10: Axiomatic Treatment of \(\mathbb{R, N, Z, Q}\) and \(\mathbb{C}\)
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There are several ways to axiomatize the standard number systems \(\mathbb{R}\), \(\mathbb{N}\), \(\mathbb{Z}\), and \(\mathbb{Q}\). One way is to start by laying down axioms for \(\mathbb{N}\) and then using \(\mathbb{N}\) and set theory to build successively the number systems \(\mathbb{Z}\), \(\mathbb{Q}\) and \(\mathbb{R}\). A quicker way is to start with axioms for \(\mathbb{R}\) and using these axioms find \(\mathbb{N}\), \(\mathbb{Z}\), and \(\mathbb{Q}\) inside of \(\mathbb{R}\). We follow the latter approach here. We begin by defining an ordered ring.
An ordered ring is a quadruple \[(R,+,\cdot, <)\] where \((R,+,\cdot)\) is a commutative ring and \(<\) is a binary relation on \(R\) which satisfies the following properties for all \(a,b,c \in R\).
- \(a < b\) and \(b < c \Longrightarrow a < c\).
- \(a < b \Longrightarrow a+c < b+c\).
- \(a < b\) and \(0 < c \Longrightarrow ac < bc\).
- Given \(a,b \in R\) one and only one of the following holds: \[a=b, \quad a < b, \quad b < a.\]
Note that we could develop some of the theory of ordered rings without the assumption of commutativity; however, this assumption will make things a little easier. All of the ordered rings we are interested in are commutative anyhow.
Terminology The binary relation \(<\) is as usual called less than. Condition 1 above is called transitivity and condition 4 is called the Law of Trichotomy. We also refer to \(<\) as an ordering or order relation on the ring \(R\). We use the following abbreviations: \[\begin{aligned} b > a &\Longleftrightarrow& a < b \\ a \le b &\Longleftrightarrow& a < b \mbox{ or } a=b\\ b \ge a &\Longleftrightarrow& a \le b \\ a < b < c &\Longleftrightarrow& a < b \mbox{ and } b < c \\ a \le b \le c &\Longleftrightarrow& a \le b \mbox{ and } b \le c\end{aligned}\] An element \(a\) is said to be positive if \(a > 0\) and, negative if \(a < 0\). Note that \(-a\) may be positive or negative, depending on whether or not \(a\) is positive or negative. Hence it is best to read \(-a\) as minus \(a\) rather that negative \(a\).
Problem 10.1 Let \(R\) be an ordered ring with identity \(1 \ne 0\). Prove that for all \(a,b,c \in R\) the following statements hold:
- \(0 < a \mbox{ and } 0 < b \Longrightarrow 0 < ab\).
- \(a < 0 \Longrightarrow 0 < -a\).
- \(0 < 1\).
- \(a \neq 0 \Longrightarrow 0 < a^2\).
- If \(a < b\) and \(c < d\) then \(a+c < b+d\).
- \(a < b \Longrightarrow -b < -a\).
- \(a < b \mbox{ and } c < 0 \Longrightarrow bc < ac\).
- If \(a\) is a unit and \(0 < a\) then \(0 < a^{-1}\).
- If \(a\) is a unit and \(0 < a < 1\) then \(1 < a^{-1}\).
- \(R\) is infinite.
Note that some rings cannot be ordered. For example, the last statement of the above problem shows that there is no way to make the rings \(\mathbb{Z}_n\) into ordered rings. As we shall see the field of complex numbers is an infinite ring that cannot be made into an ordered ring. We will give a rigorous definition of the complex numbers later. The main examples of ordered rings are \(\mathbb{Z}\), \(\mathbb{Q}\) and \(\mathbb{R}\).
Problem 10.2 Show that if a ring \(R\) has an identity \(1 \neq 0\) and contains an element \(i\) such that \(i^2 = -1\), then \(R\) cannot be an ordered ring.
If an ordered ring \(R\) is a integral domain (or field), we call \(R\) an ordered domain (or ordered field). Now we can distinguish \(\mathbb{Z}\) from \(\mathbb{Q}\) and \(\mathbb{R}\) by the fact that \(\mathbb{Z}\) is an ordered domain and not an ordered field, whereas both \(\mathbb{Q}\) and \(\mathbb{R}\) are ordered fields. The problem is how to distinguish \(\mathbb{Q}\) from \(\mathbb{R}\). This was historically a difficult thing to accomplish. The first clue was the fact that \(\sqrt{2}\) is not a rational number. To describe the difference, we need a few more definitions.
Let \(R\) be an ordered ring. Let \(S\) be a subset of \(R\). An element \(b\) of \(R\) is called an upper bound for \(S\) if \(x \le b\) for all \(x \in S\). If \(S\) has an upper bound we say that \(S\) is bounded from above.
Problem 10.3 Give examples of subsets \(S\) of \(\mathbb{R}\) satisfying the following conditions:
- \(S\) has no upper bound.
- \(S\) has an upper bound \(b \in S\).
- \(S\) is bounded from above but has no upper bound \(b \in S\).
Let \(S\) be a subset of an ordered ring \(R\) which is bounded from above. An element \(\ell \in R\) is a least upper bound (l.u.b) for \(S\) if \(\ell\) is an upper bound for \(S\) and \(\ell \le b\) for all upper bounds \(b\) of \(S\).
Problem 10.4 Give least upper bounds for the following subsets of \(\mathbb{R}\).
- \([0,1) = \{ x \in \mathbb{R}\ | \ 0 \le x < 1 \}\).
- \([0,1] = \{ x \in \mathbb{R}\ | \ 0 \le x \le 1 \}\).
An ordered field \(R\) is said to be complete if it satisfies the following:
Least Upper Bound Axiom Every non-empty subset of \(R\) which is bounded from above has a least upper bound.
There exists a complete ordered field. Any two such fields are isomorphic. \(\blacksquare\)
The proof of this is beyond the scope of this course. Many books on analysis begin by just assuming that there exists such a field. Actually we began this course by assuming familiarity with \(\mathbb{R}\) as well as \(\mathbb{N}\), \(\mathbb{Q}\) and \(\mathbb{Z}\).
The unique complete ordered field whose existence is asserted by Theorem 10.1 is called the field of real numbers and denote by \(\mathbb{R}\).
All properties of the real numbers follow from the defining properties of a complete ordered field. For example, one can prove that if \(a \in \mathbb{R}\) and \(a > 0\), then there is a unique element \(x \in \mathbb{R}\) such that \(x^2=a\) and \(x > 0\).
It can be shown that \(\mathbb{Q}\) is not complete. For example, the set \[S=\{ x \in \mathbb{Q} \, \vert \, x^2 < 2 \}\] is bounded from above but has no least upper bound in \(\mathbb{Q}\). Since we assume \(\mathbb{R}\) is complete, the set \(S\) does have a least upper bound \(\ell\) in \(\mathbb{R}\) which one can prove is positive and satisfies \(\ell^2=2\).
We also observe that just as we defined subtraction in a ring by the rule \[a-b=a+(-b),\] we define division in a field as follows:
Let \(a\) and \(b\) be elements of a field. If \(b \neq 0\) we define \[a/b = \frac a b = a \div b = a\cdot b^{-1}\] where \(b^{-1}\) is the inverse of \(b\) with respect to multiplication.
Under the assumption of the existence of a complete ordered field \(\mathbb{R}\), we can define \(\mathbb{N}\), \(\mathbb{Z}\), and \(\mathbb{Q}\) as follows. First we define \(\mathbb{N}\).
Say that a subset \(S\) of \(\mathbb{R}\) is inductive if it satisfies both of the following conditions:
- \(1 \in S\).
- If \(n \in S\), then \(n+1 \in S\).
Then we define the natural numbers \(\mathbb{N}\) to be the intersection of the collection of all inductive subsets of\(\mathbb{R}\).
Let \(1\) denote the identity of \(\mathbb{R}\). Define \(2 = 1 + 1\), \(3 = 2 + 1\), \(4 = 3+1\), \(5= 4+1\), \(6 = 5+1\), \(7 = 6 + 1\), \(8 = 7 + 1\), \(9 = 8+1\).
If we start with only the axioms for a complete ordered field, we have initially only the numbers \(0\) and \(1\). From the above definition we obtain in addition the numbers 2,3,4,5,6,7,8,9. Using the fact that for each \(a \in \mathbb{R}\) we have \(-a \in \mathbb{R}\) we get also \(-1, -2, -3, -4, -5, \dots\), as well as numbers such as \[\frac 1 2 = 2^{-1}, \quad \frac 1 3 = 3^{-1}, \quad \frac 1 4 = 4^{-1}, \quad \frac 2 3 = 2 \cdot 3^{-1}, \dots\]
Show that each of the following is an inductive subset of \(\mathbb{R}\).
- \(\mathbb{R}\).
- \(\{ x \in \mathbb{R}\, | \, x \ge 1 \}\).
- \(\{1,2 \} \cup \{ x \in \mathbb{R}\, | \, x \ge 3 \}\).
- \(\{1,2,3 \} \cup \{ x \in \mathbb{R}\, | \, x \ge 4 \}\).
From Definition 10.7 and Definition 10.8 one may prove the following two theorems:
\(\mathbb{N}\) is an inductive subset of \(\mathbb{R}\). \(\blacksquare\)
If \(S \subseteq \mathbb{N}\) and \(S\) is inductive then \(S=\mathbb{N}\). \(\blacksquare\)
Problem 10.5
(a) Prove that \(2, 3, 4,\) and \(5\) are elements of \(\mathbb{N}\).
(b) Prove that \(2+2=4\), \(2 \cdot 2 = 4\). (c) Prove that \(1 < 2 < 5\).
Here are a few examples of things that can be proved by using induction (this is short for The Principle of Mathematical Induction).
Problem 10.6 Prove that \(n \ge 1\) for all \(n \in \mathbb{N}\). Hint: Let \[S=\{ n \in \mathbb{N}\ | \ n \ge 1 \}.\] Prove that \(S \subseteq \mathbb{N}\) and \(S\) is inductive. Conclude from the Principle of Mathematical Induction that \(S=\mathbb{N}\). This is equivalent to the statement \(n \ge 1\) for all \(n \in \mathbb{N}\) and completes the proof.
Problem 10.7 Prove that \(2^n > n\) for all \(n \in \mathbb{N}\).
Problem 10.8 Prove Part 3 of Theorem 7.2. Hint: divide the problem into two parts. First prove \(f(a^n) =f(a)^n\) for all \(n \in \mathbb{N}\) using induction. Use Theorem 7.2, Part 1 to handle the case \(n=0\) and use Theorem 7.2, Part 2 and the laws of exponents to handle the case where \(n\) is negative.
Problem 10.9 Prove that \(0 < \frac 1 2 < 1\).
Problem 10.10 As noted above it may be proved that if \(a \in \mathbb{R}\) and \(a > 0\) there exists a unique number \(x \in \mathbb{R}\) satisfying \(x^2=a\) and \(x > 0\). The number \(x\) is denoted \(\sqrt{a}\). Prove that \[1 < \sqrt{2} < \sqrt{3} < 2\] and \[\frac 52 < \sqrt{8} < 3.\]
Define \(\mathbb{Z}= \mathbb{N}\cup \{ 0 \} \cup -\mathbb{N}\) where \(-\mathbb{N}= \{ -n \ | \ n\in \mathbb{N}\}\).
The set \(\mathbb{Z}\) is a subring of the ring \(\mathbb{R}\) which we call the ring of integers. All of the properties of \(\mathbb{Z}\) that we are accustomed to follow from the axioms for \(\mathbb{R}\) and the above definitions. This includes things such as there is no integer \(x\) such that \(1 < x < 2\). In this course we will not take the time to develop all the known results of this nature.
\(\mathbb{Q}= \{ n/m \ | \ n, m \in \mathbb{Z}\mbox{ and } m \neq 0 \}.\)
The set \(\mathbb{Q}\) is a subfield of \(\mathbb{R}\) called the field of rational numbers.
The field of complex numbers is the triple \((\mathbb{C}, +, \cdot)\) where \[\mathbb{C}= \{ (a,b) \ | \ a,b \in \mathbb{R}\},\] and addition and multiplication are defined as follows for \((a,b)\),\((c,d) \in \mathbb{C}\): \[\begin{aligned} (a,b) + (c,d) &=& (a+c,b+d) \\ (a,b) \cdot (c,d) &=& (ac-bd, ad+bc)\end{aligned}\]
\(\mathbb{C}\) is a field with zero given by \((0,0)\), identity given by \((1,0)\), the additive inverse of \((a,b)\) is given by \((-a,-b)\) and if \((a,b) \neq (0,0)\) then the multiplicative inverse of \((a,b)\) is given by \[(a,b)^{-1} = \left ( \frac a{a^2+b^2}, \frac {-b}{a^2+b^2} \right ).\] \(\blacksquare\)
This theorem is straightforward to prove. To save time we prove only the following:
Problem 10.11 Prove that \((0,0)\) is the zero of \(\mathbb{C}\) and the additive inverse \(-(a,b)\) of \((a,b) \in \mathbb{C}\) is given by \((-a,-b)\).
Problem 10.12 Prove that \((1,0)\) is an identity for \(\mathbb{C}\), that \((0,1)^2 = -(1,0)\) and that if \((a,b) \neq (0,0)\) then the multiplicative inverse of \((a,b)\) is given as stated in the theorem.
If we write for \(a, b \in \mathbb{R}\) \[a+bi =(a,b), \quad a = (a,0), \quad bi=(0,b), \quad i=(0,1)\] then \[i^2 = -1\] and we can consider \(\mathbb{R}\) as a subset of \(\mathbb{C}\) and the addition and multiplication on \(\mathbb{R}\) agrees with that on \(\mathbb{C}\) for elements of \(\mathbb{R}\). That is, in this notation \(\mathbb{R}\) is a subfield of \(\mathbb{C}\).
We lack the time in this course to discuss any of the many applications of complex numbers in mathematics, engineering and physics.
Problem 10.13 Using the notation above for elements of \(\mathbb{C}\), let \(z= 2 + 3i\), \(w=-2 + 4i\) and \(\theta = (-1/2) + (\sqrt{3}/2)i\). Write the following in the form \(a + bi\) where \(a\) and \(b\) are real numbers:
- \(z + w\).
- \(zw\).
- \(z^{-1}\).
- \(\theta^3\).
Let \(a,b \in \mathbb{R}\) and let \(z = a+bi \in \mathbb{C}\). The complex number \(\overline{z} = a - bi\) is called the conjugate of \(z\). \(\overline{z}\) is read “z conjugate".
Problem 10.14 Prove the the mapping \(\varphi : \mathbb{C}\to \mathbb{C}\) defined by \(\varphi(z) = \overline{z}\) is a ring isomorphism from \(\mathbb{C}\) to itself which is its own inverse. That is, for all \(z,w \in \mathbb{C}\) prove:
- \(\overline{zw} = \overline{z} \, \overline{w},\)
- \(\overline{z+w} = \overline{z} + \overline{w},\) and
- \(\overline{\overline{z}} = z\)
Another way to define \(\mathbb{C}\) is given in the next problem.
Problem 10.15 Let \[R = \left \{ \left ( \begin{array}{cr} a & -b \\ b&a \end{array} \right ) \ | \ a,b \in \mathbb{R} \right \}.\] This is a subring of the ring of all \(2 \times 2\) matrices \(M_2(\mathbb{R})\). In fact, \(R\) is a field. Prove that \(R\) is isomorphic (as a ring) to \(\mathbb{C}\).
Problem 10.16 Compare the formula in Theorem 10.4 for the inverse of a complex number to the formula for the inverse of a matrix of the form \[\left (\begin{array}{cr} a & -b \\ b&a \end{array} \right ).\]
We mention here a few interesting theorems about \(\mathbb{R}\) that we will not have time to cover in this course. Proofs may be found in introductory analysis courses and advanced algebra courses.
A set \(S\) is said to be countable if it is finite or if there is a one-to-one correspondence between \(S\) and \(\mathbb{N}\). A set which is not countable is said to be uncountable.
\(\mathbb{Q}\) is countable. \(\blacksquare\)
\(\mathbb{R}\) is uncountable. \(\blacksquare\)
A real number which is not in \(\mathbb{Q}\), that is, is not rational, is said to be an irrational number.
The set of irrational numbers is uncountable. \(\blacksquare\)
A real number is said to be algebraic if it is a root of some non-zero polynomial \(a_nx^n + \cdots + a_2x^2 + a_1x + a_0\) where the coefficients \(a_i\) are rational numbers. For example, \(\sqrt{2}\) is algebraic since it is a root of \(x^2 - 2\) and \(\sqrt[3]{(1+\sqrt{5})}\) is algebraic since it is a root of \(x^6-2x^3-4\). A rational number \(q\) is algebraic since it is a root of \(x-q\).
The set of algebraic numbers forms a countable subfield of \(\mathbb{R}\). \(\blacksquare\)
A real number which is not algebraic is said to be transcendental.
The set of transcendental numbers is uncountable. \(\blacksquare\)
However it is very difficult to prove that a particular real number is transcendental. Important examples of transcendental numbers are \(\pi\) and \(e\).
\(e\) is transcendental. \(\blacksquare\)
\(\pi\) is transcendental. \(\blacksquare\)