Sub chapter 1. Fundamentals. 1)If Z 2 = 0,1 then (Z 2, +) is group where. + is addition modulo 2. 2) Z 2 Z 2 = (0,0),(0,1)(1,0)(1,1)

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1 DISCRETE MATHEMATICS CODING THEORY Sub chapter. Fundamentals. )If Z 2 =, then (Z 2, +) is group where + is addition modulo 2. 2) Z 2 Z 2 = (,),(,)(,)(,) Written for the sake of simplicit as Z 2 Z 2 =,,, 3) Z 2 Z 2 Z 2 = (,,),(,,),(,,),(,,), (,,),(,,), (,,), (,,) Z 3 2 =,,,,,,, = (a, a 2, a 3 )/ a,a 2, a 3 Z 2 Elements belonging to Z n 2 is Word. It is also string of s and s Eample :,,, are words like a a 2 in Z 2 2 = (a,a 2 )/ a,a 2 Z 2,, etc are words like a a 2 a 3 are in Z 3 2 a i is either or Similarl Z 2 Z 2 Z 2.. Z 2 = Z n 2 = (a a 2 a 3 a 4 a 5 a 6 a 7 a 8 a n ) / a,a 2, a 3 a n Z 2 a a 2 a 3 a 4 a 5 a 6 a 7 a 8 a n is a word in Z n 2 let us observe one eample. Component wise addition (,,,,,) + (,,,,,) = (,,,,,) With addition modulo 2 operations += ; += ; + =;+= Under Component wise addition (Z n 2 + ) is a Group

2 In the above figure we see that a word c is transmitted and received incorrectl as r(i.e. with error) But we readil observe that C = r + e a relation between c,e,r. Probabilit: Suppose in transmission of signals probabilit that is received as is p and that is received as is p 2 as depicted

3 In smmetric channel p = p 2 =p that is P( is transmitted and is received)=p( is transmitted and is received) = p Eqivalentl, probabilit that a transmitted signal ( OR ) is not received correctl is p In total in BINARY Smmetric Channel probabilit that a transmitted signal ( OR ) is received Incorrectl is p AND correctl is - p Eamples :The word c = is transmitted with error (pattern) find the word r received.if probabilit that signal is received incorrectl is.5 find the probabilit that r is received Solution: transmitted signal is c =. Now p=.5 c = r = error (pattern) e = r = c+e = P( signal c is transmitted and r is received ) = P( r is received ) = (-p) p (-p) p p (-p ) p =p 4 (-p) 3 =.5 4 (-.5) 3 E 2.what is the probabilit that c is received as r

4 E 3.what is the error pattern Solution 3:As r = c+e ( )= () + (a a 2 a 3 a 4 a 5 a 6 a 7 a 8 ) Ans: e = E4. Al binar smmetric channel has probabilit,p =.6 of incorrect transmission of some signal of Z 9 2,sa,What is the probabilit that a transmitted signal has k(= to 9) errors and also ) no error 2) single error (3) double error

5 4) Triple error Solution 4: For n =9 k bit error P (occurrence k bit error ) =c n k p k (-p) n-k ) P(no error ) =P( k=)=.6 (-.6) 9-2) P(single error ) =P( k=)=.6 (-.6) 9-3)P(double error ) =P( k=2)=.6 2 (-.6) 9-2 4)P(Triple error) =P( k=3)=.6 3 (-.6) 6 E5. A binar smmetric channel has probabilit, p =.88 of correct transmission of some signal of Z 8 2,sa,What is the probabilit that a transmitted signal has k (= to 8) errors and also )eactl ero errors; (5)at least one error 2) eactl one error ; (6 at the most 7 errors 3) at the most one error; (7) not more than 2 errors 4) not more than 5 errors; (8)all the bits with error Ans 5. p =.88 = probabilit of correct transmission of some signal of Z 8 2, probabilit (a transmitted signal has k (= to 8) errors) = c n k p k (-p) n-k ) P(eactl ero errors) = c 8.2 k (.88) 8 2)P( eactl one error)= c 8.2 (.88) 7 ; 3)P(at the most one error)=p(k= OR k=) =P()+P() 4.P(more than 5 errors) =P(,,2,3,4,5)

6 = c 8.2 (.88) 8 + c 8.2 (.88) 7 +c (.88) 6 + c (.88) 5 + c (.88) 4 + c (.88) 3 ; = c 8.2 (.88) 8 + c 8.2 (.88) 7 ; (5)P(at least one error) =P(,2,3,4,5,6,7,8) =- P() (6) P( at the most 7 errors) =P(,,2,3,4,5,6,7) =-P(8) (7) P(not more than 2 errors) =P( OR OR 2) =P( + + 2) (8) P(all the bits with error) =P(8)= C (.88) ; Sub Chapter 2 Encoding and Decoding : An Encoding function is an one-to-one function E: Z m 2 Z n 2 which provides a means to detect OR correct Errors occurred during transmission of signals( s and s) The above process is known as Encoding process. A Decoding function is an onto function E: Z n 2 Z m 2 which provides a means to recapture the transmitted word The process is known as Decoding process. The word after encoding is code word. The rule given b encoding function(e) is known as CODE

7 EXAMPLES ( Coding function of e. 7 2 are known as (m,m+) parit-check code).an Encoding function E: Z 8 2 Z 9 2 is defined as each w w 2 w 3 w 4 w 5 w 6 w 7 w 8 Z 8 2,E(w) = w w 2 w 3 w 4 w 5 w 6 w 7 w 8 w 9 where w 9 = w i i = to 8.(Sum wrt modulo 2) Discuss the significance of E(w). FURTHER if p =.2 is the probabilit of incorrect transformation of signals find the probabilit that a code word (sa ) is received with at most one error Solution :the significance of E(w). E(w) = w w 2 w 3 w 4 w 5 w 6 w 7 w 8 w 9 where w 9 = w i i = to 8.(Sum wrt modulo 2) if w = w w 2 w 3 w 4 w 5 w 6 w 7 w 8 () has even number of s then w 9 is (2) has odd number of s then w 9 is In both the cases coded word contains even number of s Net we find probabilit. As p =.2 probabilit of receiving word with at the most one error = p (-p) 9 +p (-p) 8 =.2 (-.2) +.2 (-.2) 8 Q2.An Encoding function E: Z m 2 Z m+ 2 is defined as each w w 2 w 3 w 4 w 5 w 6 w 7 w 8 Z 8 2,E(w) = E(w w 2 w 3 w m ) = w w 2 w 3 w m w m+ w m+ =, if w contains even number of s =, if w contains odd number of s and Decoding function D: Z m+ 2 Z m 2 with D(w) = D(r r 2 r 3 r m r m+ ) = r r 2 r 3 r m.encode (find codeword) of,,,,,, Z Dcode,,,,,,, Ans(2)

8 B the definition of E E()= E()= E()= E()= E()= E()= E()= 2.Dcoding D()= D()= D()= D()= D()= D()= D()= D()= (3m,m) Triple Repetition code Q3.An Encoding function E: Z m 2 Z 3m 2 is defined as each w w 2 w 3 w 4 w 5 w 6 w 7 w 8 Z 8 2,E(w) = E(w w 2 w 3 w m ) = w w 2 w 3 w m w w 2 w 3 w m w w 2 w 3 w m and Decoding function D: Z 3m 2 Z m 2 D(w) = D(r r 2 r 3 r m r m+ r 2 r 3 r 2m r 2m+ r 2 r 3 r m3m ) = s s 2 s 3 s m Where s i = if majorit of r i+m r i +2m r i+3m = if majorit of r i+m r i +2m r i+3m (a) Encode (Or find Code words) of E: Z 3 2 Z 6 2,,,,,,,

9 (b) Decode E: Z 6 2 Z 2 2,,,,,,, Ans 3a)Encoding(Or find Code words) E() = ; E() = E() = ; E()= E () = ; E() = E() = ; E( ) = b) Decoding D()=; D() =; D()=; D() =; D()=; D() =;D()= ; D() = ; Q4.An Encoding function is E: Z 3 2 Z 9 2 Use triple repetition(9,3) code to Decode (a),,, (b)find THREE different received words r for which D(r ) = Ans4.D: Z 9 2 Z 3 2 triple repetition(9,3) code to Decode are s i = if majorit of r i+m r i +2m r i+3m = if majorit of r i+m r i +2m r i+3m a)d()= D() = D()= D( ) = (b) D() = D() = D() =

10 Sub Chapter 3 Hamming Metric Consider a word Z m 2. = 2 3. m Z m 2. i, The Number of s in is Weight of denoted b wt() or Eample: wt() = wt() = 5 wt() = wt() = 8 Hamming distance: = 2 3. m, = 2 3. Y m Z m 2.Then the Hamming Distance, (d (, )) between and is the NUMBER of positions in which i i d (, )) is also same as NUMBER of s in + Eample: = = d (, )= 6 ALSO + = and wt( + ) = 6 d(, ) = wt( + ) = is no.of s in + Eample2: () = () = 5 () = () = 5 d (, )= 3 ALSO + = () and wt( + ) =3 d(, ) = wt( + )

11 Theorems Theorem: For all, Z m 2. Wt(+ ) Wt( ) + Wt(+ ) Theorem 2: For all,, Z m 2. (i) d (, )= d (, ) (ii) d (, ) (iii) d (, ) = if and onl if = (iv) d (, )+ d (, ) d (, ) Function d is known as metric hence d is known as distance OR metric Or Hamming metric The pair (Z m 2,d) is known as Hamming metric space. Sphere: For a Z m 2 and positive integer k S(a,k)= Z m 2 / d (, a) k is called sphere of radius k with centre a. For a= S(a,2) =,,,,,, Minimum distance: E: Z 2 2 Z 6 2 E is Triple repetition code E()= E()= E()= E()= d(e(), E()) =6 d(e(), E()) =3 d(e(), E()) =3 d(e(), E()) =3 d(e(), E()) =6 d(e(), E()) =3 Minimum distance ORMinimum distance of code words OR Minimum E (min E ) = is 3

12 CORRECTION and DETECTION of errors Question: If E: Z 2 2 Z 6 2 with min E 5 then How man () errors can be detected (2) errors can be corrected E()= E()= E()= E()= ANS = 5 () errors of weight - = 5- =4be detected (2) errors of weight ( - )/2= 2 can be corrected Question:2 If E: Z 2 2 Z 6 2 FIND () min E then (2) error detecting capacit (3) errors correcting capacit Where E()= E()= E()= E()= Ans 2 d(e(), E()) =6 d(e(), E()) =3 d(e(), E()) =3 d(e(), E()) =3 d(e(), E()) =6 d(e(), E()) = 3 ()min E = is 3 (2) error detecting capacit: E can detect errors with weight = - = 5- =4 (3) errors correcting capacit: errors of weight ( - )/2= 2 can be corrected

13 Question:3 If E: Z 3 2 Z 6 2 FIND () min E then (2) error detecting capacit (3) errors correcting capacit Where E()= E()= E()= E()= E()= E()= E()= E()= ANS 3First Distnces are found d(e(), E()) =3 d(e(), E()) =4 d(e(), E()) =3 d(e(), E()) =4 d(e(), E()) =3 d(e(), E()) =3 d(e(), E()) =4 d(e(), E()) =3 d(e(), E()) =4 d(e(), E()) =3 d(e(), E()) =4 d(e(), E()) =3 d(e(), E()) =6 d(e(), E()) =4 d(e(), E()) =3 d(e(), E()) =3 d(e(), E()) =3 d(e(), E()) =3 d(e(), E()) =2 d(e(), E()) =3 d(e(), E()) =2 d(e(), E() =3 d(e(), E()) =4 d(e(), E()) =3 d(e(), E()) =4 d(e(), E()) =2 d(e(), E()) =2 d(e(), E()) =2 Min E =2 = error detecting capacit: =E can detect errors with weight = - = 2- = (3) errors correcting capacit = errors of weight ( - )/2= / 2 can be corrected This means E cannot correct an error

14 Sub Chapter 4 :Generator Matri: Parit Check Matri Eample ) An encoding function : Z Z E is given b the generator matri G a) Determine all the code words. What can be said about the error detection capabilit of this code? What about its error correction capabilit? b) Find the associated parit - check matri H. c)use H to decode the received words:, Soln: We know that the given G is of the form A I G / 2, where 2 I and A a) We find that ) ( G E ) ( G E () E

15 E () These matri equations show that the code words are E ( ), E(), E(), E() From these, we find that d( E(), E()) 3, d( E(), E()) 4 d( E(), E()) 3 d( E(), E()) 3 d( E(), E()) 4 d( E(), E()) 3 Thus min( E ) 3. Therefore the code can detect all errors of weight 2 and can correct all single errors. b)the Parit-check matri H associated with G is given b H A T / I 3 We observe that H does not contain a column of s and further no two columns of H are identical. Therefore, H corrects single errors in transmission. c)for r, the sndrome of r is

16 T H Since this is a ero matri, the decoded message is got b retaining the first two components of r. The decoded message is therefore. For r, the sndrome of r is T H We observe that the matri is identical with the first column of H. Therefore, we change the first component of r (from to ) to get. This is the code word. The first two components of this code word, namel, is the original message. Eample 2) The generating function of an encoding function : Z Z E is given b G a) Find the code words assigned to and. b) Obtain associated parit - check matri. c)hence decode the received words :,. d)show that the decoding of is not possible b using H.

17 Soln: We note that A I G / 3, where 3 I and A a) We find that ) ( E ) ( E Thus the required code words are ) ( E and ) ( E. b) The parit check matri associated with G is / 3 I A H T c)for r, the sndrome of r is T H We observe that this matri is identical with the second column of H. Therefore, we change the second component in r (from to ) to get the word

18 c. The first three components of this code word gives the original message w. For r, the sndrome of r is T H We observe that this matri is identical with the third column of H. Therefore, we change the third component in r (from to ) to get the word c. The first three components of this code word gives the original message w. d)for r, the sndrome of r is T H We observe this matri is not a ero matri and is not equal to an column of H. Therefore, we cannot decode r b using H. Eample 3) The parit check matri for an encoding function : Z Z E is given b H a) Determine the associated generator matri. b)does this code correct all single errors in transmission?

19 Soln: a) we have H Which is of the form / I 3 A T. Accordingl, T A so that A Hence, the associated generator matri is / I 3 A G b)we observe that two columns of H (namel the 2 nd and 5 th ) are identical. Therefore, H does not provide a decoding scheme that corrects single errors in transmission.

20 sub Chapter 5 GROUP CODES let E: Z m 2 Z n 2, n > m be encoding function, C= E (Z m 2) = E (w) w/ Z m 2 =Set of CODES. Then C is known as group code. C is subgroup of Z n 2 Eample: E: Z 2 2 Z 6 2, E() = E() = E() = E() = C =,,, C is a group under addition modulo 2 Decoding with Coset leaders Eample :A group code C is defined b the generator matri G Decode the received words,,, with Cosets of c Matri G is 25 matri.hence E is E: Z 2 2 Z 5 2 given b [E(w) }=[ w]g for all w in Z 2 2 along with E()= [] similarl E() =[] E() =[] E() =[] E (Z 2 2) = C =,,, Is a sub group of Z 5 2 Let = C We find Coset

21 +C =,,, we continue process till when union of distinct cosets along with C is Z 5 2 order of C = 4,O( Z 5 2) = 2 5 =32 b Lagrange s theorem number of cosets = O( Z 5 2) / O( Z 5 2) = (32 / 4)=8 We construct table which shows the different cosets Rowwise is known as Decoding table. Elements in the first column C Are known as COSET LEADERS Ecept Decoding Process.received word r = This belongs to THIRD column Corresponding c in that column is c= appending the last three component we get the decode as 2.received word r =

22 Second column c= decode = 3. r= decoded = 4 r= decoded = Eample2:A group code C is defined b the generator matri G. Decode the received words,,with Cosets of c,if ANS 2 E( Z 3 2)= C =,,,,,,, O(C) = 8 O(Z 6 2) = 2 6 B Lagranges theorem Number of cosets (O(Z 6 2)/ O(C) ) = ((2 6 )/8) =8

23 received decoded

24 subchapter 6 RINGS The reader is familiar with the following properties of integers: ) The Z set of all integers is closed under the usual addition and the usual multiplication operations. In other words, and are binar operations in Z. 2) Z is an abelian group under. 3)The operation is associative in Z. 4) The operation is distributive over the operation in Z. That is, (i) a ( b c) a b a c and (ii) ( a b) c a c b c, a, b, cz A generaliation of the above said properties of Z to an abstract set leads to the definition of a ring. Definition of a Ring Let R be a non- empt set which is closed under two binar operations and. Then R together with these operations is called a Ring provided the following conditions (aioms) hold: ) R is an abelian group under the operation. 2)The operation is associative in R ; that is, a ( b c) ( a b) c, a, b, cr 3) The operation is distributive over the operation in R ; that is a ( b c) a b a c and ( a b) c a c b c, a, b, cr

25 Remarks: ) The operations and in a ring R are respectivel analogous to the usual addition and multiplication operations in Z. For this reason, is referred to as addition and is referred to as multiplication in an abstract sense. 2)In a ring, the roles of and cannot be interchanged. Because, a ring is an abelian group under ; it need not be a group under. 3)Since a ring is a group under,there must eist an identit element w.r.t. This identit element is analogous to the number and is called a ero of a ring. We denote it b a the smbol. Since the identit element in a group is unique, the ero element in a ring is unique. An element of a ring other than is referred to as a non-ero element. 4)Since a ring is a group under, ever element of a ring must have an additive inverse(i.e, an inverse under ). We denote the additive inverse of an element a b a; it is referred to as the negative of a. Since the inverse of ever element of a group is unique, the negative of ever element in a ring is unique. Further, ( a) a. 5)Since a ring is a group under and since cancellation laws hold in a group, the cancellation laws for addition hold in a ring. 6)For simplicit in the notation, we write 2 Also, we write a for Eamples of Rings 3 a a, a for a a a b as ab and a ( b) 2, etc. a. as b )The algebraic properties of integers listed in the beginning of this section shows that Z is a ring under the usual addition and multiplication.2)the algebraic properties of sets Q, R and C of all rational numbers, real numbers and comple numbers(respectivel) indicate that these sets are also rings under the usual addition and multiplication. 3) The algebraic properties of matrices indicate that the set n M of all square(real or comple) matrices of order n is a ring under matri addition and matri multiplication.

26 Some Basic Results The following theorem contains some simple results valid in all rings. Theorem: Let R be a ring and a b, c, be an three elements in R. Then ) a a 2) a( b) ( a)( b) ( ab) 3) a)( b) ab ( 4) ( a b) ( a) ( b) 5) a( b c) ab ac 6) ( a b) c ac bc Proof: )Since R is a group under with as the identit, we have ( a ) a ( ). But a ( ) ( a ) ( a ), b distributive law. Thus, ( a ) ( a ) ( a ). Therefore, b using the cancellation law(which is valid because R is a group under ). We have ( a ). Similarl, a 2) Using the distributive law and the result () proved above, we have a ( b) ab a( b b) a. Therefore, a( b) ( ab).similarl, ( a)( b) ( ab) 3)From the result (2), we have ( a)( b) { ( a) b} [ ( ab)]... Since R is a group under, we have ( ), for an R. Therefore, [ ( ab)] ab. Thus ( a)( b) ab. 4) Since R is an abelian group under, we have ( a b) {( a) ( b)} ( a b) {( b) ( a)} a { b ( b) ( a)} a ( a)

27 a ( a) Therefore, ( a b) ( a) ( b). 5) a( b c) a { b ( c)} ab a( c) ab ac. 6) ( a b) c { a ( b)} c ac ( b) c ac ( bc) ac bc. Some special tpes of Rings From the definition of a ring, we note that the multiplication operation in a ring is just associative and distributive(over the addition operation). B imposing etra conditions on this operation, we obtain certain special tpes of rings. Some of these are defined below. Commutative Ring A ring R is said to be a commutative ring if the multiplication operation in R is a commutative operation; that is if ab = ba, a, br. It is trivial to note that Z Q, R, and C are all commutative rings under the usual addition and multiplication. But, the ring of square matrices (of the same order) is not a commutative ring. Because, matri multiplication is not commutative. Ring with Unit A ring R is said to be a ring with unit if there eists an identit element in R w.r.t multiplication; that is, if there an element, which we denote b the smbol, in R such that a a a, a R. Then, is called the unit of the ring.

28 Since unit in a ring (if eists) is the multiplication identit and since the identit for a binar operation is unique, the unit in a ring (if it eists) is unique. It is trivial to note that Z Q, R, and C are all commutative rings with unit. The number is the unit of all these rings. The ring of all square matrices (of the same order n) is a non-commutative ring with unit; the unit matri of order n is the unit of this ring. The commutative ring of all even integers (under the usual addition and multiplication) does not have a unit. Ring with ero divisors If two non-ero elements a and b of a ring R are such that ab =, then a and b are called ero divisors in R. A ring containing at least one pair of ero-divisors is called a ring with ero-divisors. For eample, consider the ring M 2 of all real square matrices of order 2. In this ring, A and B are two non-ero elements such that AB=. Hence, A and B are ero divisors in the ring M 2. Thus, the ring of all real square matrices of order 2 is a ring with ero divisors. Remark: A ring R is said to have no proper ero-divisors if, for an a, br, ab implies a = or b =. Evidentl, Z, Q, R and C are rings with with no proper ero divisors. Integral Domain A ring R is said to be an integral domain if (i) R is a commutative ring with unit, and (ii) a, br, ab implies a = or b =. In other words, an integral domain is a ring with unit which contain no proper ero divisors.

29 It is evident that Z, Q, R and C are all integral domains (under the usual addition and multiplication). But the ring of square matrices are (of the same order) is not an integral domain. Because, the ring is not commutative on the one hand, and has ero divisors on the other hand. Divisor Ring An element in a ring R with unit is called a unit of R if a has an inverse in R under multiplication; that is if there eists b R such that ab = ba =. Then b is called a multiplicative inverse of a and is denoted b a. Since inverse of an element under a binar operation is unique, it follows that a ring R has a unit, and is a unit of R, then the multiplicative inverse of a is unique. A ring R with unit is said to be a division ring (or askew field) if ever non-ero element of R is a unit of R ; that is if for ever a ( ) R, there eists a R such that a a a a It is evident that. Q, R and C are all divisor rings. The ring of all non- singular matrices of the same order is also a divisor ring. But Z is not a divisor ring; because multiplicative inverse of a non ero integer is not an integer. Field A commutative divisor ring is called a field. In other words: A ring R is said to be a field if (i) R is commutative, (ii) R has a unit; and (iii) ever non-ero element of R has a multiplicative inverse in R. Evidentl Q, R and C are all fields. The ring of all square matrices (of the same order) is also not a field. Note : From the definition of a field, it follows that (i) a field has at least two

30 elements and. And (ii) The set of all non-ero elements of a field is an abelian groupunder multiplication. The following is a standard result: Ever field is an integral domain. For a proof this result see eample () below. But the converse of this result is not in general true. Note that Z is an integral domain, but not a field. The converse is true onl in the case of a finite integral domain. See eample 2 given below. 2 Eample Let R { a ib / a, b Z, i } with addition and multiplication defined b ( a ib) ( c id ) ( a c) ( b d) i and ( a ib)( c id ) ( ac bd) ( bc ad) i respectivel. (a) Verif that R is an integral domain, (b) Determine all units in R. Soln: We observe that the given R is a subset of the set set C of all comple numbers and that the given and are the usual addition and the usual multiplication of comple numbers. We readil note that ( R, ) is a subgroup of the additive group of comple numbers which is abelian. Further, the multiplication of comple numbers is commutative, associative and distributive (w.r.t addition). In view of these facts, we infer that the given ( R,, ) is a commutative ring. In this ring, the number is unit. Further, for an, R, the condition and. Therefore, ( R,, ) is an integral domain. For an, R, the condition holds onl when holds when,, and i, i or vice versa. Thus,, -, i, -i are the onl units in this ring. (Consequentl, this ring is not a field). Eample Prove that the set Z with binar operations and defined b

31 , is commutative ring with unit. Is this ring an integral domain? A field? Soln: For all Z,, we note that R and R. Therefore, R is closed under the given addition and the given multiplication. Also, we note that, for an Z,,,, } ) {( ) ( ) ( ), ( )} ( { ) ( ) (. ) (2 } ) {(2 ) 2 ( ) 2 ( The above facts show that ), ( Z is a commutative group with Z as the identit element and Z 2 ) ( as inverse of Z. Further, we note that for an Z,,, ) ( ) ( ) ( ) ( ) ( } {( ) ( ) ( and ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( } {( ) ( ) ( ) ( ) ( ) ( ) ( ) (

32 The above facts show that the multiplication is associative and distributive w.r.t.. Accordingl, ( Z,, ) is a ring. Alsofor an, Z, we have This shows that ( Z,, ) is commutative. We further observe that, for an Z,. This shows that is an identit in Z under. Thus, ( Z,, ) is a commutative ring with unit, the unit being. Net, for an, Z, (ero element) ( )( ) or. This shows that ( Z,, ) has no proper ero-divisors. Thus, ( Z,, ) is a commutative ring with unit, which has no proper ero-divisors. As such, ( Z,, ) is an integral domain. If we take an (ero element) in Z, then is possible onl if ; i.e, if or 2. Thus, elements other than and 2have no multiplicative inverses. Hence ( Z,, ) is not a field. Eample Find all integers k and m for which ( Z,, ) is a ring under the binar operations k, m. Soln: For ( Z,, ) to be a ring, it is necessar that the distributive laws must hold (among other laws). Thus we should have,

33 ( ) ( ) ( ) (i) B using the definitions of and, we find that ( ) ( ) m( ) ( k) m( k) m ) k mk ( (ii) ( ) ( ) ( ) ( ) k ( m ) ( m k m ) k ( (iii) In view of (ii) and (iii), the law (i) holds if mk, or mk so that m k or m k. m k. Thus, for ( Z,, ) to be a ring, it is necessar that m k or We can check that for these values of m and k all other laws (aioms) for a ring will hold. Eample Let U (,2 ) and R P(U ). Define and on the elements of R b A B AB, A B A B. Prove that R is a commutative ring with unit but not an integral domain. Soln: The subset of the given set U are,{},{2} and U. Therefore, R { P( U ) {,{},{2}, U} It is given that the smmetric difference operation on sets in R { P( U) is the addition in R and the intersection operation is the multiplication. B using the definitions of smmetric difference and the intersection of sets, we find that the operation tables for and in R are as given below.

34 ( ) {} {2} U ( ) {} {2} U {} {2} U { } { } U {2} { } { } { } { 2} { 2} U { } { 2} {2} {2} U U {2} {} U {} {2} U B using above table, we can check that ( R,, ) is a ring in which is ero (additive identit) and ever elements is its own additive inverse. Further, is commutative and U is the multiplicative identit. Thus ( R,, ) is a commutative ring with unit. We also find that { } {2} (ero element) but } { and 2} Thus, { } and { 2} are ero -divisors in R. Since ( R,, ) contains erodivisors, it is not an integral domain. {. Eample Consider a ring ( R,, ), where R { a, b, c, d, e}, and and are defined b the following tables. a b c d e a b c d e a b c d e a b c d e b c d e a c d e a b d e a b c e a b c d a b c d e a a a a a a b c d e a c e b d a d b e c a e d c b Prove that R is a field.

35 Soln: We check that, for a given ring, a is the ero (additive identit) and b is the unit (multiplicative identit). The smmetr (with respect to the diagonal) in the multiplication table for shows that is commutative. Thus, the given ring is a commutative ring with unit. We check that ever element of R other than a has a multiplicative inverse; in fact, b and e are their multiplicative inverses and c and d are multiplicative inverses of each other. Therefore, this ring is a field. Eample For R { s, t,, }, define and, making R into a ring, b the following tables, the second of which is onl partial. s t s t s s t s s s s s t t s t s t?? s t s t? t s s? s? Using the associative and distributive laws, determine the missing entries (denoted b?) Is this ring commutative? Does it have a unit? Is it an integral domain? A field? Soln: We observe that the second table is the table for multiplication operation. The missing entries in this table are the products t, t,, t,. Using the first table (which is the operation table for addition ) and the entries available in the second table, we find that these missing entries are as given below:

36 ( t ) t t, t ( t) t t tt t t s, ( t ) t s s s, t ( ) s, t ( ) s, We note that t where as t t. Thus, not commutative. t t.therefore, the ring R is From the multiplication table, we find that R does not contain multiplicative identit. This means that R is not a ring with unit. Consequentl, R is neither an integral domain nor a field. Eample If R is a ring with unit, show that ( )( ). ( ) a a, a R, and Soln: We have ( ) a a ( ) a ( a) ( ) a a Therefore, ( ) a a. For a, this becomes ( )( ) ( ). Eample If R is a ring with unit and a, b are units in R, prove that ab is a unit in R and that ( ab ) b a. Soln: We note that ( ab )( b a a( bb ) a a() a aa This shows that ab has b a as the multiplicative inverse. Thus, ab is a unit and ( ab ) b a. Eample If the elements of a ring R from a cclic group under addition, prove that R is commutative.

37 Soln: Since R is a ring, the elements of R form an abelian group under addition. It is given that this group is cclic. Let g be a generator of this additive group so that ever element of R is of the form kg for some positive integer k. Consider an two elements a, b of R. Then some positive integer m and n. Then, a mg and b mg for ab {( mg)( ng) ( g g g... m terms )}.{ ( g g g... nterms )} { g. g g. g g. g... mn terms } ( g g g... n terms )}.{ ( g g g... mterms )} {( ng)( mg) ba Therefore, R is commutative. Eample Prove that a ring is without ero divisors if and onl if the cancellation laws hold in it. Soln: Let R be a ring. First suppose that R has no ero divisors. Then, a b, cr, with a, ab ac ab ac a( b c) b c b c. Thus a and ab ac implies b c in R. Similarl, the right cancellation law also holds in R. ; that is, the left cancellation law holds Conversel, suppose the cancellation laws hold in R. Then, for a, br, ab ab a b if, ab ab b a if b. a and Thus, in R, if a or if b. Hence R has no ero-divisors. The required result is thus proved. Eample Let R is a commutative ring with unit. Prove that R is an integral domain if and onl if, for all a, b, cr where a, ab ac b c.

38 Soln : First suppose that R is an integral domain. Then ab ac ( ab ac) a ( b c) b c, because a b c Conversel, suppose that ab ac b c when, a in R. Take an, R with. Since, we find that, when This proves that R is an integral domain. Eample Prove that ever field is an integral domain Soln: Let F be a field. Then F is a commutative ring with unit. Take an a, bf Such that ab. If a, a eists in F and we get Similarl, if b,, then a. b b ( a a) b a ( ab) a. Thus, ab in F implies a or b. Hence F is an integral domain. Eample 2)Prove that ever finite integral domain is a field. Soln: Let D be an integral domain consisting of onl a finite number of elements. Then D is a commutative ring with unit which contains no proper ero divisors. Therefore, the cancellation laws hold in D. To prove that D is a field, we have to show that ever non ero element of D has a multiplicative inverse in D. Let us take an f ( ) a, D. Then a( ) D and define the function f : D D b f ( ) f ( ) a a a a a( ).

39 Thus, f is a one to- one function from D to D itself. Since D is a finite set, it follows that f is onto as well. Consequentl, ever element of D has a unique pre-image in D under f. Since D, should have a unique pre-image, sa c in D, under f so that f ( c). But f ( c) ac b the definition of f. Therefore, ac so that c is the multiplicative inverse of a. Thus, ever non-ero a in D has a multiplicative inverse in D. Consequentl, D is a field. This completes the proof. SUB RINGS Let S be a non empt subset of a ring R. Then S is called a subring of R whenever R itself is a ring under the operations in R. Eample The set of all even integrs is a subring of the ring ( Z,, ) of all integers under the usual addition and multiplication. The subring is denoted b ( 2Z,, ). Eample The ring of integers under usual addition and multiplication is a subring of all real numbers usual addition and multiplication. Note : For an ring R, the singleton set {} is a subring of R. Also ever ring is a subring of itself. Criterion for a subring The following theorem provides a necessar and sufficient condition for a subset R of a ring R to be a subring of R. Theorem: S is a subring of R if and onlif (i) For all,, we have a b S and S a bs ab.

40 (ii) For all a S, we have a S. Proof: First, suppose that S is a subring of R. Then S is a ring in its own right. Therefore, S satisfies all the conditions of a ring, which include the two conditions specified in the theorem. Thus, if S is a subring of R, then the two conditions specified in the theorem automaticall hold. Conversel, suppose that the two conditions in the theorem hold. Since the associative and distributive laws hold for all elements of R and since S is a subset of R, these laws hold for all elements of S as well. Since the commutative law w.r.t holds in R, it holds in S as well. Also, for an a S, since a S (b condition (ii)), it follows that a ( a) S (b condition (i)); that is S. All of these show that S is an abelian group under. Consequentl, S is a ring. Since S is a subset of R, it is a subring of R. The proof of the theorem is complete. Eample Consider the ring ( Z,, ), where the addition and the multiplication are defined b a b a b and a b ab. Also, let S denote the set of all odd integers. We note that, for all a, bs, a b a b and a b ab Since a and b are odd integers, and a b and and ab are even integers and consequentl a b and ab are odd integers. Therefore, a b S and a b S. Also, for an a S, we have a a ( a ) a and a ( 2 a) (2 a) a a (2 a) These show that is the identit w.r.t and and both of these ( and 2-a) belongs to S. 2 a is the inverse of a w.r.t Therefore, in view of the theorem proved above, it follows that S is a subring of ( Z,, ).

41 Ideal of a Ring A non- empt subset I of a ring R is called an ideal of R if, for all and all r R, we have a b I and ar ra I,. a, bi Eample) The subring ( 2Z,, ) of the ring ( Z,, ) is an ideal of the ring. Eample ) The ring ( Z,, ) is a subring of ( Q,, ) but is not an ideal. Because, for eample, 3 Z and Q 2 (. 2, but 3 Z The Ring of Integers modulo n Let n be a given integer. Then, for a, bz, we sa that a is congruent to b modulo n if a b is a multiple of n; that is if a b kn some k Z. Then we write a b(mod n). For eample, 5(mod 2) 9, 6 2(mod4) or a b kn and 3(mod7). We have proved that the relation congruence modulo n is an equivalence relation. Further, we have noted the following: for () The set of all equivalence classes of integers modulo n (Residue classes modulo n), namel Z, is an abelian group under the operation called n addition modulo n, with [] as the identit and -[a]=[-a]=[n-a] as the inverse of [a]. (2) The set Z n is closed under the operation is called multiplication modulo n. (3)The operation is commutative and associative. (4) [] is the multiplicative identit under. (5) If n is a prime (sa, n = p), then ever [ a ] [] in Z p has [b] as the inverse under, where b is an integer such that a and b are relativel prime; i.e., gcd(a,b)=. * (6) If Z p is the set of all non ero elements in abelian group. Z, then ( Z *, ) is an p p

42 . We now show that the operation is distributive w.r.t the operation Let us take an in Z n, we find that [ a],[ b],[ c] Z. Then using the definitions of and n [ a] ([ b] [ c]) [ a] [ b c] [ a( b c)] [ ab ac] [ ab] [ ac] [ a][ b] [ a][ c] and ([ a] [ b]) [ c] ([ a b]) [ c] [( a b) c] [ ac bc] These show that is distributive w.r.t. In view of this result and the results () (4) mentioned above, it follows that ( Z,, ) is a commutative ring with unit. This ring is n called the Ring of integers modulo n and is denoted b Z n. In view of the result (5) indicated above, it follows that if n = p, a prime, then Z n is a field. The converse of this is also true. See following Eample (5) Eample ) Show that Z 5 is an integral domain. Soln: We first note that Z 5 is a commutative ring with unit. The multiplicative table for Z 5 is as given below: [ ] [] [2] [3] [4] [] [] [2] [3] [4] [] [] [] [] [] [] [] [2] [3] [4] [] [2] [4] [] [3] [] [3] [] [4] [2] [] [4] [3] [2] []

43 From the table, we note that for an [ a ] and an [ b] in Z 5, we have [ a ][ b]. That is Z 5 has no proper ero-divisors. Therefore, Z 5 is an integral domain. Remark: Since Z 5 5, Z 5 is a finite integral domain and is therefore a field. Eample 2) Show that Z 6 is not an integral domain. Soln: We first note that Z 6 is a commutative ring with unit. The multiplication table for Z 6 is as given below: [ ] [] [2] [3] [4] [5] [] [] [2] [3] [4] [5] [] [] [] [] [] [] [] [] [2] [3] [4] [5] [] [2] [4] [] [2] [4] [] [3] [] [3] [] [3] [] [4] [2] [] [4] [2] [] [5] [4] [3] [2] [] [ 3] in 6 From the table, we note that for an [ 2][3] although [ 2] and integral domain. Z,Thus, Z 6 has pair of ero-divisors. Therefore, Z 6 is not an Remark: Since Z 6 is not an integral domain, hence it is not a field. Eample 3)(a) Prove that, in Z n, ] [a is a unit if and onl if gcd(, n) a. (b) Find all the units in Z 2. Soln: (a) If gcd( a, n), then ar ns for some integers r and s. Thus ar ns So that ar (mod n). This ields [ a ][ r] [] which shows that [a] is a unit whose multiplicative inverse is [r].

44 Conversel, suppose that [a] is a unit of Z n and let [ a] [ r]. Then ar so that ar (mod n). This ields ar sn for some integer [ ] [ a][ r] [] s. Then, we have ar ( s) n which shows that gcd( a, n) (b) We have We note that Z 2 {[], gcd(,. [], [2], [3], [4], [5], [6], [7], [8], [9], [], []} 2) gcd(5, 2) gcd)7, 2) gcd(, 2). Therefore, [ ], [5], [7], [] are the units of Z 2. Eample 4) Find [25] in Z 72 and [777] in Z 9. Soln: We note that gcd( 25, 72). Therefore, [ 25] is a unit in Z 72. Let [25] [ a] Z in 72. Then [ 25][ a ] [25a] [] or 25a (mod72) or 25 72r, for some integer r. or 25 72r a. We check that this holds when 49 [25] a and 7 [ a ] [49]. r. Thus, in 72 Net, we note that gcd( 777, 9). Therefore, [ 777] is a unit in Z 9. Let [777] [ b] Z in 9. Then 777b (mod 9), Z. or 777b (9) s, for some integer s. we check that this holds for 735 b and 566 s. Thus, in 9 Z. [777] [ b ] 735. Eample 5) Prove that, in Z n is a field if and onl if n is prime.

45 Soln: Suppose n is a prime. Then, for an integer a such that a n, we have gcd( a, n). Accordingl, [a] is a unit in Z n. This means that ever non ero element of field. Z n has a multiplicative inverse. Therefore, Z n is a Net, suppose n is not a prime. Then n n n2 for some positive integers n and n 2. We note that [ 2 2 n ][ n ] [ n n ] [ n] [] in Z n. Thus, [ n ] and [ n 2 ] are proper ero divisors in Z n. Hence, Z n is not an integral domain. Consequentl, Z n is not a fild. Thus, if n is not a prime then Z n is not a field. Taking the contrapositive of this, we get the result that if Z n is a field then n is prime. This completes the proof. Reference.Discrete Mathematics:Geemaldi 2.Discrete Mathematics Dr.D.S.Chandrashekhar