A binary operation \(f\) on a set \(S\) is a mapping \(f\colon S\times S \to S\).
A group \(G\) is a set with a binary operation \(*\) which satisfies the following axioms.
A structure \((G, *)\) which only satisfies associativity is called a semigroup. A semigroup with an identity element is called a monoid.
An abelian group also satisfies the axiom of commutativity.
A ring is a set \(R\) with two binary operations \(+\) and \(\times\) (addition and multiplication) such that \((R, +)\) is an abelian group, \((R, \times)\) is a monoid and \(\times\) is distributive over \(+\).
The additive and multiplicative identities of \(R\) are denoted by \(0_R\) and \(1_R\) respectively.
If \(\times\) is commutative, \((R, +, \times)\) is called a commutative ring.
A field \(F\) is a commutative ring such that the non-zero elements form an abelian group under multiplication.
An ordered field \(F\) is also an ordered set such that for all \(a, b, c \in F\),
The ordered field \(\Z/p\Z\) is an important example. Not that \(\Z/n\Z\) are not fields in general.
Question: Can a nontrivial finite abelian group be ordered?
Response: No. Let \((G, +)\) be a group with \(n > 1\) elements, and select two members such that \(a > b\). We thus have the chain of terms \[ nb < a + (n-1)b < 2a + (n-2)b + \dots + (n-1)a + b < na. \] Thus, we have generated \(n+1\) distinct elements in \(G\), which is a contradiction!
Note that we simply denote \(na = \underbrace{a + \dots + a}_{n \text{ times}}\).
The real numbers are a complete, ordered field.
The irrational number \(\sqrt{2}\) can be approximated by the rationals with arbitrary precision. This is done iteratively, using the AM-GM-HM inequality to generate sequences of rationals converging to \(\sqrt{2}\). We define the sequences \(\{u_n\}_{n=0}^\infty\) and \(\{l_n\}_{n=0}^\infty\) as follows. \[ u_0 = 2, \quad\quad l_0 = 1. \] \[ u_{n+1} = \frac{u_n + l_n}{2}, \quad\quad l_{n+1} = \frac{2}{\frac{1}{u_n} + \frac{1}{l_n}}. \]
Note that \(u_{n+1}l_{n+1} = u_nl_n\), so \(u_nl_n = 2\). Thus, \(u_{n+1} > \sqrt{2} > l_{n+1}\) by \(AM > GM > HM\). Also, \(u_{n+1} < u_n\) and \(l_{n+1} > l_n\), so by iterating successively over \(n\), we construct the sequence of rational numbers \[ u_0 > u_1 > \dots > u_n > u_{n+1} > \sqrt{2} > l_{n+1} > l_n > \dots > l_1 > l_0. \]
To put strict bounds on our approximation, we calculate the difference \(R_{n+1} = u_{n+1} - l_{n+1}\). \[ R_{n+1} = u_{n+1} - l_{n+1} = \frac{u_n + l_n}{2} - \frac{2u_nl_n}{u_n + l_n} = \frac{(u_n - l_n)^2}{2(u_n + l_n)}.\]
Since \(u_n + l_n > 2l_0 = 2\), we have \(R_{n+1} < R_{n}^2 / 4\). Also, \(u_n - l_n < u_0 - l_0 = 1\), so \(R_{n + 1}/R_{n} < 1/4\). Thus, \(R_n \to 0\) as \(n \to \infty\), so we can approximate \(\sqrt{2}\) with arbitrary precision. Indeed, after \(n\) iterations of this algorithm, the error is at most \(1/4^n\).
These two properties follow from the fact that \(\R\) is an ordered set.
The real numbers are also complete, i.e., \(\R\) has the least upper bound property.
These two properties are equivalent. Let \(S\) have the least upper bound property, and let \(E \subseteq S\) be bounded below. Let \(L\) be the set of all lower bounds of \(L\). Then \(L \neq \emptyset\), and every element of \(E\) is an upper bound of \(L\). Also, from the least upper bound property of \(S\), \(\sup L\) must exist in \(S\). Note that for all \(s \in S\), if \(s < \sup L\), then there exists \(l \in L, l > s\) so \(s\) is not an upper bound of \(L\), hence \(s \notin E\). The contrapositive of this is that for every \(s \in E\), we must have \(s \geq \sup L\). This means that \(\sup L\) is a lower bound of \(E\), so \(\sup L \in L\). Also, \(\sup L\) is an upper bound of \(L\), so \(y \leq \sup L\) for all \(y \in L\). This means that \(\sup L \geq y\) for all lower bounds \(y\in L\) of \(E\). Thus, \(\inf E = \sup L\) for all non-empty \(E \subseteq S\) bounded below, so \(S\) also has the greatest lower bound property. The converse follows similarly.
Note that \(\R\) contains \(\Q\) as a subfield. The rationals \(\Q\) do not have the least upper bound property. For example, the set \(\{x\in\Q: x^2 < 2\}\) may have upper bounds in \(\Q\), but has no least upper bound in \(\Q\).
We introduce two new symbols \(\infty\) and \(-\infty\) to \(\R\). Preserving the original order, \[ -\infty < x < \infty \] for all \(x \in \R\). We also define the following conventions.