The chain of number sets \(\mathbb{N}\subseteq\mathbb{N}_0\subseteq\mathbb{Z}\subseteq\mathbb{Q}\subseteq\mathbb{R}\subseteq\mathbb{C}\) has, up to this point, been built one explicit construction at a time: the integers as equivalence classes of pairs of natural numbers, the rational numbers as equivalence classes of pairs of integers. The real numbers break this pattern. Rather than construct \(\mathbb{R}\) as a further explicit set of equivalence classes, it is characterized axiomatically: a list of properties is written down, and \(\mathbb{R}\) is simply defined to be a set with two operations and an order relation satisfying exactly these properties.
Most of the axioms below are already familiar, because \(\mathbb{Q}\) satisfies them too. What singles out \(\mathbb{R}\) is one additional property, the supremum axiom, which \(\mathbb{Q}\) does not have. That failure is not abstract: \(\sqrt{2}\) has already been shown to have no rational value, yet a right triangle with legs \(1\) still has a hypotenuse of some definite length. The supremum axiom is exactly what closes this gap, and its main consequence, the existence of square roots for every positive real number, is proved in full near the end of this chapter.
The set-theoretic notation \(\in,\subseteq,\cup,\cap,\setminus,\times\) and the logical connectives \(\Rightarrow\) ("implies") and \(\Leftrightarrow\) ("if and only if") are used freely below in the same sense as in the rest of this book.
The Field Axioms
On \(\mathbb{R}\) two operations are defined, addition \(+\) and multiplication \(\cdot\) (the multiplication dot is usually omitted, \(xy:=x\cdot y\)), assigning to any \(x,y\in\mathbb{R}\) a sum \(x+y\in\mathbb{R}\) and a product \(xy\in\mathbb{R}\). These operations satisfy the following rules for all \(x,y,z\in\mathbb{R}\):
\[ \begin{array}{rll} \text{(A1)} & (x+y)+z=x+(y+z) & \text{associativity of addition}\\ \text{(A2)} & \text{there is } 0\in\mathbb{R} \text{ with } x+0=x & \text{additive identity}\\ \text{(A3)} & \text{for every } x \text{ there is } (-x)\in\mathbb{R} \text{ with } x+(-x)=0 & \text{additive inverse}\\ \text{(A4)} & x+y=y+x & \text{commutativity of addition}\\ \text{(M1)} & (xy)z=x(yz) & \text{associativity of multiplication}\\ \text{(M2)} & \text{there is } 1\in\mathbb{R}\setminus\{0\} \text{ with } x\cdot 1=x & \text{multiplicative identity}\\ \text{(M3)} & \text{for every } x\neq 0 \text{ there is } x^{-1} \text{ with } xx^{-1}=1 & \text{multiplicative inverse}\\ \text{(M4)} & xy=yx & \text{commutativity of multiplication}\\ \text{(D)}\ \ & x(y+z)=xy+xz & \text{distributivity} \end{array} \]
A set with at least two elements, together with two operations satisfying (A1)–(A4), (M1)–(M4), and (D), is called a field. \(\mathbb{Q}\) is a field. \(\mathbb{Z}\) is not, since (M3) fails: \(2\) has no integer \(x\) with \(2x=1\).
A field need not be infinite. The set \(\mathbb{Z}_2=\{0,1\}\) with addition and multiplication taken modulo \(2\),
\[0+0=0,\quad 0+1=1+0=1,\quad 1+1=0,\qquad 0\cdot 0=0\cdot 1=1\cdot 0=0,\quad 1\cdot 1=1,\]
satisfies all nine axioms above and is therefore a field, even though it has only two elements. This example is used again below to show that not every field can be ordered.
The Order Axioms
\(\mathbb{R}\) is in addition ordered: there is a relation "\(>0\)" ("greater than zero", or positive) satisfying, for all \(x,y\in\mathbb{R}\):
\[ \begin{array}{rll} \text{(O1)} & \text{exactly one of } x>0,\ x=0,\ -x>0 \text{ holds} & \text{trichotomy}\\ \text{(O2)} & x>0 \text{ and } y>0 \implies x+y>0 & \text{closure under addition}\\ \text{(O3)} & x>0 \text{ and } y>0 \implies xy>0 & \text{closure under multiplication} \end{array} \]
A field with such a relation satisfying (O1)–(O3) is called an ordered field. \(\mathbb{Q}\) is ordered. \(\mathbb{Z}_2\) from above cannot be ordered at all: (O1) would force either \(1>0\) or \(-1=1>0\), and either way (O2) then gives \(1+1>0\), i.e. \(0>0\), contradicting (O1).
From "\(>0\)" the familiar order relations are defined for all \(x,y\in\mathbb{R}\):
\[ x>y :\iff x-y>0,\qquad x<y :\iff y-x>0,\qquad x\geq y :\iff (x>y\text{ or }x=y),\qquad x\leq y :\iff (x<y\text{ or }x=y), \]
where \(x-y:=x+(-y)\) and, for \(y\neq0\), \(\frac{x}{y}:=xy^{-1}\) abbreviate subtraction and division. Note (O1) applied to \(x-y\) says exactly one of \(x>y\), \(x=y\), \(x<y\) can hold: the order is total.
The Supremum Axiom and Completeness
Before the supremum axiom itself can be stated, boundedness needs a name. For \(M\subseteq\mathbb{R}\), \(M\neq\emptyset\):
- \(M\) is bounded above if there is \(C\in\mathbb{R}\) with \(x\leq C\) for all \(x\in M\); such a \(C\) is called an upper bound of \(M\).
- \(M\) is bounded below if there is \(D\in\mathbb{R}\) with \(D\leq x\) for all \(x\in M\); such a \(D\) is a lower bound.
- \(M\) is bounded if it is bounded both above and below.
- \(C\in\mathbb{R}\) is the supremum of \(M\), written \(C=\sup M\), if \(C\) is an upper bound of \(M\) and no smaller number is: for every \(\varepsilon>0\) there is \(x\in M\) with \(x>C-\varepsilon\). If in addition \(C\in M\), then \(C\) is the maximum of \(M\).
- Dually, \(C\in\mathbb{R}\) is the infimum of \(M\), written \(C=\inf M\), if \(C\) is a lower bound and for every \(\varepsilon>0\) there is \(x\in M\) with \(x<C+\varepsilon\). If in addition \(C\in M\), then \(C\) is the minimum of \(M\).
With this vocabulary in place, the last defining property of \(\mathbb{R}\) can be stated:
\[\text{(S)}\qquad\text{Every nonempty, bounded-above subset of } \mathbb{R} \text{ has a supremum.}\]
This is the supremum axiom, and it is the one property from this section that \(\mathbb{Q}\) does not share — a fact that will be put to direct use later in this chapter.
A supremum need not be a maximum: the set \(M=\left\{\frac{x}{1+x}:x>0\right\}\) has \(\sup M=1\), but \(1\notin M\). Indeed \(\frac{x}{1+x}<1\) for every \(x>0\), since \(\frac{x}{1+x}<1\iff x<1+x\iff 0<1\), which always holds; so \(1\) is an upper bound and no element of \(M\) equals it. To see that no smaller number is an upper bound either, fix \(\varepsilon>0\); solving \(\frac{x}{1+x}>1-\varepsilon\) for \(x\) gives \(x>\frac{1-\varepsilon}{\varepsilon}\) whenever \(0<\varepsilon\leq1\), and any such \(x\) is a witness in \(M\) exceeding \(1-\varepsilon\). So \(1\) is indeed the least upper bound, yet it is never attained.
Basic Algebraic Consequences
A short list of everyday algebraic facts already follows from (A1)–(A4), (M1)–(M4), and (D) alone, before order or completeness are used at all.
First, \(x\cdot 0=0\) for every \(x\in\mathbb{R}\): by (A2), \(x\cdot 0=x\cdot(0+0)\), and by (D) this equals \(x\cdot 0+x\cdot 0\); adding \(-(x\cdot 0)\) to both sides leaves \(0=x\cdot 0\).
Second, \((-x)y=-(xy)\) for all \(x,y\in\mathbb{R}\): using the fact just proved, \(0=0\cdot y=(x+(-x))y\), and by (D) this equals \(xy+(-x)y\); adding \(-(xy)\) to both sides gives \(-(xy)=(-x)y\).
Third, for \(a,b,c\in\mathbb{R}\) with \(a\neq0\), the equation \(ax+b=c\) has exactly one solution, \(x=\frac{c-b}{a}\). That this \(x\) solves the equation is a direct computation using (M3), (M1), (M4), (A1), and (A2) to unwind \(a\left(\frac{c-b}{a}\right)+b\) back to \(c\). For uniqueness, if \(x\) is any solution, adding \(-b\) to both sides gives \(ax=c-b\), and multiplying by \(a^{-1}\) gives \(x=a^{-1}(c-b)=\frac{c-b}{a}\).
Two familiar facts fall out of this uniqueness statement as special cases. Taking \(a=1,b=a_0,c=0\) shows the equation \(x+a_0=0\) has a unique solution, so the additive inverse \(-a_0\) of any \(a_0\) is unique; since both \(a_0\) and \(-(-a_0)\) satisfy \(x+(-a_0)=0\), uniqueness forces \(-(-a_0)=a_0\). Taking \(a=a_0,b=0,c=1\) (for \(a_0\neq0\)) shows \(a_0x=1\) has a unique solution, so the multiplicative inverse \(a_0^{-1}\) is unique and, by the same argument, \((a_0^{-1})^{-1}=a_0\).
One more fact of this kind, that \(xy=0\) exactly when \(x=0\) or \(y=0\), is exactly the zero-product property already proved elsewhere from (M3) alone; it is not repeated here.
Consequences of the Order Axioms
The order axioms (O1)–(O3) combine with the algebraic facts above to give the inequality rules used constantly throughout analysis. Let \(a,b,c,d\in\mathbb{R}\).
Adding or subtracting the same real number preserves a strict inequality:
\[ a<b \quad\Longrightarrow\quad a+c<b+c, \qquad a-c<b-c. \]
Multiplication and division preserve the direction for a positive factor and reverse it for a negative factor:
\[ \begin{aligned} a<b,\ c>0 &\quad\Longrightarrow\quad ac<bc, &\frac{a}{c}&<\frac{b}{c},\\ a<b,\ c<0 &\quad\Longrightarrow\quad ac>bc, &\frac{a}{c}&>\frac{b}{c}. \end{aligned} \]
The division statements are multiplication by \(c^{-1}\), which has the same sign as \(c\). No division rule is stated for \(c=0\), where the quotients are undefined.
Squares of nonzero numbers are positive: \(a\neq0\implies a^2>0\). If \(a>0\), (O3) gives \(a\cdot a>0\) directly. If instead \(-a>0\), then (O3) gives \((-a)(-a)>0\), and \((-a)(-a)=-(a(-a))=-(-(a\cdot a))=a\cdot a\) by the two facts from the previous section, so again \(a^2>0\). Taking \(a=1\) gives \(1=1\cdot1>0\), and hence \(-1<0\).
Multiplying by a negative number reverses an inequality: if \(a<b\) and \(c<0\), then \(b-a>0\) and \(-c>0\), so (O3) gives \((b-a)(-c)>0\), i.e. \(ac-bc>0\), i.e. \(ac>bc\).
Positive numbers have positive inverses: for \(a>0\), \(a^{-1}\) cannot be \(0\) (else \(1=aa^{-1}=a\cdot0=0\), contradicting (M2)), and if \(-a^{-1}>0\) then (O3) would give \(a(-a^{-1})=-1>0\), contradicting the previous paragraph; so \(a^{-1}>0\) is the only possibility left by (O1).
Order reverses under taking inverses: for \(0<a<b\), both \(a^{-1},b^{-1}>0\) by the previous fact, and \(b-a>0\) multiplied by \(a^{-1}b^{-1}>0\) (via (O3)) gives \(b^{-1}-a^{-1}>0\), i.e. \(\frac{1}{a}>\frac{1}{b}\).
Inequalities combine under multiplication when all quantities are positive: if \(a<b\) with \(b>0\), and \(0<c<d\), then \((b-a)c>0\) gives \(ac<bc\), and \(b(d-c)>0\) gives \(bc<bd\); chaining the two gives \(ac<bd\).
Squaring is monotonic on the positive numbers: for \(a,b>0\), \(a<b\iff a^2<b^2\). If \(a<b\), the previous rule with \(c=a,d=b\) gives \(a^2=aa<ab<bb=b^2\). Conversely, if \(a^2<b^2\) but \(a\geq b\), then either \(a=b\) (giving \(a^2=b^2\), a contradiction) or \(a>b\) (giving \(a^2>b^2\) by the forward direction with the roles of \(a,b\) swapped, again a contradiction); so \(a<b\) after all.
Absolute Value on the Real Numbers
For \(a\in\mathbb{R}\), the absolute value is defined exactly as it was for the integers,
\[ |a| := \begin{cases} a, & a\geq 0,\\ -a, & a<0, \end{cases} \]
and the basic properties — \(|a|\geq0\) with equality exactly when \(a=0\), multiplicativity \(|ab|=|a||b|\), and the triangle inequality \(|a+b|\leq|a|+|b|\) — follow from the same sign case split already carried out for \(\left|\cdot\right|\) on \(\mathbb{Z}\), and are not repeated here.
One characterization is used so often in the chapters on sequences and limits that it is worth proving explicitly: for \(\varepsilon>0\),
\[|a|<\varepsilon \iff -\varepsilon<a<\varepsilon.\]
If \(a\geq0\), then \(|a|=a\), so \(|a|<\varepsilon\) reads \(a<\varepsilon\); combined with \(a\geq0>-\varepsilon\), this is exactly \(-\varepsilon<a<\varepsilon\). If \(a<0\), then \(|a|=-a\), so \(|a|<\varepsilon\) reads \(-a<\varepsilon\), i.e. \(-\varepsilon<a\); combined with \(a<0<\varepsilon\), this again gives \(-\varepsilon<a<\varepsilon\). Both directions of the case split are reversible, which gives the equivalence.
The Natural Numbers Are Unbounded in R
The supremum axiom (S) has an immediate consequence for how \(\mathbb{N}\) sits inside \(\mathbb{R}\): \(\mathbb{N}\) is not bounded above. Suppose it were. Then, being nonempty and bounded above, \(\mathbb{N}\) has a supremum \(s=\sup\mathbb{N}\) by (S). Applying the definition of supremum with \(\varepsilon=1\), there is some \(n\in\mathbb{N}\) with \(n>s-1\), hence \(n+1>s\). But \(n+1\in\mathbb{N}\), so \(s\) fails to be an upper bound of \(\mathbb{N}\) after all — a contradiction. So no upper bound exists.
This fact, known as the Archimedean property, is exactly what underlies the density of \(\mathbb{Q}\) in \(\mathbb{R}\) shown earlier: choosing \(n\in\mathbb{N}\) with \(n(y-x)>1\) is only possible because \(\mathbb{N}\) has no upper bound to begin with.
Intervals
For \(a,b\in\mathbb{R}\) with \(a<b\), the order on \(\mathbb{R}\) picks out eight standard subsets called intervals:
\[ \begin{array}{ll} [a,b]:=\{x\in\mathbb{R}:a\leq x\leq b\} & \text{closed, bounded}\\ (a,b):=\{x\in\mathbb{R}:a<x<b\} & \text{open, bounded}\\ [a,b):=\{x\in\mathbb{R}:a\leq x<b\},\qquad (a,b]:=\{x\in\mathbb{R}:a<x\leq b\} & \text{half-open, bounded}\\ [a,\infty):=\{x\in\mathbb{R}:x\geq a\},\qquad (a,\infty):=\{x\in\mathbb{R}:x>a\} & \text{unbounded above}\\ (-\infty,a]:=\{x\in\mathbb{R}:x\leq a\},\qquad (-\infty,a):=\{x\in\mathbb{R}:x<a\} & \text{unbounded below} \end{array} \]
A closed bounded interval \([a,b]\) always has \(\min=a\) and \(\max=b\); the corresponding open interval \((a,b)\) has \(\inf=a\) and \(\sup=b\), by the same argument used for \(M=\left\{\frac{x}{1+x}:x>0\right\}\) above, but attains neither bound.
Existence of Square Roots
The gap between \(\mathbb{Q}\) and \(\mathbb{R}\) that motivated the supremum axiom can now be closed rigorously. A short lemma is needed first.
Lemma. If \(a,b\in\mathbb{R}\) and \(a<b+\varepsilon\) for every \(\varepsilon>0\), then \(a\leq b\). Indeed, if instead \(a>b\), setting \(\varepsilon=a-b>0\) gives \(a<b+\varepsilon=b+(a-b)=a\), i.e. \(a<a\), which is absurd.
Theorem. For every \(a\in\mathbb{R}\) with \(a>0\), there exists exactly one \(x>0\) with \(x^2=a\); it is written \(x=\sqrt{a}=a^{1/2}\).
Uniqueness among positive reals is immediate from the squaring-is-monotonic fact above: if \(x,y>0\) both satisfy \(x^2=y^2=a\) but \(x\neq y\), then either \(x<y\) or \(y<x\), and either way monotonic squaring would force \(x^2\neq y^2\), a contradiction.
For existence, consider \(M:=\{r\in(0,\infty):r^2\leq a\}\).
\(M\) is nonempty: if \(0<a<1\), then \(a^2<a\) (multiplying \(a<1\) by \(a>0\)), so \(a\in M\); if \(a\geq1\), then \(1^2=1\leq a\), so \(1\in M\).
\(M\) is bounded above by \(1+a\): if some \(r\in M\) had \(r>1+a\), squaring (both sides positive) would give \(r^2>(1+a)^2=1+2a+a^2\geq1+2a>a\) (since \(a^2\geq0\) and \(1+2a>a\iff1+a>0\)), contradicting \(r^2\leq a\). So every \(r\in M\) satisfies \(r\leq1+a\).
By (S), \(x:=\sup M\) exists, and \(x>0\) since \(M\) contains a positive element and \(x\) is at least as large as it.
Step 1: \(x^2\leq a\). Fix \(\varepsilon>0\) and \(0<\delta\leq1\). By definition of supremum there is \(r\in M\) with \(r>x-\delta\), so \(x<r+\delta\), and since both sides are positive,
\[x^2<(r+\delta)^2=r^2+\delta(2r+\delta)\leq a+\delta(2r+\delta),\]
using \(r^2\leq a\). The extra term is bounded uniformly in \(r\): if \(r\leq1\), then \(\delta(2r+\delta)\leq\delta(2+1)=3\delta\); if \(r>1\), then \(r<r^2\leq a\), so \(\delta(2r+\delta)\leq\delta(2a+1)\). Either way \(\delta(2r+\delta)\leq C\delta\) with \(C:=\max\{3,2a+1\}\), so \(x^2<a+C\delta\). Choosing \(\delta=\min\left\{1,\frac{\varepsilon}{C}\right\}\) gives \(x^2<a+\varepsilon\). Since \(\varepsilon>0\) was arbitrary, the Lemma gives \(x^2\leq a\).
Step 2: \(x^2\geq a\). Suppose instead \(x^2<a\), and set \(\varepsilon:=a-x^2>0\). Exactly the estimate from Step 1, now with \(x\) itself in the role of \(r\), gives \((x+\delta)^2=x^2+\delta(2x+\delta)\leq x^2+C\delta\) for \(0<\delta\leq1\) and the same constant \(C\). Choosing \(\delta=\min\left\{1,\frac{\varepsilon}{C}\right\}\) gives \((x+\delta)^2<x^2+\varepsilon=a\), so \(x+\delta\in M\); but \(x+\delta>x\) then contradicts \(x\) being an upper bound of \(M\). So \(x^2\geq a\).
Combining both steps, \(x^2=a\), completing the proof.
The same construction goes through essentially unchanged for any \(n\in\mathbb{N}\) in place of the exponent \(2\), giving the existence of an \(n\)-th root \(a^{1/n}\) for every \(a>0\); only the algebra of Step 1 and Step 2 becomes more involved and is not repeated here. Together with the definition of rational exponents \(a^{p/q}=(a^p)^{1/q}\) already given for \(\mathbb{Q}\), this shows that \(a^{p/q}\) exists as a real number for every \(a>0\) and every rational exponent \(\frac{p}{q}\), regardless of whether the result itself happens to be rational.
Properties of Real Radicals
For \(n,m\in\mathbb{N}\), the principal real \(n\)-th root of a nonnegative number is its reciprocal power:
\[ \sqrt[n]{a}=a^{\frac{1}{n}}, \qquad a\geq0. \]
For \(a,b\geq0\), and additionally \(b>0\) in the quotient, principal roots satisfy
\[ \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}, \qquad \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}, \qquad \sqrt[m]{\sqrt[n]{a}}=\sqrt[mn]{a}. \]
Raising a real number to the \(n\)-th power and then taking the principal \(n\)-th root recovers either the number itself or its absolute value, depending on the parity of \(n\):
\[ \sqrt[n]{a^n} = \begin{cases} a, & n\text{ odd},\\ |a|, & n\text{ even}. \end{cases} \]
The absolute value in the even case is essential: both \(a\) and \(-a\) have the same even power, while the principal root is nonnegative.
The Real Numbers as a Complete Ordered Field
Collecting the sections above: \(\mathbb{R}\) is a field under (A1)–(A4), (M1)–(M4), (D); it is ordered under (O1)–(O3), compatibly with the field structure; and it additionally satisfies the supremum axiom (S), which \(\mathbb{Q}\) does not. This last property is called completeness, and it is precisely what the rest of analysis is built on: the existence of limits of monotonic bounded sequences, the algebra of limits, and the intermediate value theorem for continuous functions all trace back, at some point, to the supremum axiom proved here to guarantee square roots.