In the previous section, we saw the basics of binary operations. We saw some solved examples also. In this section, we will see a few more solved examples. We will also see some properties of binary operations.
Solved example 17.33
Show that the ∨: R × R → R given by (a,b) → max {a,b} and the ∧: R × R → R given by (a,b) → min {a,b} are binary operations.
Solution:
Part (i): ∨: R × R → R given by (a,b) → max {a,b}
1. R ×
R will contain ordered pairs of the form (a,b), where a and b are two
real numbers.
2. (a,b) is the input. Output will be the largest among a and b.
• So the output will be a real number. It will be a unique number in the codomain R.
3. So we can write:
♦ The given operation,
♦ carries the element (a,b) of the domain R×R,
♦ to an unique element of the codomain R
• Therefore, the given operation is a binary operation.
Part (ii): ∨: R × R → R given by (a,b) → min {a,b}
1. R ×
R will contain ordered pairs of the form (a,b), where a and b are two
real numbers.
2. (a,b) is the input. Output will be the smallest among a and b.
• So the output will be a real number. It will be a unique number in the codomain R.
3. So we can write:
♦ The given operation,
♦ carries the element (a,b) of the domain R×R,
♦ to an unique element of the codomain R
• Therefore, the given operation is a binary operation.
• Let us write some examples related to the above solved example:
♦ ∨(5,9) = 9
♦ ∧(5,9) = 5
♦ ∨(5,-9) = 5
♦ ∧(5,-9) = -9
• The above examples help us to get a better understanding about the max and min functions.
Now we will learn about operation table. We will learn it by analyzing an example. It can be written in 7 steps:
1. Consider the following function:
∨: A × A → A given by ∨(a,b) → max {a,b}
♦ Where A = {1,2,3}
2. At the top of the table, in the red cell, we write the function.
3. In the magenta cell, we write the symbol of the binary operation. In our present case, it is ‘∨’.
• This is shown in the fig.17.11(I) below:
 |
| Fig.17.11 |
4. Next we fill up the yellow row and yellow column.
• In the yellow column, we fill up the elements of A. They are filled from top to bottom.
♦ These elements are taken as ‘a’ of the function.
♦ In the table, these elements are denoted as ‘ai’.
• In the yellow row also, we fill up the elements of A. They are filled from left to right.
♦ These elements are taken as ‘b’ of the function.
♦ In the table, these elements are denoted as ‘aj’.
5. Now we fill up the green rows and columns.
♦ The row numbers are denoted by the letter i.
♦ The column numbers are denoted by the letter j.
• Let us see an example: We want to fill the green cell with i = 1 and j = 2.
♦ When the row number i is 1, we see that, ai = a1 = 1.
♦ When the column number j is 2, we see that, aj = a2 = 2.
♦ When we have both ai and aj, we can find the largest among them. It is 2.
♦ So the cell should be filled with 2.
This is shown in fig.17.11(II) above. In this way, all green cells can be filled.
6. Note that, if set A has n elements, then the green cells will be arranged in n rows and n columns.
7. Now we will write about the general case. It can be written in 3 steps:
(i) Consider the binary operation *: A × A → A
where A = {a1, a2, a3, . . . an}
(ii) Since there are n elements in A, the green cells will be arranged in n rows and n columns.
(iii) The green cell at the intersection of the ith row and jth column will have the value: (ai * aj)
• The converse can be written in 3 steps:
(i) Given any binary operation table with n rows and n columns.
(ii) Based on that operation table, we can define a binary operation as follows:
*: A × A → A given by: (ai * aj) = value in the ith row and jth column.
(iii) Set A will have n elements.
Commutative binary operation
This can be explained in 6 steps:
1. We know that, two numbers a and b can be added in any order. We can write (a+b) or (b+a). Both will give the same answer.
• For example: (5+7) = (7+5) = 12
2. But for subtraction, order is important. (a-b) need not be equal to (b-a).
• For example: (5-7) ≠ (7-5)
3. We know that, two numbers a and b can be multiplied in any order. We can write (ab) or (ba). Both will give the same answer.
• For example: (5 × 7) = (7 × 5) = 35
4. But for division, order is important. (a÷b) need not be equal to (b÷a).
• For example: (5÷7) ≠ (7÷5)
5. We can write a summary:
• ‘Addition of 5 and 7’ is meaningful.
• ‘Subtraction of 5 and 7’ is meaningless.
♦ We must write:
✰ Subtract 7 from 5.
✰ or
✰ Subtract 5 from 7.
• ‘Multiplication of 5 and 7’ is meaningful.
• ‘Division of 5 and 7’ is meaningless.
♦ We must write:
✰ Divide 7 by 5.
✰ or
✰ Divide 5 by 7.
6.Based on the above steps, we can write:
A binary operation ∗ on the set X is called commutative, if a ∗ b = b ∗ a, for every a, b ∈ X
Let us see some solved examples:
Solved example 17.34
Show that +: R × R → R and ×: R × R → R are commutative binary operations, but –: R × R → R and ÷: R × R → R are not commutative.
Solution:
Part (i):
1. Let (a,b) ∈ R × R
We can write: (a+b) = (b+a) for all a, b ∈ R.
So ‘+’ is a commutative binary operation.
2. Let (a,b) ∈ R × R
We can write: (a × b) = (b × a) for all a, b ∈ R.
So ‘×’ is a commutative binary operation.
Part (ii):
1. Consider (5,7) ∈ R × R
We have: (5 - 7) ≠ (7 – 5).
So ‘-’ is not a commutative binary operation.
2. Consider (5,7) ∈ R × R
We have: (5÷7) ≠ (7÷5).
So ‘÷’ is not a commutative binary operation.
Solved example 17.35
Show that ∗ : R × R → R defined by a ∗ b = a + 2b is not commutative.
Solution:
1. Consider (5,7) ∈ R × R
2. We are going to calculate (a∗b). We get:
(a∗b) = (5∗7) = [5 + (2×7)] = [5 + 14] = 19
3. Next, we are going to calculate (b∗a). We get:
(b∗a) = (7∗5) = [7 + (2×5)] = [7 + 10] = 17
4. From (2) and (3), we see that: (a∗b) ≠ (b∗a).
So the given ‘∗’ is not a commutative binary operation.
Associative binary operation
This can be explained in 6 steps:
1. We know that, three numbers a, b and c can be added in any convenient groups.
• We can write (a+b) +c or a + (b+c). Both will give the same result. Rearranging the parentheses will not change the result.
• For example: (5+7) + 2 = 5 + (7+2) = 14
2. But for subtraction, we cannot rearrange the parentheses.
(a-b) - c need not be equal to a - (b-c).
• For example: (5-7) - 2 ≠ 5 - (7-2)
3. We know that, three numbers a, b and c can be multiplied in any convenient groups.
• We can write (ab)c or a(bc). Both will give the same result. Rearranging the parentheses will not change the result.
• For example: (5 × 7) × 2 = 5 × (7 × 2) = 70
4. But for division, we cannot rearrange the parentheses.
(a÷b) ÷ c need not be equal to a ÷ (b÷c).
• For example: (5÷7)÷2 ≠ 5÷(7÷2)
5. We can write a summary:
• For addition and multiplication, rearranging the parentheses will not change the result.
• For subtraction and division, rearranging the parentheses will change the result.
6. Based on the above steps, we can write:
A binary operation ∗ on the set X is called associative,
if (a ∗ b) ∗ c = a ∗ (b ∗ c), for all a, b, c ∈ X
Let us see some solved examples:
Solved example 17.36
Show
that +: R × R → R and ×: R × R → R are associative binary operations, but – : R × R → R and ÷ : R × R → R are not associative.
Solution:
Part (i):
1. Let a, b, c ∈ R
We can write: (a+b) + c = a + (b+c) for all a, b, c ∈ R.
So ‘+’ is an associative binary operation.
2. Let (a,b) ∈ R × R
We can write: (ab)c = a(bc) for all a, b, c ∈ R.
So ‘×’ is an associative binary operation.
Part (ii):
1. Consider 5,7,2 ∈ R
We have: (5 - 7) - 2 ≠ 5 - (7 – 2).
So ‘-’ is not an associative binary operation.
2. Consider 5,7,2 ∈ R
We have: (5÷7)÷2 ≠ 5÷(7÷2).
So ‘÷’ is not an associative binary operation.
Solved example 17.37
Show that ∗ : R × R → R defined by a ∗ b = a + 2b is not associative.
Solution:
1. Let a= 5, b = 7 and c = 3.
2. We are going to calculate (a∗b)∗c.We get:
(a∗b) = (5∗7) = [5 + (2×7)] = [5 + 14] = 19
So (a∗b)∗c = 19∗3 = [19 + (2×3)] = [19 + 6] = 25
3. Next, we are going to calculate a∗(b∗c). We get:
(b∗c) = (7∗3) = [7 + (2×3)] = [7 + 6] = 13
So a∗(b∗c) = (5∗13) = 5 + (2×13)] = [5 + 26] = 31
4. From (2) and (3), we see that: (a∗b)∗c ≠ a∗(b∗c).
So the given ‘∗’ is not an associative binary operation.
Importance of associative property of binary operation
This can be written in 3 steps:
1. Suppose that, we are given the following expression:
a1 ∗ a2 ∗ a3 ∗ . . . ∗ an.
2. We see that parentheses are not present.
•
But no doubts will arise because, associative property is valid for ‘∗’.
•
Since no parentheses are present, we can straight away evaluate the expression by substituting the appropriate symbol in the place of ‘∗’
3. Due to the validity of the associative property, all algebraic expressions will contain parentheses where ever they are required.
4. Parentheses are compulsory in the following two situations:
(i) When subtraction or division occur.
(ii) When more than one type of operation occur.
In the next section, we will see identity of a binary operation.
No comments:
Post a Comment