Integrity Domains and Integer Numbers

Summary:
This class introduces the concept of Integrity Domain, explains its relevance in the study of general algebra, and demonstrates some of its most important properties through formal proofs.

Learning Objectives:
By the end of this class, the student will be able to:

1. Understand the purpose of studying general algebra.
2. Understand the concept of an integrity domain.
3. Explain the basic common aspects between integrity domains and integer numbers.
4. Demonstrate the basic properties of integrity domains through formal proofs.

CONTENT INDEX
THE PURPOSE OF GENERAL ALGEBRA AND PRIOR KNOWLEDGE
FROM INTEGER NUMBERS TO INTEGRITY DOMAINS
BASIC COMMON ASPECTS BETWEEN INTEGRITY DOMAINS AND INTEGER NUMBERS
PROPERTIES OF INTEGRITY DOMAINS AND INTEGER NUMBERS
EXERCISES

The Purpose of General Algebra and Prior Knowledge

The main objective of general algebra is the study of the entire variety of possible mathematical systems. Here we will study several such systems, among the most important of which are natural and integer numbers, and through the latter, we will reach integrity domains.

\mathbb{N}= \{1,2,3,4,\cdots\}

\mathbb{Z}= \{0,\pm 1,\pm 2,\pm 3,\pm 4,\cdots\}

From Integer Numbers to Integrity Domains

We will begin our study with integer numbers, and the reason for proceeding this way is that they share the most similarities with the majority of the numerical systems we will review in this study.

Instead of trying to define what integer numbers are, we will start by assuming that, whatever they may be, they satisfy certain properties. To do this, a set of axioms is chosen so that it is possible to infer all the properties that we intuitively associate with integers.

All these things are done through the Peano axioms of the Naturals by introducing the basic operations of arithmetic. Following this axiomatic method and expanding the various operations on natural and integer numbers, new numerical sets emerge, such as rationals, irrationals, reals, complex numbers, quaternions, octonions, and many more.

Then, if we observe integer numbers, we will see that they have properties that will repeat across all other numerical sets, such as the existence of a multiplicative identity, an additive identity, and distributive laws. By referring to these concepts, we can establish a language that allows us to talk about all these sets simultaneously. It is in this context that words like

• Integrity Domain
• Ring
• Group
• Vector Space

And many more terms of this kind emerge. We will focus our efforts on studying Integrity Domains first.

Basic Common Aspects of Integrity Domains and Integer Numbers

To explain what an integrity domain is, we will rely on the properties that we understand very well from integer numbers. In this context, if a, b, and c are integers, then the following laws hold:

1. Commutative Laws:
   • a+b = b + a
   • ab = ba
2. Associative Laws:
   • a+(b+c) = a+b+c = (a+b)+c
   • (ab)c = abc = a(bc)
3. Distributive Law:
   • a+(b+c) = a(b+c) = ab+ac

In addition, there exist certain special elements known as identity elements:

• Additive Identity: a+ c = a \leftrightarrow c=0
• Multiplicative Identity: ac = a \leftrightarrow c=1

The object denoted by 0 is the additive identity, while the one denoted by 1 is the multiplicative identity.

Integers also possess additive inverses. Each integer has an additive inverse that, when added to it, yields the additive identity.

• Additive Inverse: a+ c = 0 \longleftrightarrow c=-a

Additive inverses are recognized by the “-” sign that accompanies them.

Finally, there exists a simplification law expressed through the relation:

(c\neq 0 \wedge ca = cb) \longleftrightarrow (a=b)

These properties we have reviewed hold for many other sets: real numbers, complex numbers, polynomials, etc. Thus, we call any set that satisfies these properties an Integrity Domain.

DEFINITION: An Integrity Domain is any set D equipped with an addition and multiplication operation such that:

• a,b\in D \longrightarrow a+b \in D
• a,b\in D \longrightarrow ab \in D

Additionally, the associative, commutative, and distributive laws hold, D contains additive and multiplicative identities (each of these being unique), and finally, the simplification law holds.

Example of an Integrity Domain

Consider the set A=\{a+b\sqrt{3}\; |\; a,b\in \mathbb{Z}\}. This set, equipped with the usual operations of addition and multiplication, is an integrity domain because it satisfies the laws of commutativity, associativity, and distribution, has additive and multiplicative identities, and finally, an additive inverse.

• Additive Identity: 0+0\sqrt{3}
• Multiplicative Identity: 1+0\sqrt{3}
• Additive Inverse: Every element a+b\sqrt{3} has an additive inverse -a-b\sqrt{3}

Most importantly, this set A is closed under addition and multiplication, meaning that if we take x,y\in A, then x+y\in A and xy\in A. This is easy to verify: If a_1 + b_1\sqrt{3} and a_2 + b_2\sqrt{3} are elements of A, then:

\begin{array}{rl} (a_1 + b_1\sqrt{3}) + (a_2 + b_2\sqrt{3}) &=(a_1+a_2) + (b_1 + b_2)\sqrt{3} \in A\\ \\ (a_1 + b_1\sqrt{3}) (a_2 + b_2\sqrt{3}) &= a_1a_2 + a_1b_2\sqrt{3}+b_1a_2\sqrt{3} + 3b_1b_2 \\ &=(a_1a_2 + 3b_1b_2) + (a_1b_2 + b_1a_2)\sqrt{3} \in A \end{array}

Properties of Integrity Domains and Integer Numbers

The Additive Identity of an Integrity Domain is Unique

This can be proven by contradiction: Suppose that there exist two additive identities, namely 0 and 0^\prime. Then, we would have:

\begin{array}{rll} (1) & 0\neq 0^\prime & \text{; Premise}\\ (2) & a+0 = a & \text{; Premise: $0$ is the additive identity}\\ (3) & b+0^\prime = b & \text{; Premise: $0^\prime$ is the additive identity}\\ (4) & 0^\prime + 0 = 0^\prime & \text{; Substituting $a=0^\prime$ in $(2)$}\\ (5) & 0 + 0^\prime = 0 & \text{; Substituting $b=0$ in $(3)$}\\ (6) & 0 = 0^\prime & \text{; From $(4,5)$ and the commutativity of addition}\\ (7) & \bot &\text{; From $(1,6)$} \end{array}

From this reasoning, we conclude that:

\{0 \neq 0^\prime, a + 0 = a, b + 0^\prime = b\}\vdash \bot.

Thus, by contradiction, we obtain:

\{a + 0 = a, b + 0^\prime = b\}\vdash 0 = 0^\prime.

That is, if there are two additive identities, then they must be the same, and therefore, the additive identity is unique.

The Multiplicative Identity is Also Unique

The proof is practically analogous to the previous one. If there were two: 1 and 1^\prime, then we could follow this reasoning:

\begin{array}{rll} (1) & 1\neq 1^\prime & \text{; Premise}\\ (2) & 1\cdot a = a & \text{; Premise: $1$ is the multiplicative identity}\\ (3) & 1^\prime \cdot b = b & \text{; Premise: $1^\prime$ is the multiplicative identity}\\ (4) & 1\cdot 1^\prime = 1^\prime & \text{; Substituting $a=1^\prime$ in $(2)$}\\ (5) & 1^\prime \cdot 1 = 1 & \text{; Substituting $b=1$ in $(3)$}\\ (6) & 1 = 1^\prime & \text{; From $(4,5)$ and the commutativity of multiplication}\\ (7) & \bot &\text{; From $(1,6)$} \end{array}

Therefore, we reach the conclusion that:

\{1 \neq 1^\prime, 1a= a, 1b = b\}\vdash \bot.

Thus, by contradiction, we obtain:

\{1a= a, 1b= b\}\vdash 1 = 1^\prime.

That is, if there are two multiplicative identities, then they must be the same, and therefore, the multiplicative identity is unique.

The Simplification Law for Addition Holds

This is what we do when we eliminate terms in an equality:

a+b = a+c \longleftrightarrow b = c

It is not difficult to prove this situation; we only need to follow the reasoning below:

\begin{array}{rll} (1) & a+b = a+c & \text{; Premise} \\ (2) & a+b-a = a+c-a & \text{; From $(1)$, subtracting $a$ from both sides} \\ (3) & (a-a)+b = (a-a)+c & \text{; From $(2)$, using commutativity and associativity} \\ (4) & 0+b = 0+c & \text{; From $(3)$ and Additive Inverse} \\ (5) & b = c & \text{; From $(4)$ and Additive Identity} \\ \end{array}

Since this reasoning can be applied in both directions using the same steps, we obtain:

a+b=a+c \dashv \vdash b=c

Which is equivalent to saying:

\vdash a+b=a+c \longleftrightarrow b=c

The Additive Identity is Also a Multiplicative Absorber

This simply means that for every a in the integrity domain, the following holds:

a\cdot 0 = 0

This is also easy to prove; we only need to follow the reasoning below:

\begin{array}{rll} (1) & a\cdot a + a\cdot 0 = a\cdot (a+0) & \text{; Distributive Laws}\\ (2) & a\cdot a + a\cdot 0 = a\cdot (a+a-a) & \text{; From $(1)$ and Additive Inverse}\\ (3) & a\cdot a + a\cdot 0 = a\cdot a + a\cdot a - a\cdot a & \text{; From $(2)$ and Distributivity}\\ (4) & a\cdot 0 = a\cdot a - a\cdot a & \text{; From $(3)$ and Addition Simplification}\\ (5) & a\cdot 0 = 0 & \text{; From $(4)$ and Additive Inverse}\\ \end{array}

Sign Rule:

The product of numbers with the same sign is always positive; the product of numbers with opposite signs is always negative. The proof of this property is also simple:

\begin{array}{rll} (1) & a\cdot b = a\cdot b + 0 & \text{; Additive Identity}\\ (2) & a\cdot b = a\cdot b + (a)\cdot(-b) - (a)\cdot(-b) & \text{; From $(1)$ and Additive Inverse}\\ (3) & a\cdot b = a\cdot (b -b) - (a)\cdot(-b) & \text{; From $(2)$ and Additive Inverse}\\ (4) & a\cdot b = a\cdot 0 + (-a)\cdot(-b) & \text{; From $(3)$ and Additive Inverse}\\ (5) & a\cdot b = (-a)\cdot(-b) & \text{; From $(4)$ and Multiplicative Absorber}\\ \end{array}

Therefore: ab = (-a)(-b)

For opposite signs, the reasoning is similar:

\begin{array}{rll} (1) & a\cdot(-b) = a \cdot (-b) + 0 & \text{; Additive Identity} \\ (2) & a\cdot(-b) = a \cdot (-b) + a \cdot b - a \cdot b & \text{; From $(1)$ and Additive Inverse} \\ (3) & a\cdot(-b) = a \cdot (b-b) - a \cdot b & \text{; From $(2)$ and Distributivity} \\ (4) & a\cdot(-b) = a \cdot 0 - a \cdot b & \text{; From $(3)$ and Additive Inverse} \\ (5) & a\cdot(-b) = - a \cdot b & \text{; From $(4)$ and Multiplicative Absorber} \\ \end{array}

Therefore: a(-b) = -a(b)

If the Product of Two Numbers is Zero, at Least One of Them is Zero

Another property that is also widely used is the following:

ab=0 \leftrightarrow (a=0 \vee b=0)

Its proof is also simple:

\begin{array}{rll} (1) & \{a=0\} \models a\cdot b = 0 & \textbf{; Multiplicative Absorber} \\ (2) & \models a=0 \rightarrow a\cdot b = 0 &\text{; TD$(1)$} \\ (3) & \models \neg (a\cdot b = 0 ) \rightarrow \neg(a=0) &\text{; CPI$(2)$} \\ (4) & \{\neg (a\cdot b = 0 ) \}\models \neg(a=0) &\text{; RTD$(3)$} \\ (5) & \{\neg (a\cdot b = 0 ) \}\models \neg(b=0) &\text{; Analogous to $(4)$} \\ (6) & \{\neg (a\cdot b = 0 ) \}\models \neg(a=0) \wedge \neg(b=0) &\text{; $\wedge$-int$(4,5)$} \\ (7) & \models (\neg (a\cdot b = 0 )) \rightarrow \neg(a=0) \wedge \neg(b=0) &\text{; TD(6)} \\ (8) & \models \neg(\neg(a=0) \wedge \neg(b=0) ) \rightarrow (a\cdot b = 0 ) &\text{; CPI(7)} \\ (9) & \models (a=0 \vee b=0) \rightarrow (a\cdot b = 0 ) &\text{; DM(8)} \\ (10)& \{a\neq 0 , a\cdot b=0\} \models b=0 & \textbf{; Multiplicative Absorber}\\ (11)& \{a\cdot b=0\} \models a\neq 0 \rightarrow b=0 & \text{; TD(10)}\\ (12)& \{a\cdot b=0\} \models \neg(a\neq 0) \vee b=0 & \text{; $\rightarrow$-Def(11)}\\ (13)& \{a\cdot b=0\} \models a=0 \vee b=0 & \text{; DN(12)}\\ (14)& \models (a\cdot b=0) \rightarrow (a=0 \vee b=0) & \text{; TD(13)}\\ (15)& \models (a\cdot b=0) \leftrightarrow (a=0 \vee b=0) & \text{; From (9,14)} \end{array}

Exercises

Let a, b, and c be any elements of an integrity domain D. Prove that the following properties hold:

1. (-a)=(-1)a [SOLUTION]
2. -(a+b)=(-a) + (-b) [SOLUTION]
3. a(-b)=-(ab) [SOLUTION]
4. -(-a)=a [SOLUTION]
5. a(b-c) = ab - ac [PROPOSED]
6. (a-b)+(b-c) = a-c [PROPOSED]
7. For all a\in D, there exists a unique 1 such that a\cdot 1 = a [SOLUTION]
8. xx = x \leftrightarrow (x=1 \vee x=0) [PROPOSED]