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1. Def: A set A along with at least one binary operation written (A,*) where * is the binary operation is called an Algebraic Structure. Examples of algebraic structures (N,+), (N,x), (W,+), (W,x), (Z,+), (Z,x), (Z,+,x), (Q,+), (Q,x), (Q^*,x) and some properties of the structures: Commutative law for +, for x, Associative law for +, for x, the Identity property for + (0 in (W,+), (Z,+), (Q,+)) and for x (1 in (N,x), (W,x), (Z,x), (Q,x) and (Q^*,x)): the Inverse property in (Z,+) and (Q,+) every element has an additive inverse, in (Q^*,x) every element has a multiplicative inverse. Although (N,x) and (W,+) are both algebraic structures with associative, commutative binary operations and both satisfy the identity property these two structures are different algebraically: In (Z,+,x) the operation x distributes over the operation +

· Definitions. A binary operation on a set A is a function from AXA to A. If * is a binary operation on A we write *((a,b)) = a*b. If (A,*) is an algebraic structure then * is said to be associative provided the following statement is true (TFSIT): For all a,b,c ÎA, (a*b)*c = a*(b*c). In this case we write a*b*c unambiguously. If (A,*) is an algebraic structure then * is said to be commutative provided TFSIT): For all a,b ÎA, a*b = b*a. If (A,*) is an algebraic structure then (A,*) has an identity with respect to * provided TFSIT: There exists an element eÎA so that for all aÎA, a*e = a and e*a = a.

Creating binary operations with specific properties on A = {a,b} and on A = {a,b,c}.

HW:

q Is the operation of subtraction on the set of all integers (a) commutative? (b) associative? In each case explain your answer.

q Show that division by non-zero elements is right distributive, but not left distributive over addition in the set of rational numbers.

q For the set of integers, define the operation * as follows: a*b = a + b – ab where and b are any integers and + and - are the usual addition and subtraction and ab is the usual product of integers. Find the identity element in the set relative to this operation. Let n be an arbitrary integer and find the inverse of n relative to the operation *.

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2. Define left and right distributivity. Prove that various algebraic structures have or do not have an identity. In (Z,+), 0 is an identity and 3 + x = 0 is solvable. The solution is called the (additive) inverse of 3 and we know it by the name -3.

Definitions. Let (A,*) be an algebraic structure with identity e. An element aÎA has an inverse (or -*inverse) in A (with respect to *) provided TFSIT: There exists an element xÎA so that a*x = e and x*a=e. We denote the inverse of a by the symbol a ^–1. We say the structure (A,*) with identity satisfies THE INVERSE PROPERTY provided every element of A has an inverse. Let (A,*) be an algebraic structure and let SÍA. The set S is closed (with respect to * ) provided TFSIT: If s, tÎS then s*tÎS.

EXAMPLES. The additive inverse of 3 in (Z,+ ) is 3 ^–1 which we denote by –3. The multiplicative inverse of 3 in (Q,x) is 3 ^–1 which we denote by 1/3 and which is also called the reciprocal of 3. The algebraic structure (Q*,x) has identity 1 and satisfies THE INVERSE PROPERTY. The set of odd numbers is not closed in (Z,+). The set of odd numbers is closed in (Z,x). Let S={nÎZ|GCD(n,100)=20}. S is not closed in (Z,+) and S is not closed in (Z,x).

The abstract case of the 8x8 table defined algebraic structure.

HW:

q Let subtraction be defined by a-b=c means c is an element of the set and a = b+c. Determine whether each set below is closed with respect to subtraction.

a) the set of all integers

b) the set of positive integers

c) the set of all integers which are divisible by three

d) In each case explain your answer.

q For the set of rational numbers, we define two binary operations $ and # (in terms of ordinary arithmetic operations) as follows: a $ b = 2ab and a#b = a + 2b for any two rational numbers a and b.

a) Is $ commutative? If so, explain your answer and if not, give a counter example.

b) Is # commutative? If so, explain your answer and if not, give a counter example.

c) Prove or disprove that $ is associative.

d) Prove or disprove that # is associative.

e) Prove or disprove that $ is left distributive over #.

f) Prove or disprove that # is left distributive over $.

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3. Laboratory One with Mrs. Buracker. Topics included Place Value, Order of Operations, Integers, One step equations and operations on whole numbers using various games, manipulatives and models accessible to middle school students.

4. Discuss form of proof. Example. Define # on Q by (p/q)#(s/t) = ps/t where each of p/q and s/t are in reduced form and ps/t is expressed in reduced form. Can you solve (a/b)#X = (c/d) for all choices of a/b and c/d in reduced form? Suppose X = p/q is a solution. Then (ap)/q = c/d. This is easily solvable for p/q as the reduced form of c/(ad) for all choices of a/b except for a = 0. So the answer is yes it is solvable as long as a is unequal to 0. For homework see if you can solve X#(a/b) = (c/d) for all choices of a/b and c/d in reduced form.

· Definitions. A matrix is a rectangular array of objects. Each object’s position in the array is uniquely determined by its row (®) location and its column (¯) location. If a matrix has m rows and n columns we say it is an m by n matrix. Let A be such a matrix then we denote A by [a_ij] where i = 1,2,3,…,m and j = 1,2,3,…,n and a_ij denotes the entry in position (i,j) where i and j denote, respectively, the row and column locations. Let S denote an algebraic structure (S,*). An m by n matrix where each a_ijÎS is called an m by n matrix over S. The set of all m by n matrices over S is denoted by M_m,n(S). Let A = [a_ij] and B = [b_ij] be elements of M_m,n(S). We extend * a binary operation of S to a binary operation on M_m,n(S) by the following.

AB = C = [c_ij] where c_ij = a_ij * b_ij for i = 1,2,3,…,m and j = 1,2,3,…,n. This is called the pointwise extension of *. (M_m,n(S), ) is an algebraic structure and shares some of the properties of (S,*). Two matrices A and B in M_m,n(S) are equal provided TFSIT a_ij = b_ij for all i = 1,2,3,…,m and j = 1,2,3,…,n. Matrices and systems of equations motivate the following definition. Let A be an m by n matrix over Q and let B be an m by t matrix over Q. The product matrix AB = C =[c_ij] where c_ij = a_i1 * b_1j + a_i2 * b_2j + a_i3 * b_3j + … + a_im * b_mj for i = 1,2,3, …, m and j = 1,2,3,…,t. This multiplication takes the m elements in A from row i (®) and multiplies each by the corresponding m elements in B from column j (¯) and adds these products together to get c_i,j producing the m by t matrix C.

HW:

q Given the matrices find (a) A+B, (b) AB, (c) BA, (d) 5A, (e) A-B, (f) B-A.

q Given the matrices find (a) X(YZ), (b) (XY)Z, (c) X(Y+Z), (d) XY + XZ, (e) (X+Y)Z, (f) XZ + YZ

q Find a nonzero divisor of zero in M the set of all 2x2 matrices with integer entries.

q Let E be the set of all English words and let S be the set of all “letter strings” [finite lists of letters, possibly repeated] of our alphabet. Notice that E is a subset of S. In each instance below a process * that makes sense on each of these two sets is described. [The Greek letters a and b represent arbitrary elements of these sets.] Answer each of the following questions for each exercise:

1. Give two more examples to show how * works.

2. Is * a binary operation on S? on E? Justify each answer with a reason or counterexample.

3. If * is a binary operation on the set, say whether * is associative and/or commutative, supporting your answers with reasons or counterexamples.

a) a*b is formed by putting a and b next to each other, forming a single string. For example, moon*glow = moonglow.

[This is like the password process used by CompuServe and other electronic services.]

b) a*b is the word or string that in alphabetical order comes first. For example, trip*trap = trap.

[This is a basic part of any word-sorting algorithm.]

c) a*b is the string of letters, including repeated letters, that are common to a and b, arranged in alphabetical order. For example, beaker*knee = eek.

d) a*b is the number of letters in the longer word or string, or if they have the same number of letters, it’s that number. For example, movie*theater = 7.

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5. Retraction #1 from day 2.

· Definitions. A group is an algebraic structure (G, ¨) in which the following properties hold.

(1) ¨ is associative. [For all x, y, z in G, x¨ (y¨z) = (x¨y) ¨z.]

(2) There is an identity in G. [There exists an element e in G so that for all x in G, e¨x=x and x¨e=x.]

(3) (G, ¨) satisfies the inverse property. [ For every x in G there exists an element in G such that ¨x=e and x¨=e. The element is called the inverse of x.]

Examples: (Z,+) is a group.

(N,+) is not a group because property (2) fails to hold.

(Q,+) is a group.

(W,+) is not a group because property (3) fails to hold.

(M_m,n(Z),+) is a group.

(M_2,2(Z),x) is not a group because property (3) does not hold.

(Q*,x) is a group.

(Q,x) is not a group because the element 0 does not have an inverse element.

· Theorems

1. For any elements a, x and y in a group G, if a¨x = a¨y then x = y.

2. For any elements a, x and y in a group G, if x¨a = y¨a then x = y.

3. For any elements a and b in a group G, the equation a¨x = b has one and only one solution (for x) in G.

4. For any elements a and b in a group G, the equation x¨a = b has one and only one solution (for x) in G.

The proof of Theorem 1. is motivated by scratch paper work. [Assume x = y and try to rewrite each of x and y using the group properties and try to introduce a factor of a to the left of each of x and y.]

A proof of Theorem 1. Let a, x and y be elements of G. Assume a¨x = a¨y. Since G is a group and a is an element of G, by property (3) there is an inverse for a in G, namely . Also since ¨is a binary operation it follows that ¨(a¨x)= ¨(a¨y). The associative law guarantees that (¨a)¨x =(¨a)¨y so e¨x=e¨y by property (3). Finally, x=e¨x=e¨y=y by property (2).

The proof of Theorem 3. is motivated by scratch paper. [Assume there is an element x in G so that a¨x = b. Solve for x using Theorem 1 by rewriting b with a factor of a on the left. a¨x=b=e¨b=(a¨)¨b and Theorem 1 gives x =¨b.]

A proof of Theorem 2. Let a and b be elements of G. Since G is a group, the element a has inverse and the element ¨b is an element of G. Now, a¨(¨b) = (a¨)¨b by the associative law and (a¨)¨b = e¨b = b by properties (3) and (2) respectively. Therefore the element x = ¨b is a solution to the equation a¨x=b for any a and b in G. Suppose y were any other solution to the equation. Then a¨x = b = a¨y so a¨x = a¨y and Theorem 1. implies x = y. So, there is only one solution to the equation and it is ¨b

HW:

q Prove Theorems 2,4

q The converse of theorem 1 is: For any elements a,x and y in a group G, if x = y then a¨x = a¨y.

a) Does this require proof as a result in group theory? Why or why not? (Hint: consider the definition of a binary operation and use this in your discussion.)

b) Euclid stated this same idea by saying “If equals are multiplied by equals, the results are equal.” Discuss the appropriateness of saying “multiply both sides of an equation by the same thing” or “multiply one equation by another” in the context of Group theory.

q Discuss what each of theorems 1-4 says about elements of (a) the integers and (b) the rational numbers in the context of the kinds of equations with (a) integer and (b) rational number coefficients that are guaranteed to have unique solutions for x.

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6. Show how to prove the uniqueness of an object. Rework the material from Theorems 2 and 4.

· Theorems

5. In any group G, there is only one identity.

6. In any group G, for each element x in G, there is only one inverse element

A proof of Theorem 5. Let e and f be identities in a group G. By the identity property we know that e¨g=g¨e=g for all elements g in G. In particular, since f is in G we have that e¨f = f. Similarly, since f is an identity in G, f¨g=g¨f=g for all elements g in G. So e¨f = e. Therefore we have e = e¨f = f. So the identity of G is unique.

A proof of Theorem 6. Let x be an element in G. We know is an element of G and x¨=e. Thus the element