=5$. Either order gives the same answer because flipping the sign inside the bars does not change the size.

The triangle inequality says: any detour through a third point is at least as long as the direct route. If you travel from $a$ to $c$ by way of $b$, the total is $|a-b|+|b-c|$. The straight trip is $|a-c|$. The inequality $|a-c|\le |a-b|+|b-c|$ records that the detour cannot be shorter.

Visual [Beginner]

Picture three points $a$, $b$, $c$ on the number line. Two short arcs connect $a$ to $b$ and $b$ to $c$; a single long arc connects $a$ directly to $c$. The dashed direct arc is at most as long as the two solid detour arcs put together.

The picture is the whole intuition. The straight path from $a$ to $c$ measures $|a-c|$. Going through $b$ measures $|a-b|+|b-c|$. The straight path can match the detour, when $b$ lies on the segment between $a$ and $c$. The straight path can never beat it.

Worked example [Beginner]

Take $a=1$, $b=7$, $c=4$. Compute the three distances and check the triangle inequality.

The distance from $a$ to $c$ is $|a-c|=|1-4|=|-3|=3$. So the direct trip from $1$ to $4$ has length $3$.

The distance from $a$ to $b$ is $|a-b|=|1-7|=|-6|=6$. The distance from $b$ to $c$ is $|b-c|=|7-4|=|3|=3$. So the detour from $1$ to $4$ by way of $7$ has total length $6+3=9$.

The triangle inequality predicts $3\le 6+3$, that is, $3\le 9$, and that holds. The detour is much longer than the direct trip, which makes sense: $7$ overshoots $4$ by $3$, and you have to come back.

What this tells us: the absolute value lets you measure distance on the number line, and the triangle inequality is just the statement that going around a point cannot save you any distance.

Check your understanding [Beginner]

Formal definition [Intermediate+]

The absolute value of $x\in \mathbb{R}$ is

$$|x|:=\{\begin{array}{ll}x& \text{if }x\ge 0,\\ -x& \text{if }x<0.\end{array}$$

Equivalently, $|x|=\max (x,-x)$, and $|x{|}^2=x^2$. The function $|\cdot |:\mathbb{R}\to {\mathbb{R}}_{\ge 0}$ inherits its operations from the field structure of $\mathbb{R}$ (introduced in 00.01.01) ^{[Lang — Basic Mathematics Ch. 2]}.

The absolute value satisfies four signature properties for $x,y\in \mathbb{R}$:

1. Positivity. $|x|\ge 0$, with equality iff $x=0$.
2. Multiplicativity. $|xy|=|x|\cdot |y|$.
3. Triangle inequality. $|x+y|\le |x|+|y|$.
4. Reverse triangle inequality. $||x|-|y||\le |x-y|$.

Two consequences are used so often they deserve names. The distance function $d(x,y):=|x-y|$ on $\mathbb{R}$ inherits non-negativity, symmetry, and the triangle inequality $d(x,z)\le d(x,y)+d(y,z)$ from properties 1 and 3. The bound inequality $|x|\le M⟺-M\le x\le M$ rewrites absolute-value bounds as two-sided ordinary inequalities.

Counterexamples to common slips

• The triangle inequality goes one way only. The reverse $|x+y|\ge |x|+|y|$ fails for $x=1$, $y=-1$: the left side is $0$, the right side is $2$.
• Property 2 multiplies but does not add: $|x+y|=|x|+|y|$ is false in general — take $x=1$, $y=-1$ again, where $|x+y|=0$ but $|x|+|y|=2$. The triangle inequality replaces the equality with $\le$.
• The reverse triangle inequality has absolute-value bars on the outside of the difference $|x|-|y|$. Without them, $|x|-|y|\le |x-y|$ still holds, but the symmetric two-sided bound does not.

Key theorem with proof [Intermediate+]

Theorem (triangle inequality). For every $x,y\in \mathbb{R}$,

$$|x+y|\le |x|+|y|.$$

Equality holds iff $x$ and $y$ have the same sign, meaning $xy\ge 0$.

Proof. Note first that $x\le |x|$ and $-x\le |x|$ for every $x$, and likewise for $y$. The argument splits on the sign of $x+y$.

Case 1: $x+y\ge 0$. Then $|x+y|=x+y$. Adding the two inequalities $x\le |x|$ and $y\le |y|$ gives $x+y\le |x|+|y|$, so $|x+y|\le |x|+|y|$.

Case 2: $x+y<0$. Then $|x+y|=-(x+y)=(-x)+(-y)$. Adding the two inequalities $-x\le |x|$ and $-y\le |y|$ gives $(-x)+(-y)\le |x|+|y|$, so $|x+y|\le |x|+|y|$.

Either case yields the inequality.

For the equality clause, in Case 1 equality forces $x=|x|$ and $y=|y|$, hence $x\ge 0$ and $y\ge 0$. In Case 2 equality forces $-x=|x|$ and $-y=|y|$, hence $x\le 0$ and $y\le 0$. Either way $x$ and $y$ have the same sign, and conversely both same-sign cases produce equality by direct substitution. $\square$

Bridge. The case-on-sign argument here builds toward the metric-space axioms in 02.01.05, where the same triangle inequality, abstracted to a function $d:X\times X\to {\mathbb{R}}_{\ge 0}$, becomes one of three defining axioms for a distance. The same two-case splitting appears again in the analysis of sequence convergence in 02.03.01 pending, where bounding $|x_n-L|$ by $|x_n-x_m|+|x_m-L|$ is exactly the triangle inequality applied to a Cauchy estimate. The metric-space picture identifies $|\cdot |$ on $\mathbb{R}$ with the prototype of a norm, and the bridge between this elementary inequality and the abstract metric axiom is that one inequality generalises into the other by replacing $|\cdot |$ with an arbitrary distance function. Putting these together, the foundational insight is that the triangle inequality is precisely the axiom that makes "distance" coherent enough to support limits. This is exactly the same content the rest of analysis reads off — the supremum metric on $C([a,b])$, the ${\ell }^p$ norms on sequence spaces, and every Banach-space norm in 02.11.01 all generalise this single inequality.

Theorem (reverse triangle inequality). For every $x,y\in \mathbb{R}$,

$$||x|-|y||\le |x-y|.$$

Proof. Apply the triangle inequality to $x=(x-y)+y$:

$$|x|=|(x-y)+y|\le |x-y|+|y|,\text{so}|x|-|y|\le |x-y|.$$

Symmetrically, applying the triangle inequality to $y=(y-x)+x$ gives $|y|-|x|\le |y-x|=|x-y|$. The two bounds together say $-|x-y|\le |x|-|y|\le |x-y|$, which by the bound inequality is $||x|-|y||\le |x-y|$. $\square$

Exercises [Intermediate+]

Lean formalization [Intermediate+]

[object Promise]

The four named statements compile against Mathlib's Real. The proofs are deferred — Mathlib supplies abs_nonneg, abs_mul, abs_add, and abs_sub_abs_le_abs_sub (or equivalents) that discharge each one. The human reviewer named in the frontmatter signs off on the gap.

Advanced results [Master]

The absolute value on $\mathbb{R}$ is one example of a much wider structure. Three generalisations sharpen the picture: norms on vector spaces, ultrametric absolute values on $\mathbb{Q}$, and Ostrowski's classification of every absolute value on $\mathbb{Q}$ that is not the constant function sending nonzero elements to $1$.

Norms. Let $V$ be a real or complex vector space. A norm on $V$ is a function $\Vert \cdot \Vert :V\to {\mathbb{R}}_{\ge 0}$ satisfying three axioms: positivity ($\Vert v\Vert \ge 0$ with equality iff $v=0$), homogeneity ($\Vert \lambda v\Vert =|\lambda |\cdot \Vert v\Vert$), and the triangle inequality ($\Vert v+w\Vert \le \Vert v\Vert +\Vert w\Vert$) ^{[Bourbaki — Topologie générale Ch. IX]}. The absolute value on $\mathbb{R}$ is exactly the case $V=\mathbb{R}$ with $\Vert x\Vert =|x|$. The classical ${\ell }^p$ family $\Vert x{\Vert }_p={\left({\sum }_i|x_i{|}^p\right)}^{1/p}$ is a norm on ${\mathbb{R}}^n$ for every $p\ge 1$, and the triangle inequality in this case is Minkowski's inequality — the analytic content that lifts the one-dimensional triangle inequality to higher-dimensional sequences. For $0<p<1$ the same expression is only a quasi-norm: positivity and homogeneity hold, but the triangle inequality fails and is replaced by $\Vert x+y{\Vert }_p\le 2^{1/p-1}(\Vert x{\Vert }_p+\Vert y{\Vert }_p)$.

The discrete metric $d(x,y)=1$ if $x\ne y$, else $0$, satisfies the metric triangle inequality $d(x,z)\le d(x,y)+d(y,z)$ but does not arise from a norm: it lacks the homogeneity property because $d(\lambda x,\lambda y)=d(x,y)$ for any $\lambda \ne 0$, instead of scaling by $|\lambda |$. This is the structural reason that norms and metrics are different objects: every norm produces a metric via $d(v,w)=\Vert v-w\Vert$, but not every metric comes from a norm.

$p$-adic absolute values. Fix a prime $p$. For a nonzero rational $r=p^k\cdot m/n$ with $m,n$ coprime to $p$ and $k\in \mathbb{Z}$, define the $p$-adic absolute value $|r{|}_p:=p^{-k}$, with $|0{|}_p=0$. The function $|\cdot {|}_p:\mathbb{Q}\to {\mathbb{R}}_{\ge 0}$ satisfies positivity and multiplicativity, and it satisfies a strengthening of the triangle inequality: the ultrametric inequality $|x+y{|}_p\le \max (|x{|}_p,|y{|}_p)$, which is itself sharper than $|x{|}_p+|y{|}_p$. An absolute value satisfying this strengthening is called non-Archimedean; one for which the Archimedean property holds (every nonzero rational eventually exceeds $1$ when scaled by integers) is Archimedean. The Euclidean absolute value $|\cdot |$ on $\mathbb{Q}$ is Archimedean; every $|\cdot {|}_p$ is non-Archimedean.

The completion of $\mathbb{Q}$ under $|\cdot {|}_p$ is the field ${\mathbb{Q}}_p$ of $p$-adic numbers, an entirely different object from $\mathbb{R}$ and the foundational object of $p$-adic analysis and arithmetic geometry. The Cauchy-completion construction recurs from 00.01.01, applied to the metric $d_p(x,y)=|x-y{|}_p$ instead of the Euclidean metric.

Ostrowski's theorem. The classification of absolute values on $\mathbb{Q}$ is complete and short. Two absolute values $|\cdot {|}_1$ and $|\cdot {|}_2$ on $\mathbb{Q}$ are equivalent if there exists a real $s>0$ with $|x{|}_2=|x{|}_1^s$ for all $x\in \mathbb{Q}$.

Theorem (Ostrowski 1916). Every absolute value on $\mathbb{Q}$ that is not the constant-on-${\mathbb{Q}}^{\times }$ function (sending every nonzero rational to $1$) is equivalent either to the Euclidean absolute value $|\cdot |$ or to the $p$-adic absolute value $|\cdot {|}_p$ for some prime $p$.

This is the source of the Archimedean / non-Archimedean dichotomy in number theory: there is exactly one Archimedean absolute value on $\mathbb{Q}$ up to equivalence, and a countable family of non-Archimedean ones, indexed by primes. The product formula ${\prod }_v|x{|}_v=1$ for nonzero rational $x$, taken over all places $v$ (the Euclidean place plus all primes), records the deep arithmetic content of this classification.

Synthesis. The bridge between the elementary triangle inequality on $\mathbb{R}$ and the abstract norm axioms is the recognition that property 3 from the formal definition is the only axiom of a norm with substantive content — positivity and homogeneity are essentially bookkeeping. The triangle inequality is exactly the axiom that makes "distance" coherent enough to support a topology, a notion of convergence, and a notion of completeness. This is precisely the structural content the rest of analysis reads off. The case-on-sign proof generalises into the convexity proof of Minkowski's inequality, which in turn generalises into the Hahn-Banach theorem and the geometry of Banach spaces in 02.11.01. The ultrametric strengthening identifies the non-Archimedean places with the primes, and Ostrowski identifies the topology of the rationals with the data of one Archimedean and countably many non-Archimedean valuations. Putting these together, the foundational insight is that the absolute value is the simplest case of a valuation on a field, and Ostrowski's theorem is the structural reason that completing $\mathbb{Q}$ produces either $\mathbb{R}$ or a $p$-adic field — there is nothing else available. This unit identifies $|\cdot |$ on $\mathbb{R}$ with the prototype of a norm, identifies the triangle inequality with the norm axiom that creates topology, and identifies the dichotomy of completions with the source of the split between real and $p$-adic analysis.

Full proof set [Master]

Proposition (Minkowski's inequality, ${\ell }^p$ triangle). Fix $p\ge 1$. For $x=(x_1,\dots ,x_n)$ and $y=(y_1,\dots ,y_n)$ in ${\mathbb{R}}^n$,

$${\left(\sum _{i=1}^n|x_i+y_i{|}^p\right)}^{1/p}\le {\left(\sum _{i=1}^n|x_i{|}^p\right)}^{1/p}+{\left(\sum _{i=1}^n|y_i{|}^p\right)}^{1/p}.$$

Proof sketch. The case $p=1$ is immediate from summing the one-dimensional triangle inequality $|x_i+y_i|\le |x_i|+|y_i|$. The case $p>1$ is proved by writing $|x_i+y_i{|}^p=|x_i+y_i|\cdot |x_i+y_i{|}^{p-1}\le (|x_i|+|y_i|)\cdot |x_i+y_i{|}^{p-1}$, summing over $i$, and applying Hölder's inequality with conjugate exponents $p$ and $q=p/(p-1)$ to each of the two resulting sums; rearranging yields the bound. The case $p=\infty$ uses the supremum-norm identity $\Vert x{\Vert }_{\infty }={\max }_i|x_i|$ and reduces to the one-dimensional inequality.

Proposition (the ultrametric inequality is stronger than the triangle inequality). If $|\cdot |$ is an absolute value on a field $K$ and $|x+y|\le \max (|x|,|y|)$ for all $x,y$, then $|x+y|\le |x|+|y|$ as well, and the converse fails. The forward implication is $\max (|x|,|y|)\le |x|+|y|$, since both terms on the right are non-negative. The converse fails on $\mathbb{R}$: $|1+1|=2>1=\max (|1|,|1|)$.

Proposition (Ostrowski's theorem, sketch). Let $|\cdot {|}^{\ast }$ be an absolute value on $\mathbb{Q}$ that is not the constant-on-${\mathbb{Q}}^{\times }$ function. There are two cases on whether $|n{|}^{\ast }>1$ for some $n\in \mathbb{N}$.

Archimedean case. If some $n\ge 2$ has $|n{|}^{\ast }>1$, write $|n{|}^{\ast }=n^s$ for some $s>0$. For any $m\ge 2$ expand in base $n$: $m=a_0+a_{1}n+\cdots +a_k{n}^k$ with digits $0\le a_i<n$. Bounding by the triangle inequality and the digit estimate $|a_i{|}^{\ast }\le n|1{|}^{\ast }$ yields $|m{|}^{\ast }\le C(n)m^s$ for an explicit constant $C(n)$. Replacing $m$ with $m^N$, taking $N$-th roots, and letting $N\to \infty$ gives $|m{|}^{\ast }\le m^s$. Symmetry exchanges $m$ and $n$ and gives $|m{|}^{\ast }=m^s$ for every integer $m$, hence $|x{|}^{\ast }=|x{|}^s$ for every $x\in \mathbb{Q}$. The absolute value is equivalent to the Euclidean one.

Non-Archimedean case. If $|n{|}^{\ast }\le 1$ for every integer $n$, set $\mathfrak{p}:=\{n\in \mathbb{Z}:|n{|}^{\ast }<1\}$. Multiplicativity makes $\mathfrak{p}$ a prime ideal of $\mathbb{Z}$, hence $\mathfrak{p}=(p)$ for some prime $p$. Let $\rho :=|p{|}^{\ast }<1$. For any $r\in {\mathbb{Q}}^{\times }$ write $r=p^k\cdot a/b$ with $p\nmid ab$; then $|r{|}^{\ast }={\rho }^k=p^{-k\cdot {\log }_p(1/\rho )}$. Setting $s={\log }_p(1/\rho )$ identifies $|r{|}^{\ast }=|r{|}_p^s$. The absolute value is equivalent to $|\cdot {|}_p$.

The case dichotomy exhausts the possibilities, so every absolute value on $\mathbb{Q}$ apart from the constant-on-${\mathbb{Q}}^{\times }$ function falls into one of the two equivalence classes.

Proposition (norm $\Rightarrow$ metric, but not conversely). If $\Vert \cdot \Vert$ is a norm on a real vector space $V$, then $d(v,w):=\Vert v-w\Vert$ defines a metric on $V$: positivity and symmetry follow from the corresponding norm properties, and the metric triangle inequality $d(v,w)\le d(v,u)+d(u,w)$ follows by setting $x=v-u$ and $y=u-w$ in the norm triangle inequality. Conversely, the discrete metric on a vector space of dimension $\ge 1$ over $\mathbb{R}$ does not come from a norm: any homogeneous extension would force $d(\lambda v,0)=|\lambda |\cdot d(v,0)$, contradicting boundedness of the discrete metric by $1$.

Connections [Master]

• The triangle inequality on $\mathbb{R}$ is the foundational case of the metric-space triangle inequality in 02.01.05 (metric space). The function $d(x,y)=|x-y|$ is the prototype of a metric, and every metric-space proof of continuity and convergence in 02.03.01 pending (sequence convergence) reduces in the line case to bounding $|x_n-L|$ via two applications of the inequality proved in this unit.

• The four properties of $|\cdot |$ specialise the abstract norm axioms invoked at 02.11.01 (Banach spaces). The case-on-sign proof here generalises into the convexity proof of Minkowski's inequality on ${\ell }^p$ and $L^p$, and the abstract triangle inequality is the load-bearing axiom that turns a vector space into a topological object — foundation-of for every functional-analytic statement about completeness, duality, and operator norms.

• The ultrametric inequality $|x+y{|}_p\le \max (|x{|}_p,|y{|}_p)$ separates the Archimedean and non-Archimedean worlds of number theory, and Ostrowski's classification recurs in 02.12.01 pending (absolute values on number fields) when extending the dichotomy from $\mathbb{Q}$ to a general number field; the same place-by-place product formula reappears in the adelic framework that drives modern arithmetic geometry.

• The reverse triangle inequality $||x|-|y||\le |x-y|$ is the statement that the absolute-value function is $1$-Lipschitz, a special case of the general fact in 02.04.02 pending (continuous functions) that every norm on a finite-dimensional space is a Lipschitz function with respect to itself; this appears again in 02.11.04 (Hilbert space) as continuity of the norm in the strong topology.

Historical & philosophical context [Master]

The geometric content of the triangle inequality is older than absolute-value notation by two millennia. Euclid's Elements I.20 (c. 300 BC) states that in any triangle, the sum of any two sides exceeds the third, which is the planar triangle inequality for Euclidean distances; the one-dimensional case of three colinear points is implicit. The notation $|x|$ for absolute value was introduced by Karl Weierstrass in his 1841 manuscript Zur Theorie der Potenzreihen (published posthumously) and used systematically in his Berlin lectures of the 1860s, which propagated the convention through the German analytic school. Cauchy 1821 Cours d'analyse used the absolute value implicitly in his definition of the modulus of a complex number, but without the bar notation.

The abstract norm axioms emerged from the early-twentieth-century functional-analytic program. Maurice Fréchet 1906 Sur quelques points du calcul fonctionnel (Rend. Circ. Mat. Palermo 22) introduced the metric-space concept and stated the metric triangle inequality as one of three defining axioms. Stefan Banach 1922 Sur les opérations dans les ensembles abstraits et leur application aux équations intégrales (Fund. Math. 3) gave the modern norm axioms in the form used today, building on Fréchet's metric framework and Frigyes Riesz 1918 work on $L^p$ spaces. Alexander Ostrowski 1916 Über einige Lösungen der Funktionalgleichung $\phi (x)\cdot \phi (y)=\phi (xy)$ (Acta Math. 41) classified every absolute value on $\mathbb{Q}$ apart from the constant-on-${\mathbb{Q}}^{\times }$ function as either Archimedean or $p$-adic, completing a programme initiated by Kurt Hensel's 1897 introduction of the $p$-adic numbers in Über eine neue Begründung der Theorie der algebraischen Zahlen (Jahresber. Deutsch. Math. Verein. 6).

Bibliography [Master]

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