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XII Mathematics

XII Mathematics. DISCRETE MATHEMATICS. DISCRETE MATHEMATICS. PREPARED BY: R.RAJENDRAN. M.A., M . Sc., M. Ed., K.C.SANKARALINGA NADAR HR. SEC. SCHOOL, CHENNAI-21. Definition of group.

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XII Mathematics

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  1. XII Mathematics DISCRETE MATHEMATICS DISCRETE MATHEMATICS PREPARED BY: R.RAJENDRAN. M.A., M. Sc., M. Ed., K.C.SANKARALINGA NADAR HR. SEC. SCHOOL, CHENNAI-21

  2. Definition of group A non-empty set G with an operation * i.e (G,*) is said to be a group if it satisfies the following axioms • CLOSURE AXIOM:For all a, b  G  a*b  G • ASSOCIATIVE AXIOM:For all a, b, c  G  (a * b) * c = a * (b * c) • IDENTITY AXIOM:There exists an element e  G, such that a * e = a = e * a, for all a  G • INVERSE AXIOM:For all a G, there exists an element a– 1 G such that a * a – 1 = e = a – 1 * a e is called identity element and a – 1 is called inverse of a

  3. Show that the cube roots of unity forms a finite abelian group under multiplication. From the Cayley’s table • All the entries are in G. So, the closure axiom is true. • Multiplication is always associative. • The identity element is 1 and it satisfies the identity axiom Let G = {1, , 2} The Cayley’s table is

  4. (iv) The inverse of 1 is 1 The inverse of  is 2 The inverse of2 is The inverse axiom is true. (G, •) is a group. (v) The commutative axiom is true (G, •) is an abelian group. Since G is finite (G, •) is a finite abelian group.

  5. Show that the set of all fourth roots of unity forms a finite abelian group under multiplication. From the Cayley’s table • All the entries are in G. So, the closure axiom is true. • Multiplication is always associative. • The identity element is 1G and it satisfies the identity axiom Let G = {1, –1, i, –i} The Cayley’s table is

  6. (iv) The inverse of 1 is 1 The inverse of i is –i The inverse of–1is–1 The inverse of–i is i The inverse axiom is true. (G, •) is a group. (v) The commutative axiom is true (G, •) is an abelian group. Since G is finite (G, •) is a finite abelian group

  7. Show that (Z7 – {[0], .7}) forms group Let G = {[1], [2],[3],[4],[5],[6]} The Cayley’s table is

  8. From the Cayley’s table • All the entries are in G. So, the closure axiom is true. • Multiplication modulo 7 is always associative. • The identity element is [1]G and it satisfies the identity axiom • (iv) The inverse of [1] is [1], The inverse of [2]is [4] • The inverse of [3] is[5] The inverse of [4] is [2] • The inverse of [5] is[3] The inverse of [6] is[6] • The inverse axiom is true. • (G, •) is a group. • (v) The commutative axiom is true • (G, •) is an abelian group.

  9. (iv) The inverse of 1 is 1 The inverse of i is –i The inverse of–1is–1 The inverse of–i is i The inverse axiom is true. (G, •) is a group. (v) The commutative axiom is true (G, •) is an abelian group.

  10. Show that (Z, *) is an abelian group where * is defined as a*b = a + b + 2. (1) Closure axiom:Since a, b, 2 are integers a + b + 2 is also integer. a*b  Z  a, b  Z.  The closure axiom is true (2) Associative axiom:Let a, b, c Z (a*b)*c = (a + b + 2)*c = (a + b + 2) + c + 2 = a + b + c + 4 a*(b*c) = a*(b + c + 2) = a + (b + c + 2) + 2 = a + b + c + 4  Associative axiom is true.

  11. (3) Identity axiom:Let e be the identity element By definition of e, a*e = a a + e + 2 = a e = –2 Z  Identity axiom is true (4) Inverse axiom:Let a  Z and a–1 be the inverse of a By definition of inverse a*a–1 = e = – 2 a + a–1 + 2 = – 2 a–1 = – 2 – 2 – a = – 4 – a Z  Inverse axiom is true.

  12. (5)commutative axiom:Let a, b  Z a*b = a + b + 2 = b + a + 2 = b*a  commutative axiom is true (Z, *) is an abelian group Z is an infinite set  (Z,*) is an infinite abelian group.

  13. Let G be the set of all rational numbers except 1 and * be defined on G by a*b = a + b – ab for all a, b G. Show that (G, *) is an abelian group. Let G = Q – {1}. Let a, b G Then a, b are rational numbers and a  1, b  1 Closure axiom: a*b = a + b – ab is a rational number. To prove a*b  G, We have to prove a*b  1 Assume that a*b = 1 a + b – ab = 1  b – ab = 1 – a b(1 – a) = 1 – a  b = 1 since a  1, 1 – a  0  our assumption is wrong That is a*b  1 and hence a*bG  The closure axiom is true

  14. Associative axiom: Let a,b,cG a*(b*c) = a*(b + c – bc) = a + b + c – bc – a(b + c – bc) = a + b + c – bc – ab – ac + abc (a*b)*c = (a + b – ab)*c = a + b – ab + c – (a + b – ab)c = a + b + c – bc – ab – ac + abc a*(b*c) = (a*b)*c Associative axiom is true

  15. (3) Identity axiom:Let e be the identity element By definition of e, a*e = a a + e – ae = a e(1 – a) = 0  e = 0 G since a  1  Identity axiom is true (4) Inverse axiom:Let a  Z and a–1 be the inverse of a By definition of inverse a*a–1 = e = 0 a + a–1 – a a–1 = 0 (1 – a)a–1 = – a  Inverse axiom is true. (G, *) is a group

  16. (5)commutative axiom:Let a, b  Z a*b = a + b – ab = b + a – ba = b*a  commutative axiom is true (G, *) is an abelian group G is an infinite set  (G,*) is an infinite abelian group.

  17. Let G be the set of all rational numbers except –1 and * be defined on G by a*b = a + b + ab for all a, b G. Show that (G, *) is an abelian group. Let G = Q – {– 1}. Let a, b G Then a, b are rational numbers and a  – 1, b  – 1 Closure axiom: a*b = a + b + ab is a rational number. To prove a*b  G, We have to prove a*b  – 1 Assume that a*b = – 1 a + b + ab = – 1  1 + a + b + ab = 0 (1 + a) + b(1 + a) = 0 (1 + a)(1 + b) = 0  a = – 1, b = – 1 But a  – 1, b  –1 our assumption is wrong That is a*b  1 and hence a*bG  The closure axiom is true

  18. Associative axiom: Let a,b,cG a*(b*c) = a*(b + c + bc) = a + b + c + bc + a(b + c + bc) = a + b + c + bc + ab + ac + abc (a*b)*c = (a + b + ab)*c = a + b + ab + c + (a + b + ab)c = a + b + c + bc + ab + ac + abc a*(b*c) = (a*b)*c Associative axiom is true

  19. (3) Identity axiom:Let e be the identity element By definition of e, a*e = a a + e + ae = a e(1 + a) = 0  e = 0 G since a  – 1  Identity axiom is true (4) Inverse axiom:Let a  Z and a–1 be the inverse of a By definition of inverse a*a–1 = e = 0 a + a–1 + a a–1 = 0 (1 + a)a–1 = – a  Inverse axiom is true. (G, *) is a group

  20. (5)commutative axiom:Let a, b  Z a*b = a + b + ab = b + a + ba = b*a  commutative axiom is true (G, *) is an abelian group G is an infinite set  (G,*) is an infinite abelian group.

  21. Prove that the set of four functions f1, f2, f3, f4 on set of non-zero complex numbers C – {0} defined by f1(z) = z, f2(z) = –z, f3(z) = 1/z, f4(z) = –1/z  z C – {0} forms an abelian group with respect to the composition of functions. Let G = {f1, f2, f3, f4} (f1°f1)(z) = f1(z) = z = f1 (f1°f2)(z) = f1(–z ) = –z = f2 (f1°f3)(z) = f1(1/z) = 1/z = f3 (f1°f4)(z) = f1(-1/z) = -1/z = f4 (f2°f1)(z) = f2(z) = -z = f2

  22. (f2°f2)(z) = f2(-z) = z = f1 (f2°f3)(z) = f2(1/z) = -1/z = f3 (f2°f4)(z) = f2(–1/z ) = 1/z = f3 (f3°f1)(z) = f3(z) = 1/z = f3 (f3°f2)(z) = f3(-z) = -1/z = f4 (f3°f3)(z) = f3(1/z) = z = f1 (f3°f4)(z) = f3(-1/z) = -z = f2 (f4°f1)(z) = f4(z) = -1/z = f4 (f4°f2)(z) = f4(-z) = 1/z = f3 (f4°f3)(z) = f4(1/z) = -z = f2 (f4°f4)(z) = f4(-1/z) = z = f1

  23. Using the above results, we have the composition table as follows From the table

  24. All the entries in the composition table are the elements of G. Closure axiom is true • Since the composition of functions is always associative. Associative axiom is true • Clearly f1 is identity element of G.  Identity axiom is true. • From the table, the inverse of f1 is f1the inverse of f2 is f2the inverse of f3 is f3the inverse of f4 is f4Inverse axiom is true. (G, °) is a group • From the table the commutative axiom is true (G, °) is an abelian group.

  25. Cancellation lawsLet G be a group. then for all a, b, cG(1) a*b = a*c  b = c (Left cancellation law)(2) b*a = c*a  b = c (Right cancellation law) Proof: (1) a*b = a*c  a– 1 * (a * b) = a– 1 * (a * c)  (a– 1 * a) * b = (a– 1 * a) * c)  e*b = e*c  b = c (2) b*a = c*a  (b*a) * a– 1 = (c*a) * a– 1  b * (a * a– 1) = c * (a * a– 1)  b * e = c * e  b = c

  26. Reversal lawLet G be a group. then for all a, bG, then(a*b)– 1 = b– 1 * a– 1 Proof: It is enough to prove b– 1* a– 1 is the inverse of (a*b) That is to prove (i) (a*b)* (b– 1* a– 1 ) = e (ii)(b– 1 * a– 1 )(a * b) = e (i) (a * b) * (b– 1 * a– 1 ) = a * (b– 1 * b) * a– 1 = a * (e) * a– 1 = a * a– 1 = e (ii) (b– 1 * a– 1)(a * b) = b– 1 * (a– 1 * a) * b = b– 1 * (e) * b = b– 1 * b = e b– 1* a– 1 is the inverse of(a*b). That is (a*b)–1 = b– 1* a– 1

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