Ordered Fields

From the Real Numbers to Ordered Fields

In the previous tutorial, we used the real numbers’ order together with their field operations to compare numbers and control signs. The same ideas are not unique to \(\mathbb{R}\). They can be described as conditions on a field and an order that work together. This abstraction lets us distinguish properties that follow from the field and order axioms from properties specific to the real numbers.

An ordered field is a field equipped with a strict total order that is compatible with addition and multiplication. “Strict total order” means that the order is transitive, no element is less than itself, and for distinct elements \(x\) and \(y\), exactly one of \(x<y\) or \(y<x\) holds. The compatibility conditions specify how the order behaves under the field operations.

Definition (Ordered field). An ordered field is a field \(F\) with a strict total order \(<\) such that, for all \(x,y,z\in F\):

If \(x<y\), then \(x+z<y+z\).
If \(0<x\) and \(0<y\), then \(0<xy\).

We write \(x\leq y\) when \(x<y\) or \(x=y\), and use \(>\) and \(\geq\) for the reversed comparisons.

The first condition says that translating both sides of a strict inequality preserves it. The second says that the product of two positive elements is positive. These are the order-compatibility rules; many other familiar sign rules follow from them and the field axioms. In particular, the positivity set \(P=\{x\in F:0<x\}\) is closed under addition as well as multiplication. For addition, if \(0<x\) and \(0<y\), translation of \(0<x\) by \(y\) gives \(y<x+y\); together with \(0<y\), transitivity gives \(0<x+y\).

Consequences of the Ordered-Field Axioms

The order axioms do not separately require that \(1\) be positive. Instead, positivity of \(1\) follows from the field and order together. This matters because the sign of \(1\) determines the signs of the integer multiples of \(1\), and hence the signs of the rational numbers represented inside the field.

Theorem (Basic signs in an ordered field). In an ordered field \(F\), \(0<1\). Every positive integer multiple of \(1\) is positive. Moreover, \(x^2\geq0\) for every \(x\in F\), and \(x^2>0\) whenever \(x\ne0\).

Proof. Since \(F\) is a field, \(1\ne0\). By totality, either \(0<1\) or \(1<0\). Suppose, for contradiction, that \(1<0\). Translating this inequality by \(-1\) gives \(0<-1\). The product of two positive elements is positive, so \((-1)(-1)>0\). The field identity \((-1)(-1)=1\) then gives \(0<1\), contradicting \(1<0\). Therefore \(0<1\).

For every positive integer \(n\), the element \(n\cdot1\), meaning the sum of \(n\) copies of \(1\), is positive. This follows by induction. It holds for \(n=1\). If \(n\cdot1>0\), then both \(n\cdot1\) and \(1\) are positive, so their sum \((n+1)\cdot1\) is positive by closure of the positive elements under addition.

Finally, let \(x\in F\). If \(x=0\), then \(x^2=0\). If \(x\ne0\), totality gives either \(0<x\) or \(x<0\). In the first case, \(x^2=xx>0\), because it is a product of positive elements. In the second case, translation by \(-x\) gives \(0<-x\). Thus \((-x)(-x)>0\), and the field identity \((-x)(-x)=x^2\) gives \(x^2>0\). This proves both claims about squares. \(\square\)

The proof uses only the ordered-field axioms and field identities. In particular, its conclusions hold in any ordered field, not just in the real numbers. The statement about squares also shows why an element whose square is negative cannot belong to an ordered field.

Worked Example: The Rational Numbers as an Ordered Field

The rational numbers \(\mathbb{Q}\), with their usual addition, multiplication, and order, form an ordered field. The field operations and field axioms were established in the earlier discussion of the rational numbers. It remains to check that the usual order has the required compatibility.

If \(a<b\) for rational numbers \(a,b\), then \(b-a>0\). For any rational \(c\),

$$ (b+c)-(a+c)=b-a>0. $$

Thus \(a+c<b+c\), so translating preserves the strict inequality. If \(a>0\) and \(b>0\), write \(a=m/n\) and \(b=r/s\), where \(m,n,r,s\) are positive integers. Then

ab=\frac{mr}{ns}>0,

because \(mr\) and \(ns\) are positive integers. Thus the product of positive rationals is positive, as required. For a concrete check, \(2/5<3/5\) remains true after adding \(7/4\), since both resulting sides differ by \(3/5-2/5=1/5>0\); and \((2/5)(3/7)=6/35>0\).

Why the Rational Numbers Occur in Every Ordered Field

The positivity of \(1\) has a further consequence: an ordered field cannot have characteristic \(p\) for any positive integer \(p\). In fact, every positive integer multiple of \(1\) is positive, so it is not zero. Every negative integer multiple is the additive inverse of a positive one and is not zero either. Consequently, the field has characteristic zero, and its integer multiples of \(1\) behave like the ordinary integers.

Theorem (The rational numbers embed in every ordered field). Every ordered field \(F\) contains a subfield that is an order-preserving copy of \(\mathbb{Q}\). In particular, its characteristic is zero.

Proof. Define \(n\cdot1\) for each integer \(n\) by repeated addition when \(n>0\), by \(0\) when \(n=0\), and by the additive inverse of \((-n)\cdot1\) when \(n<0\). The field laws imply that these integer multiples respect addition and multiplication. By the preceding theorem, \(n\cdot1>0\) for \(n>0\). For \(n<0\), \((-n)\cdot1>0\), so \(n\cdot1<0\). Thus \(n\cdot1\ne0\) for every nonzero integer \(n\). This proves characteristic zero.

For a rational number \(m/n\), where \(m,n\in\mathbb{Z}\) and \(n\ne0\), define

\phi\left(\frac{m}{n}\right)=(m\cdot1)(n\cdot1)^{-1}.

The inverse exists because \(n\cdot1\ne0\). To check that this definition does not depend on the chosen fraction representation, suppose \(m/n=p/q\). Then \(mq=np\) in the integers. The rules for integer multiples give

(m\cdot1)(q\cdot1)=(n\cdot1)(p\cdot1).

Since \(n\cdot1\) and \(q\cdot1\) are nonzero, division by their product gives \(\phi(m/n)=\phi(p/q)\). The field laws also show that \(\phi\) preserves addition, multiplication, and \(1\), using the usual fraction formulas. If \(\phi(m/n)=0\), then \((m\cdot1)(n\cdot1)^{-1}=0\), which implies \(m\cdot1=0\) because \((n\cdot1)^{-1}\ne0\). The integer-multiple sign result forces \(m=0\). Hence \(\phi\) is injective, so its image is a subfield isomorphic to \(\mathbb{Q}\).

It remains to check that this copy preserves order. First, if \(a>0\), then \(a^{-1}>0\): the inverse is nonzero, so it is either positive or negative. If it were negative, then \(-a^{-1}>0\), and the product \(a(-a^{-1})=-1\) would be positive. But \(0<1\) implies \(-1<0\), a contradiction. Now a positive rational can be represented as \(m/n\) with positive integers \(m,n\). Both \(m\cdot1\) and \(n\cdot1\) are positive, as is \((n\cdot1)^{-1}\), so \(\phi(m/n)>0\). Applying this to positive differences shows that \(\phi\) preserves strict order. \(\square\)

Worked Example: A Field Containing the Square Root of Two

Consider \(F=\{a+b\sqrt{2}:a,b\in\mathbb{Q}\}\), with its usual operations as a subset of \(\mathbb{R}\). It is a subfield of \(\mathbb{R}\). Addition and multiplication stay in \(F\), since

(a+b\sqrt{2})+(c+d\sqrt{2}) =(a+c)+(b+d)\sqrt{2}

and

(a+b\sqrt{2})(c+d\sqrt{2}) =(ac+2bd)+(ad+bc)\sqrt{2}.

Both displayed coefficients are rational. For a nonzero \(a+b\sqrt{2}\), its inverse is

\frac{1}{a+b\sqrt{2}} =\frac{a-b\sqrt{2}}{a^2-2b^2}.

The denominator cannot be zero unless \(a=b=0\): if \(b\ne0\) and \(a^2=2b^2\), then \((a/b)^2=2\), contradicting the irrationality of \(\sqrt{2}\); if \(b=0\), the equation forces \(a=0\). Thus the inverse is defined and belongs to \(F\). Since \(F\) is a subfield of the ordered field \(\mathbb{R}\), the real order restricted to \(F\) makes it an ordered field: translation and positive products in \(F\) obey the same rules as in \(\mathbb{R}\). For instance, \(1+\sqrt{2}>0\), and its square is \(3+2\sqrt{2}>0\).

Worked Example: The Complex Numbers Cannot Be an Ordered Field

Suppose the complex numbers could be given an order making them an ordered field. The basic-sign theorem would then imply that every square is nonnegative. But the imaginary unit \(i\) satisfies \(i^2=-1\). The theorem would require \(-1\geq0\), while positivity of \(1\) and translation of \(0<1\) by \(-1\) give \(-1<0\). These conclusions are incompatible. Therefore no order can make \(\mathbb{C}\) an ordered field.

This obstruction is not a matter of choosing a different way to arrange complex numbers on a line. The multiplication rule \(i^2=-1\) itself conflicts with the sign consequences required of any ordered field.

What the Abstraction Does—and Does Not—Say

An ordered field must satisfy both the field axioms and the order-compatibility conditions. Neither part can be omitted. A field has addition, multiplication, and inverses for nonzero elements, but those operations alone do not supply an order. Conversely, an order chosen without regard to multiplication may fail the positive-product condition.

The ordered-field axioms guarantee familiar algebraic and sign behavior, including \(0<1\), positivity of positive integer multiples of \(1\), and positivity of nonzero squares. They also guarantee a rational subfield. They do not, by themselves, say that every nonempty set bounded above has a least upper bound. That is an additional completeness property of the real numbers, not part of the definition of an ordered field.

Key takeaway. An ordered field is a field with a total order compatible with translation and multiplication of positive elements. These conditions force familiar sign rules and an order-preserving copy of \(\mathbb{Q}\), but they do not include every special property of \(\mathbb{R}\).

Check Your Understanding

Use the definition and proved consequences of ordered fields to answer the following questions.

What are the two compatibility conditions that an order on a field must satisfy to make it an ordered field?
Why must \(1\) be positive in every ordered field?
Why does an ordered field have characteristic zero?
Why must the square of every nonzero element of an ordered field be positive?
Which field operation involving \(i\) prevents \(\mathbb{C}\) from being an ordered field?
Does the definition of ordered field include the least-upper-bound property? Explain.

What You'll Learn