Transcript discrete mathematicsx

XII Mathematics
DISCRETE MATHEMATICS
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
(1) CLOSURE AXIOM:
For all a, b  G  a*b  G
(2) ASSOCIATIVE AXIOM:
For all a, b, c  G  (a * b) * c = a * (b * c)
(3) IDENTITY AXIOM:
There exists an element e  G, such that
a * e = a = e * a, for all a  G
(4) 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
Show that the cube roots of unity forms a finite
abelian group under multiplication.
2}
Let G = {1, ,
The Cayley’s table is
.
1

2
1
1

2


2
1
2
2
1

From the Cayley’s table
(i) All the entries are in G. So, the closure axiom is
true.
(ii) Multiplication is always associative.
(iii) The identity element is 1 and it satisfies the
identity axiom
(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.
Show that the set of all fourth roots of unity forms a
finite abelian group under multiplication.
Let G = {1, –1, i, –i}
The Cayley’s table is
.
1
–1
i
1 –1
1 –1
–1 1
i
i
–i
–i
–i
i
–i
–1
1
i
i
1 –1
From the Cayley’s table –i –i
(i) All the entries are in G. So, the closure axiom is
true.
(ii) Multiplication is always associative.
(iii) The identity element is 1G and it satisfies the
identity axiom
(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
Show that (Z7 – {[0], .7}) forms group
Let G = {[1], [2],[3],[4],[5],[6]}
The Cayley’s table is
.7
[1]
[2]
[3]
[1]
[1]
[2]
[3]
[2]
[2]
[4]
[6]
[3]
[3]
[6]
[2]
[4]
[4]
[1]
[5]
[5]
[5]
[3]
[1]
[6]
[6]
[5]
[4]
[4] [4] [1] [5] [2] [6] [3]
[5] [5] [3] [1] [6] [4] [2]
[6] [6] [5] [4] [3] [2] [1]
From the Cayley’s table
(i) All the entries are in G. So, the closure axiom is
true.
(ii) Multiplication modulo 7 is always associative.
(iii) 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.
(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.
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.
(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.
(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.
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
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
(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
a
1
a 
G
(1 – a)a–1 = – a
a 1
 Inverse axiom is true.
(G, *) is a group
(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.
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
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
(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
a
1
a 
G
(1 + a)a–1 = – a
a 1
 Inverse axiom is true.
(G, *) is a group
(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.
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
(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
Using the above results, we have the composition
table as follows
°
f1
f2
f3
f4
f1
f1
f2
f3
f4
f2
f2
f1
f4
f3
f3
f3
f4
f1
f2
f4
f4
f3
f2
f1
From the table
(1) All the entries in the composition table are the
elements of G. Closure axiom is true
(2) Since the composition of functions is always
associative. Associative axiom is true
(3) Clearly f1 is identity element of G.  Identity
axiom is true.
(4) From the table, the inverse of f1 is f1
the inverse of f2 is f2
the inverse of f3 is f3
the inverse of f4 is f4
Inverse axiom is true. (G, °) is a group
(5) From the table the commutative axiom is true
(G, °) is an abelian group.
Cancellation laws
Let 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
Reversal law
Let 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