Math Enrichment

Edition Date 12/2004

Developed for The Association For Conceptual Studies

Presented by: Richard Singer

website: www.fractions-plus.com email: richard-acs@worldnet.att.net

Purpose: The initial focus of most secondary school mathematics courses is on ordinary algebra. Ordinary algebra focuses on the study of laws for various numerical systems and the application of these laws in solving equations and in transforming algebraic expressions. More advanced secondary courses use ordinary algebra with functions whose domain is the field of real numbers. The emphasis in all of these courses tends to be more on techniques than on concepts and conceptual relationships. This gives a very limited appreciation of contemporary mathematics. This present resource is a Type 4 resource that briefly indicated a slightly broader perspective by introducing some structures whose algebra is similar in some ways but different in others from those usually studied in secondary mathematics. For an application of these structures you can consult the resource Using Remaider Concepts.

The intended audience is secondary teachers of mathematics, home schooling parents, as well as anyone else who thinks algebra is fun. So the presentation is often more abstract than the presentation found in most books on ordinary algebra. Some of the presentation may seem aimed at a reader with less sophistication than indicated in the prerequisites. This is because some readers may want to adapt some of these ideas for use with students. For more such ideas, as well as an additional perspective on the ideas from this unit, I recommend The Theory of Remainders by Andrea Rothbart: rothbart@webster.edu

Prerequisites: A good understanding of ordinary algebra.

Main Concepts: When dividing whole numbers, we can think 25 divided by 6 as having a quotient of 4 and a remainder of 1, and 18 divided by 6 as having a quotient of 3 and a remainder of 0. Etc. Altho many applications of whole number focus on quotients, there are also reasons to focus on remainders. We use the notation Â_nx for the remainder when x is divided by n. The two main remainder concepts in this resource are the concept of a cyclic ring and the concept of a morphism. These concepts will be introduced after defining the concept of a ring.

Rings: The laws below are satisfied not only by various numerical structures, such as the integers, but also by a variety of other algebraic structures have similar operations. Structures that satisfy such laws are called commutative rings. The ring of integers is denoted as Z.

A cyclic ring is a finite commutative ring whose elements can be generated from 1 by addition. This means that a finite list of the form 1, 1+1, 1+1+1, 1+1+1+1, … gives all elements of the ring.

The Cyclic Ring Z₆: The cyclic operations {+,·} for the set {0,1,2,3,4,5} are indicated in the tables below. We will later show that this set with these operations is a ring, which we denote as Z₆. Z₆ is a cyclic ring, since 2 = 1+1, 3 = 1+1+1, 4 = 1+1+1+1, 5 = 1+1+1+1+1, 0 = 1+1+1+1+1+1.

One way to think of these operations is to imagine Z₆ arranged as in a clock. To obtain x+y (mod 6), start at x and move y positions clockwise. To obtain x·y (mod 6), start at 0 and move y positions clockwise x times.

Notation: The convention of ordinary algebra for omitting the multiplication signs and parenthesis will be used for ring operations. For instance a·b is abbreviated as ab, and the distributive law as a(b+ c) = ab+ac..

As a minor illustration using the cyclic Z₆ ring, we will solve the puzzle below.

Tea Party Puzzle: Ms Army, Ms Banjo, Ms Clay, Ms Dumont, Ms Far, Ms Grey had a tea party. They were seated at a circular table. One of these women was pretty, one realistic, one slim, one talkative, one unreliable, one quiet. Ms Banjo sat opposite the unreliable one. The pretty one sat opposite Ms Clay, who sat between the quiet one and the unreliable one. The slim one sat opposite Ms Army, next to the pretty one, to the left of the unreliable one. Ms Far sat between the realistic and the slim one. Ms Grey sat to the right of Jan who was opposite the talkative one. Who is Jan?

Comment: A manifest way to solve this the Tea Party Puzzle is to use tokens {a,b,c,d,f,g} for their last names and tokens {p,q,r,s,t,u} for their characteristics. We can then use the clues to arrange them in a circle, starting with the slim one at the top. For a pictorial account of this way of solving the puzzle you can go to the manifest solution at the end of this paper.

Solution Using Z₆: Let {a,b,c,d,f,g} be the positions occupied by the women whose last names start with those letter. Let j be the position of the woman whose first name is Jan. Let {p,q,r,s,t,u} be the positions occupied by the women whose characteristics start with those letter. Clues involving the relations ‘left of’ or ‘opposite’ are easily translated using mod 6 addition. That s is left of u translates as s = u+1, that b is opposite u as b = u+3, etc. The translation of ‘x is between y and z’ is more subtle. Check some instances to see that this implies y+z = 2x. The tea party information gives the following equations, along with a multitude of inequalities such as g ¹ c.

b = u+3, p = c+3, q+u = 2c, s = a+3, s = u+1, p = s+1, r+ s = 2f, j = g+1, j = t+3

Since s occurs in more equations than any other variable, it is convenient to label the seat occupied by s as 0. This gives s = 0, u = 5, p = 1, r = 2f, a = 3, b = 2, c = 4, q = 3. If t = 2 then g = 4. However c = 4, so t ¹ 2. This leaves t = 4 & r = 2. From this, j =1, f = 1, j = f. Thus Jan is Ms Far. You can check to see that using any other value for s also gives j = f.

Modular Operations: One way to calculate x· y (mod 6) is to multiply as usual and then take the remainder, i.e. to obtain 5· 5 = 1 (mod 6), use  ₆25 = 1. A similar comment applies to calculating
x+y (mod 6). This generalizes easily to give a cyclic ring Z_n for each positive integer n. To focus attention on an operation being cyclic you can use a subscript of n with the operation symbol. This can be tedious, so when it is clear from context that we are using a specific cyclic operation subscripts are omitted, including subscripts with  .

The cyclic operation +_n and ·_n are defined as follows: x+_ny = Â_n(x+y) & x·_ny = Â_n(x·y)

In general, to check the results of dividing m by n, take the quotient times n and add the remainder, checking to see that the result is m. Also make sure the remainder is a natural number less than n.
This idea can be used to define the remainder (mod n) function for any positive integer n.

Â_nm = m-qn, where qn is the largest multiple of n that is less than or equal to m.

The remainder concept also applies to negative integers. For example, to see that Â_6- 25 = 5, observe that
^-25 = ^-5· 6+5. Note that 5+1 = 0, so 5 = ^-1 in Z₆, so Â₆^-25 = -Â₆25. In general, the remainder of the opposite of an integer is the Z_n opposite of its remainder: Â_n(-m) = -Â_nm in Z_n.

Remainder Equality Law: Two integers have the same remainder (mod n) if and only if n is a factor of their difference. This law is stated compactly below using the symbol ‘½ ’ for ‘is a factor of’. If this is not intuitively obvious, look as some examples.

Â_ny = Â_nx Û n½(y-x)

Morphisms Law: Suppose we sum two integers and take the remainder (mod n) of the result. We will obtain the same answer if we first take their remainders (mod n) and then sum (mod n). Likewise, suppose we multiply two integers and take the remainder (mod n) of the result. We will obtain the same answer if we first take their remainders (mod n) and then take the product (mod n). In this sense a remainder function maps the ring of integers to a cyclic ring in a way that preserves the ring structure, and we call the remainder function a morphism. This morphism law is stated compactly below, abbreviating Â_n as Â. You may want to check some example.

Â(x+y) = Âx+_nÂy Â(x·y) = Âx·_nÂy

Ring Theorem: Each Z_n is a commutative ring with respect to the cyclic operations.

This follows from the fact Z is a ring by using the morphism laws. For instance, to prove the distributive law, let a,b,cÎ Z_n and apply Â_n to both sides of the distributive law for Z. A wide variety of laws for ordinary algebra are satisfied in any ring.

Terminology: If c is an element of some Z_n, and there is an element b in Z_n such that bc = 1, the we call b a multiplicative inverse of c. Appendix contains a proof that if c has a multiplicative inverse then it is unique. I denote the multiplicative inverse of c as /c. When c does not have a multiplicative inverse in Z_n, the usual practice is to treat /c as undefined. However we can allow / to be a function on Z_n, by arbitrarily letting /c be 0 if c does not have a multiplicative inverse. I adopt this unusual practice as a matter of personal taste. It makes certain result easier to state. For instance the formula we do not need to qualify the formula /(a b) = (/a)(/b). It will still hold even if either a or b has no inverse.

Units in Z_n: If c has a multiplicative inverse in Z_n then c is called a unit of Z_n.

Annihilators in Z_n: If c ¹ 0 & ac = 0 then a is called an annihilator of c. So 0 is an annihilator of all other elements of Z_n. Elements of Z_n may have other annihilators. For example, all even numbers are annihilators of 6 in Z₁₂.

Remark: A field is a special type of commutative ring, namely a ring in which 0 is the only annihilator.

Example: Looking at the table given earlier, you can see that Z₇ is a field but Z₆ is not a field.

Roots of Equations: An equation such as 2x+ 4 = 2 in Z₇ can be solved in the same way it would be solved in ordinary algebra. First add 3 (i.e. the additive inverse of 4) to both side to obtain 2x = 5. Then multiply both sides by 4 (i.e. the multiplicative inverse of 2) to obtain x = 6. For the equation 2x+ 4 = 2 in Z₈, we could first add 4 to obtain 2x = 6. Since 2 does not have a multiplicative inverse in Z₈, we must use some other strategy. By inspection we see that x must be either 3 or 7. On the other hand, we can solve the equation 5x+ 4 = 2 in Z₈, by first obtaining we 5x = 6 and then multiplying by 5 to obtain x = 6. Note that in Z₈ each odd number is its own multiplicative inverse and each even number is an annihilator.

We can use annihilators to find root of some equations in Z₈. For 2x+4 = 2 we can add 6 and the factor to obtain 2(x+ 1) = 0. Since 2 only annihilates 0 and 4 we again have x = 3 or x = 7. To use annihilators to for 2x+4 = 3 in Z₈, add 5 to obtain 2x+1 = 0.Since 2x+1 does not appear to factor, multiply giving 4 = 0, so this equation has no roots in Z₈.

Notational Remark: The same names used for integers will be used as names for elements of cyclic rings. This gives each of the elements of a cyclic ring an infinite number of names, each of which is the standard name of the integer having that element as its remainder. The elements three in Z₃ can be named as follows.

¼ , ^-9, ^-6,^-3, 0, 3, 6, 9,¼ ¼ , ^-8, ^-5,^-2, 1, 4, 7, 10,¼ ¼ ^-10, ^-6, ^-4,^-1, 0, 2, 5, 8,¼