algebra: table of contents

Algebra: table of contents

Topic SlideThe concept of algebra ADefinitions and example BFundamental properties of ops. CBoolean algebra DQ. following from definition ETheorems of BA F1-7BA: one outstanding issue GSet algebra H1-4

Exercises

© 1991-1997 by A.ShafarenkoAll rights reserved.

Copying any material referenced directly orindirectly on this page is permitted only aspart of a degree course at Surrey University,England and only for the duration of thecourse. Any form of distribution of the audio,graphic or typeset data constituting the fileslinked up, directly or indirectly, to this page isprohibited without the prior written permissionof Dr Alex Shafarenko.

Welcome!

Here is a compact introduction into abstract algebra.The material presented here presupposes goodknowledge of logic, especially axiomatic theory andproof techniques. Section “Logic” would be a sufficientbackground for this section provided that the learnerhas developed reasonable skills of symbolicmanipulation.

We shall use Boolean algebra as an example of abstractalgebra. This will be a real example in the sense ofbuilding up a complete theory of an abstract object fromfoundations to applications. In addition it addressespractical needs as Boolean algebra describes variouscomputations with sets which would otherwise require aseparate study of set theory. In this sections set theory isintroduced painlessly as we borrow the notional fabricfrom Boolean algebra using what mathematicians callhomomorphism (never mind what this means).

Study the proofs of all theorems presented here verycarefully. Following a proof through brings significantbenefits to the learner as it not only gives the reason tobelieve the statement of the theorem, but also showswhat the truth of this statement hinges upon. Thismakes one’s understanding of the meaning andlimitations of the theorem so much deeper.

The tutorial problems found here are slightly morecomplex than the ones offered in “Logic”. Still thelectured part provides enough guidance for a successfulsolution. Remember that only by solving problems canone realise the level of one’s knowledge —remembering facts in mathematics is necessary, butalas, not nearly sufficient!

The Author

click here for the instruction leaflet

2

A.Shafarenko, 1995. All rights reserved

AlgebraAlgebra

• The objectives of algebra– to formalise objects on which some operations are defined– to define primary properties of such operations as an

axiomatic theory– to derive and prove further properties of operations as

theorems

• The use of algebra– in computer programs– in logical circuits– in specifications– as an abstraction mechanism

A

3


Definition and ExampleDefinition and Example

Algebra = Set + Operations + Axiomatic Theory

• An algebra is a triple ( S, O, A), where S is a set of someobjects, O is a set of operations on objects in S yieldingresult objects from S (it is said that they are closed in S) andA is an axiomatic theory defining some properties of O.

• Axioms and theorems of algebra are often used as rewriterules

Example:Define an abstract algebra of integer numbers as follows:S is a set of objectsO is {+,×}A is (∃Z∀x) Z+x = x; (∀x,y) x+y = y+x;

(∃U∀x) U×x = x; (∀x,y) x×y = y×x;(∀x,y,z) (x+y)+z = x+(y+z); (∀x,y,z) (x×y)×z = x×(y×z);(∀x∃(−x)) x+(−x)= Z; (∀x,y,z) x×(y+z) = (x×y)+(x×z);

B

As usual we can only deal with objects we have defined previously. In the case of algebra,the definition is displayed on the slideand needs little comment. One thing that does need tobe said here is that there are two meanings to the word algebra. First there is a notion ofabstract algebra, which is a triplet of a set of objects, a collection of abstract operations onthem and an axiomatic theory constraining the properties of those operations. If you definethe algebra of numbers as shown in the example box, then the “numbers” could be anythingyou like, the “plus” and “times” could be any actions on pairs of those yielding a member ofthe same class as long as the axioms remain satisfied. An abstract algebra is a powerful toolof abstraction which enables us to study properties of mathematical models in general withthe results of those studies applicable to any particular realisation of the model.

This brings us close to the second meaning of algebra, which is concrete algebra. It is nolonger a model but is the same triplet except now the operations are procedurally defined.This means for every pair of objects in S there is a procedure that generates the result. Thisprocedure can be any mathematically acceptable method of obtaining the answer providedthat it gives the same answer for the same pair of objects whenever we use it. The axiomsshould still be satisfied, although they are no longer regarded as constraints. There is nofreedom in the choice of operations: the procedure should be guaranteed to conform to theaxioms. We therefore can use them as additional information about the algebra, which canbe useful for simplifying expressions, showing two expressions to be equal, etc.

In this section we shall discuss abstract algebras and will demonstrate that a lot ofinformation can be gleaned within an abstract framework irrespective of any proceduresused in a concrete algebra afterwards. The abstract setting can be assumed to haveprediction power, as it allows one to determine the properties of abstract algebras evenbefore they are formulated --- as long as they confrom to the abstract algebra that they aremodelled after. On the next few slides this power will become more apparent.

4


Fundamental Properties of OperationsFundamental Properties of Operations

• Most operations are associative• Distributivity is a property of a pair of operations

Commutativity: (∀x,y) x ⊗ y = y ⊗ x

Associativity: (∀x,y,z) (x ⊗ y) ⊗ z = x ⊗ (y ⊗ z)

Identity i of ⊗ : (∀x) i ⊗ x = x

Inverse x−1 of x with respect to ⊗ : x−1⊗ x = i

Distributivity of ⊗ over ⊕ : (∀x,y,z) x ⊗ (y ⊕ z)= (x ⊗ y) ⊕ (x ⊗ z)

C

These properties of operations are considered fundamental because theygenerate a rich set of identities in an algebra that advance abstract analysisvery far without any need for concrete procedural information. There is nodoubt as to their origin either: every one of them occurs in arithmeticalready. Abstract algebras often use these properties as an axiomatic basis.

5


Boolean AlgebraBoolean Algebra

Boolean Algebra = Set S of any objects+ Set O of 2 operations: ∧ (conjunction) and ∨ (disjunction)

closed in S+ axiomatic theory

axiom 1: both ∧ and ∨ are commutativeaxiom 2: ∧ and ∨ are mutually distributiveaxiom 3: each of ∧ , ∨ has an identity, denoted as

T (top ) and ⊥ (bottom ), respectively axiom 4: (∀ x ∃ x′) x ∧ x′=⊥⊥ and x ∨ x′= T

x′ is said to be a complement of x

• Duality: exchanging ∧ T and ∨ ⊥ results in the same algebra

D

George Boole1815-1864

This is a famous symmetrical definition of abstract Boolean algebra. Notethat there is no mention of ones and naughts in it, no recourse to logicalgates or any procedural information defining the effects of conjunction anddisjunction.

6


BA: Questions following from definitionBA: Questions following from definition

• Can I use a set of any size for Boolean Algebra?• Is the identity of conjunction (disjunction) unique?• Is the complement of x unique?• What about associativity?• What can we tell about the complement of top (bottom)?• What if bottom (top) is used with conjunction

(disjunction)?• If I write an expression in BA, what rules can I use to

simplify it?

E

These questions owe their existence to the abstract nature of Boolean algebradefined earlier. Every one of them can be answered in a concrete Booleanalgebra quite easily, even by direct test. However, if we could find theabstract answers we would benefit from these answers being applicable toall concrete Boolean algebras. Thence comes the analysis on the followingslides.

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T1: Identity is uniqueT1: Identity is unique

Theorem 1: The bottom (top) elements is unique.

Proof (by contradiction):

F1

8


T2: Complement is uniqueT2: Complement is unique

Theorem 2: For all x, the complement of x is unique.

Proof (by contradiction):

F2

9


T3,4: Properties of complementT3,4: Properties of complement

Theorem 3: Double complement law (∀ x) (x′)′= x

Proof (th.3):

Proof (th.4):

Theorem 4: T′= ⊥

F3

10


T5,6: Idempotence and cross-identityT5,6: Idempotence and cross-identity

Theorem 5: Idempotent law (∀ x) x ∧ x = x ∨ x = x

Proof (th.5):

Proof (th.6):

Theorem 6: (∀ x) x ∨ T= T and x ∧ ⊥= ⊥

F4

11


T7,8: Absorbtion and cancellation lawsT7,8: Absorbtion and cancellation laws

Theorem 7: Absorption law (∀x,y ) x ∧(x ∨ y) = x ∨ (x ∧ y)= x

Proof (th.7):

Proof (th.8):

Theorem 8: Cancellation law

(∀ a, x, y) (a ∧ x = a ∧ y and a′ ∧ x = a′ ∧ y ) → x = y

F5

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T9. AssociativityT9. AssociativityTheorem 9: Associativity (∀ x, y, x) (x ∧ y) ∧ z = x ∧ (y ∧ z)

Proof:

F6

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T10. De Morgan’s LawsT10. De Morgan’s Laws

Theorem 10: De Morgan’s Laws (∀ x, y) (x ∧ y)′ = x ′ ∨ y ′ and the dual statement

Proof:

F7

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BA: One outstanding issueBA: One outstanding issue

• How many elementsin S?– Can be one element

T = ⊥, but this isn’tinteresting

– For two elementsthere is a nontrivialalgebra…BA(2): T ≠ ⊥ (prove!)

S = { T, ⊥ }– More elements?

Can’t tell just yet,require theory ofrelations...

P Q P ∨ Q Proof

⊥ ⊥ ⊥ Axiom 3

⊥ T T Axiom 3

T ⊥ T Axioms 1,3

T T T Theorem 6

or?

and?

G

P Q P ∧ Q Proof

⊥ ⊥ ⊥ Theorem 6

⊥ T ⊥ Axioms 1,3

T ⊥ ⊥ Axiom 3

T T T Axiom 3

Now that all major theorems have been proven, let us see where this gets us.We know quite a lot about BA already, despite the total lack of proceduralinformation. Now, if there was a procedurally defined BA, what size wouldit be? What is the smallest BA?

It follows from the BA axioms that the top and bottom elements must exist.The axioms suggest nothing at all as to whether these should be different orthe same. It is easy to see however that if there is just one element to BA, i.e.if top=bottom, then the procedural definition is obtained immediately fromthe fact that the operations are closed in S. The operans and result of eitheroperator has to be that bottom/top element. This satisfies all the otheraxioms of abstract BA. However, a one-element algebra is somewhatpathological and has no practical purpose: it will satisfy many sets of axiomsbesides those of BA.

If there are at least two elements in the BA, top is not equal to bottom. Atwo-element BA is therefore a different story: using the axioms and onetheorem we have proven previously, we can provide just one proceduraldefinition! In a sense there is exactly one concrete BA of 2 elements and thatis what is known as the Boolean Algebra in most textbooks.

Of course BAs may exist in larger sets and they don’t have to be uniquelydefined procedurally either. Do they exist? It is interesting to note that weare still unable to provide the answer, since it does not seem to follow fromthe available axioms and theorems. In fact the whole new topic, theory ofrelations, will have to be introduced before this issue is resolved.

15


Set algebraSet algebra

• A set algebra can be viewed as a concrete Boolean algebra

The powerset 33(A) of set A is the set of all subsets of A. 33(A)={ x | x ⊂ A }

The (concrete) Set Algebra on set U is Set S = 33(U) with some finite set U+Set O of operations: {∪, ∩} (set-theoretical union and intersection )A ∪ B = { x | x ∈ A or x ∈ B }A ∩ B = { x | x ∈ A and x ∈ B }

{z|P(z)} means “the set of z such that P(z)”

Example: A={a, b}; 33(A)={∅, {a},{b},{a, b}}

H1

Here is a concrete algebra, defined procedurally. It has an independent valuesince we concern ourselves with sets in many areas of mathematics andapplications. It also happens to be a Boolean algebra, which we shalldemonstrate presently.

16


SA is an instance of BASA is an instance of BA

A

C

B A

C

B

=

A

C

B A

C

B

=

A ∪ (B ∩ C) =(A ∪ B ) ∩ (A ∪ C)

A ∩ (B ∪ C) =(A ∩ B ) ∪ (A ∩ C)

Consider SA on some set U1. ∪ and ∩ are (obviously) comutative2. ∪ and ∩ are mutually distributive

3. ∅ is the identity of ∪, and U of ∩4. Define set complement of X in U as

the set of elements of U that donot belong to X; then

X′ ∩ ∩ X = ∅ and X′ ∪ X = U

U

X

∪ and ∩ satisfy all BA axioms ∪ and ∩ satisfy all BA axioms

H2

The box on the right-hand side shows a constructive proof of mutualdistributivity diagrammatically. We introduce what is known as a Venndiagram, which is the schematic view of sets as intersecting blobs (usuallycircles). The purpose of the diagram is to show all combination of setmemberships, e.g. the elements that belong to only one given set, two givensets but not the third one, none of the sets, etc. The diagram therefore revealsthe structure of a number of intersecting sets with respect to thoseintersections. Complete Venn diagrams could only be drawn for no morethan three sets on a plane (why?). Inside the domains of the Venn diagram,all elements can be treated as one since there is no membership property thatcould distinguish them. So if any property which only involves membershipcan be shown to be had by a triplet of sets on a Venn diagram that should beregarded as proof. However, one must also bear in mind that all statementsthat can be made about membership of elements can be regarded aspredicates and any proof that results can be re-formulated in terms of thoseto the complete exclusion of the diagram as such. The diagram is thereforeno more than a useful device for making proofs more comprehensible forlearners.

As an excercise, give a purely predicative proof of mutual distributivity bydefining the membership predicates for both right- and left-hand sides of theidentities and showing them to be equivalent.

What have we achieved? We have shown that union plays the role ofdisjunction and intersection of conjunction, with set complement being aproper complement in terms of abstract BA. Set algebra is hence proven tobe a concrete BA.

17


SA: Reaping the harvestSA: Reaping the harvest

• Since SA is an instance of BA, all BA theorems apply

1. Empty set ∅, U and the complement of any subset X of U are unique

2. ∅′ = U and X ′′ = X

3. X ∪ X = X and X ∩ X = X

4. X ∪ (X ∩ Y) = X ∩ (X ∪ Y) = X

4. (A ∩ B = A ∩ C and A′ ∩ B = A ′ ∩ C) → B = C

5. Set operators ∪ and ∩ are associative

6. De Morgan’s laws: (X ∩ Y) ′ = X ′ ∪ Y ′

H3

The nice thing about abstraction is that it is usually very potent. As soon aswe have established the relationship between the concrete set algebradefined procedurally and abstract BA, we can immediately assert theproperties inherent in the latter as being ones of the former. The slide showsa list of those, which are proven by matching them with the BA theorems wehave proven before.

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Subtle points about setsSubtle points about sets• Relations between sets• Other operations on set: difference and symmetric difference• Cardinality : the number of elements in a set• Sets defined recursively: paradoxes• Cartesian product of sets

Difference: A−B = A ∩ B′

Symmetric difference: A ∆ B = (A− B) ∪ (B−A)

Equality of sets: A = B means (∀ x) (x ∈ A → x ∈ B) and (x ∈ B→ x ∈ A)

Cartesian product: A × B = {(x, y ) | x ∈ A and y ∈ B}

Cardinality laws: |A × B| = |A||B|; |A ∪ B| = |A|+|B|− |A∩B|;

| 33(A) | = 2|A|

H4

A final note. Göthe’s Faustus said “Theory is dead, my friend, but the tree oflife is green forever” or something to this effect in perfect German. There isusually more than one aspect to a concrete object, so any theory orabstraction is bound to be incomplete (at least in this sense). Concrete setshave more procedurally defined operations on them than abstract BA cancouch. Of course equality on sets is such a primary thing that we oughtreally to have introduced it much earlier. The rest are useful concreteoperations and properties dealt with in applications. The only thing thatrequires some caution is Cartesian product. It looks like an ordinaryoperation, but observe that it may not be closed in the set it draws itsoperands from, and it is not even associative!

Algebra. Tutorial problems

1. Give examples of noncommutative and nonassociative operations. Can an associative operation be noncommutative?Give examples.

2. Prove that element eL such that (∀x) eL ⊕ x = x and element eR such that (∀x) x ⊕ eR = x are equal no matter what otherproperties of the operation ⊕ are.

3. Prove that if an operation ⊕ is associative and has the identity e then the definitions(∀x) x−1 ⊕ x = e and (∀x) x ⊕ x−1 = e define the same inverse element x−1.

4. For a Boolean algebra R, define the relation ≤ as follows: for every x,y∈ R, x ≤ y if x \/ y = y.a) Prove the same could be defined by demanding that for every x,y∈ R, x ≤ y if x /\ y = x.b) Prove that if x ≤ y and y ≤ x then x = y.c) Prove that if algebra R has more than two elements, there exist x and y such that neither x ≤ y nor y ≤ x.d) If for some x,y and z, x ≤ y and y ≤ z then x ≤ z

5. Given that |A∪B| = 20 and |A∩B| =10, what are the minimum and maximum values of |B|? Find these values assumingthat |P(A)| = 4096.

6. Which of the following statements are true and which are false?a) the empty set is an element of any setb) the empty set is a subset of any setc) {{}}={}d) {{},{}}={{}}

Algebra. Solutions1. Noncomputative operations are: subtraction, division, concatenation (of character strings), matrix multiplication,cross-product of vectors, etc. Associative operations can be noncommutative, for example concatenation of strings.

2. Consider eL ⊕ eR . According to the first equation this is equal to eR , according to the second one, to eL . If thesame expression is equal to both eL and eR then obviously eL = eR .

3. Suppose they are different. Let us use different letters for these variables: (∀x) x−1 ⊕ x = e and (∀x) x ⊕ X−1 = e.Now consider the expression x−1 ⊕ x ⊕ X−1. Since ⊕ is associative, (x−1 ⊕ x ) ⊕ X−1 = x−1 ⊕ (x ⊕ X−1) ore ⊕ X−1 = x−1 ⊕ e and so, by virtue of the property proven in the previous exercise, X−1 = x−1. This contradicts ourassumption and so proves the theorem.

4.a) We need to prove the equivalency x \/ y = y if and only if x /\ y = x. Prove it one way first: x /\ y = x impliesx \/ y = y. Indeed, join the premise and rewrite the left-hand side of the corollary:

x \/ y = (x /\ y) \/ y = y\/( y/\ x) = y by virtue of the absorption law.

The other half of the prove: x \/ y = y implies x /\ y = x , is fully analogous: join the premise and rewrite the left-handside of the corollary: x /\ y = x /\ (x \/ y) = x again by virtue of the absorption law.

b) Indeed, since x /\ y = x and y /\ x = y, we notice that the left-hand sides are equal by virtue of commutativity, hencethe right-hand sides are equal too.

c) If there are more than 2 elements in the algebra , then there exists x≠⊥,T. For every x there exists the complementx′, x ≤ x′ means x \/ x′ = x′ and x′ ≤ x means x′ \/ x = x. However, according to the definition of the complementelement, x \/ x′ = T≠ x,x′.

d) We need to prove that if x /\ y = x and y /\ z = y, then x /\ z = x. Direct proof:x /\ y = x ⇒ x /\ (y/\ z) = x ⇒ (x /\ y)/\ z = x ⇒ x /\ z = x

5. Since the union of the sets consists of 20 elements, neither set can be larger than 20. (Indeed if it was, the unionwould be at least as large.) Since the intersection of the sets has 10 elements, neither set could be smaller than 10, forthe intersection cannot be larger than either set. We conclude that both A and B are between 10 and 20 elements insize. Now on to the second question. Using the formula |A∪B| = |A| + |B| − |A∩B|, we find that |A| + |B| = 30. Sincethe size of the power-set |P(A)| = 4096 = 212, then |A| = 12 and |B| = 18.

6. a) false; b) true; c) false; d) true.

Using Recorded Lectures

Welcome!

This leaflet will help you use your audiographic lectures. Please make aprinted copy of it for your reference and read it carefully before you start.

The first page of the browser shows the contents of the module. When youclick the mouse on the topic of your choice you will be presented with apage on which a copy of the relevant slide and some additional notes aredisplayed. Unless you have got it already, make a hard copy of this pageand have it handy.

Every slide has an icon located just above the title, which looks like this:

By clicking the mouse on it, an audiographic playback of the lecturedmaterial associated with the slide is initiated. Be patient as the playersoftware loads up, this may take some time especially if you are using ashared computer system.

Now you can see the player window (it could be part of your Internetbrowser window or a separate one) with a scroll bar at the bottom. Whenthe slide is fully loaded, click anywhere on it to start the presentation. If youare using a shared facility, make sure you are wearing your headphoneswhich should be plugged in to the audio socket on your computer. Pleaseavoid sounding the presentation through the built-in speakers when thismay disrupt the work of any other users.

As the presentation goes on, you may see some hand-written notesappearing on the slide. Various items may be highlighted to draw yourattention to them and there may be some animated drawings to illustratevarious points the lecturer makes. Use your hard copy to take any notesyou feel important. The presentation may appear to you to go too fast (infact it will do in most cases initially). Clicking on the slide will suspend theplayback to give you more time to comprehend the material. A further clickwill resume the presentation. Even though a slide rarely has more than 5minutes of audio associated with it, you may wish to repeat or skip variousparts of the show as you go along. The horizontal scroll bar is intendedprecisely for this purpose. By clicking on the bar, you may skip to the next(previous) audio fragment and clicking at either end of the bar moves youup and down the lecture by one step at a time. You can also grab it with themouse and position it as you want, taking the visual cue from the graphicson the slide. Experiment with it for a while until you feel comfortable,then proceed to your work. Do not forget to read the additional notes on theprint-out when you have finished with the lecture.

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algebra: table of contents

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