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Composition table or Operation Table or Cayley's Table,Examples of Binary operations

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Composition table or Operation Table

Consider a finite set G with a binary operation denoted by addition (+) or multiplication (X). Then the binary operation on the elements of G can be specified or represented by a table known the composition table or the operation table. Such table was first prepared by Arthur Cayley, this table is also known as Cayley’s table by his name.

We can visualise this situation by studying the composition which is constructed with the help binary operation defined on the set. The construction is convenient only when the set has only a small finite number of elements.

  • Binary Arithmetic: Let G={0,1}. Computation under binary arithmetic is carried out under the following way: 0 + 0 = 0; 0 + 1 = 1; 1 + 0 = 1 ; 1 + 1 = 1

Its tabular or Cayley’s form is:

+

0

1

0

0

1

1

1

0

  • Set Relation: Let G denote the set of subsets of the set {0,1}. Here, G= { phi,{0},{1},{0,1}}. Consider the union relation as our binary operation. Then it can be represented in the following form:

$$\cup$$

$$\phi$$

{0}

{1}

{0,1}

$$\phi$$

$$\phi$$

{0}

{1}

{0,1}

{0}

{0}

{0}

{0,1}

{0,1}

{1}

{1}

{0,1}

{1}

{0,1}

{0,1}

{0,1}

{0,1}

{0,1}

{0,1}

  • Congruent modulo m: If a positive integer a is divided by a positive integer giving the positive integers k and b(<m) as the quotient and remainder respectively then we call ‘a congruent b’ and is written as $$\;a\equiv\;b\;(mod.m)\;.$$ Also, if a and b are two integers and a-b is divisible by positive integer m, then $$\;a\equiv\;b\;(mod.m)\;.$$.

$$For\;example\;:\;If\;20\;is\;divided\;by\;3\;the\;quotient\;is\;6\;and\;the\;remainder\;is$$

$$2\;then\;we\;say\;that\;20\;congruent\;2\;modulo\;3\;i.e.\;20\equiv\;2(mod.\;3\;)\;.$$

$$\;Thus,\;4=16=22\equiv\;1\;(mod\;3\;)\;because\;when\;each\;nubers\;4,\;10\;16\;22\;is\;divided\;by\;3,\;the\;remainder\;in\;each\;case\;is\;1.\;$$

Examples:

Example 1: Name any two binary operations except addition and multiplication:

Solution: Subtraction and Division operation.

Example 2 : A binary operation * is defined on the set of integers by a)m*n=m – n b) m*n = 2mnk

Find m*n , if m = 3, n = 2.

Solution:

  1. m*n = m – n = 3 – 2 = 1
  2. m*n = 2mn = 2. 3. 2 = 12

Example 3:$$\;The\;set\;G\;=\;\{\;1\;\omega\;,\;\omega^2\;\}\;where\omega\;represents\;the\;cube\;root\;of\;unity.$$

$$Prepare\;Cayley’s\;table\;representing\;the\;binary\;operation\;of\;ordinary\;multiplication\;\times\;.$$

Solution:

$$Since\;\omega\;\times\;\omega^2\;=\;\omega^3\;=\;1,Cayley,s\;table\;looks\;like\;$$

Cayley’s Table

$$\times$$

1

$$\omega$$

$$\omega^2$$

1

1

$$\omega$$

$$\omega^2$$

$$\omega$$

$$\omega$$

$$\omega^2$$

1

$$\omega^2$$

$$\omega^2$$

1

$$\omega$$

Example 4: Test for closure, associativity and commutativity properties in each case:

$$a)\;The\;operation\;defined\;by\;m*n\;=\;\frac{1}{2}\;(m\;+\;n)\;on\;Z,m,\;n\;\in\;Z.$$

$$b)\;The\;operation\;defined\;by\;m\;\oplus\;n\;=\;m\;on\;Z,\;m,\;n\;\in\;Z.$$

Solution:

a)

  • For closure: $$Consider\;1,\;2,\;\in\;Z\;$$ $$Then,\;m\;*\;n\;=\;\frac{1}{2}\;(m\;+\;n\;)$$ $$\rightarrow\;\;1\;*\;2\;=\;\frac{1}{2}\;(1\;+\;2)\;=\;\frac{3}{2}\;\notin\;Z$$ $$\therefore\;Z\;is\;not\;closed\;$$
  • For associativity:$$Consider\;2,\;4,\;8\;\in\;Z\;$$ $$Then,\;2\;*\;(4\;*\;8)\;=\;2\;*\;\{\;\frac{1}{2}\;(4\;+\;8)\;\}\;$$ $$\;=\;2\;*\;6\;=\;\frac{1}{2}\;(2\;+\;6)=4$$ $$Again,\;(2\;*\;4)\;*\;8\;=\;\{\;\frac{1}{2}\;(2\;+\;4)\;\}\;*\;8\;$$ $$\;=\;3\;*\;8\;=\;\frac{1}{2}\;(3\;+\;8)\;=\;\frac{11}{8}\;\notin\;Z.\;$$ Hence the operaton * on Z is not associative.

  • For Commutative: $$For\;m,\;n\;\in\;Z\;m\;*\;n\;=\;\frac{1}{2}\;(\;m\;+\;n)$$ $$\;=\;\frac{1}{2}\;(\;n\;+\;m)\;$$ $$=\;n\;*\;m\;$$ Thus, m * n = n * m but may not belong to Z. $$\therefore\;the\;operation\;*\;is\;commutative\;but\;may\;not\;be\;closed.\;$$

b)

  • $$Since,\;for\;m,\;n\;\in\;Z\;,\;m\oplus\;n\;=\;m\in\;Z.\;The\;operation\;\oplus\;is\;closed\;$$
  • $$Since,\;for\;m,\;n,\;p\;\in\;Z\;,\;(m\;\oplus\;n)\;\oplus\;p\;=\;m\oplus\;p\;=\;m\in\;Z\;$$ $$and\;m\oplus\;(n\;\oplus\;p)=\;m\;\oplus\;n\;=\;m\in\;Z $$ $$\therefore\;The\;operation\;\oplus\;is\;associative\;$$
  • $$Since,\;for\;m,\;n\;\in\;Z\;,\;m\oplus\;n\;=\;m\;\in\;Z\;and\;n\oplus\;m\;=\;n\;\in\;Z\;but\;n\;\neq\;m\;The\;operation\;\oplus\;is\;ccommutative\;$$

Example 5:Show that the multiplication is binary operation on the set S = {-1, 0, 1} but the subtraction is not.

Solution:

First, we consider the operation of multiplication. So, we examine whether

$$xy\;\in\;S\;or\;not\;for\;all\;x,\;y\;\in\;S\;$$

$$\;Since\;\;\;\;\;\;\;\;\;-1.0=0\;\in\;S\;\;\;\;0.1=0\;\in\;S$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;1.(-1)=-1\;\in\;S\;\;\;\;1.1=1\;\in\;S$$

$$\;And\;\;\;\;\;\;\;\;\;\;\;-1.(-1)=1\;\in\;S\;\;\;\;0.0=0\;\in\;S$$

So, multiplication in a binary operation.

Secondly, we consider the operation is of subtraction. So, we examine whether

$$x-y\;\in\;S\;or\;not\;for\;all\;x,\;y\;\in\;S\;$$

$$\;\;Since\;\;\;\;\;\;\;\;-1-0=-1\;\in\;S,\;\;\;\;\;1-0=1\;\in\;S$$

$$\;\;but,\;\;\;\;\;\;\;\;-1-1=-2\;\notin\;S$$

$$\therefore\;subtraction\;is\;not\;the\;binary\;operation.\;$$

Example 6: If the binary operation * on Q the set of rational numbers is defined by

$$a*b=a+b-ab\;\;\;\;for\;every\;a,\;b\;\in\;Q$$

Show that * is commutative and associative.

Solution:

$$i)\;\;'*'\;is\;commutative\;in\;Q\;because\;if\;a,b\;\in\;Q,\;then\;$$

$$a*b=a+b-ab=b+a-ba=b*a$$

$$ii)\;\;'*'\;is\;commutative\;in\;Q\;because\;if\;a,b\;\in\;Q,\;then\;$$

$$a*(b*c)=a*(b+c-bc)$$

$$=a+(b+c-bc)-a(b+c-bc)$$

$$=a+b-bc+c-(a+b-ab)c$$

$$=(a*b)*c$$

Taken reference from

( Basic mathematics Grade XII and A foundation of Mathematics Volume II and Wikipedia.com )



  • Consider a finite set G with a binary operation denoted by addition (+) or multiplication (X). Then the binary operation on the elements of G can be specified or represented by a table known the composition table or the operation table
  • Congruent modulo m: If a positive integer a is divided by a positive integer giving the positive integers k and b(<m) as the quotient and remainder respectively then we call ‘a congruent b’ and is written as $$\;a\equiv\;b\;(mod.m)\;.$$ Also, if a and b are two integers and a-b is divisible by positive integer m, then $$\;a\equiv\;b\;(mod.m)\;.$$.
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