In the previous section, we saw commutative property and associative property. We saw some solved examples also. In this section, we will see identity element and inverse element for binary operations.
Identity element
This can be explained in 6 steps:
1. We know that, any number 'a' can be added to the number zero in any order. We can write (a+0) or (0+a). Both will give the same answer.
• For example: (8+0) = (0+8) = 8
2. But for subtraction, 'any order' is not possible. (a-0) need not be equal to (0-a).
• For example: (8-0) ≠ (0-8)
3. We know that, any number 'a' can be multiplied to the number 1 in any order. We can write (a × 1) or (1 × a). Both will give the same answer.
• For example: (8 × 1) = (1 × 8) = 8
4. But for division, 'any order' is not possible. (a÷1) need not be equal to (1÷a).
• For example: (8÷1) ≠ (1÷8)
5. We can write a summary:
• For binary addition involving any number 'a' and zero, order is not important.
• For binary subtraction involving any number 'a' and zero, order is important.
♦ We must write:
✰ Subtract 0 from a.
✰ or
✰ Subtract a from 0.
• For binary multiplication involving any number 'a' and 1, order is not important.
• For binary division involving any number 'a' and one, order is important.
♦ We must write:
✰ Divide 8 by 1.
✰ or
✰ Divide 1 by 8.
6. Based on the above steps, we can write:
Given a binary operation ∗: A × A → A,
an element e ∈ A, if it exists, is called identity for the operation ∗, if:
a ∗ e = a = e ∗ a, ∀ a ∈ A.
Let us see a solved example:
Solved example 17.38
Show that zero is the identity for addition on R and 1 is the identity for multiplication on R. But there is no identity element for the operations
–: R × R → R and ÷: R × R → R.
Solution:
Part (i)
1. If a number e has to qualify as the identity element, it must satisfy two conditions:
(i) a ∗ e = a, ∀ a ∈ R.
(ii) e ∗ a = a, ∀ a ∈ R.
2. We want to prove that zero is the identity for addition on R.
•
We see that, zero satisfies both conditions because:
(i) a + 0 = a, ∀ a ∈ R.
(ii) 0 + a = a, ∀ a ∈ R.
•
So zero is the identity for addition on R.
3. Next we want to prove that 1 is the identity for multiplication on R.
•
We see that, 1 satisfies both conditions because:
(i) a × 1 = a, ∀ a ∈ R.
(ii) 1 × a = a, ∀ a ∈ R.
•
So 1 is the identity for multiplication on R.
Part (ii)
1. We want to prove that, there is no identity element for subtraction on R.
• Let us check whether the element ‘x’ satisfies the two conditions:
(i) a - x = a, ∀ a ∈ R.
(ii) x - a = a, ∀ a ∈ R.
• No real number x can simultaneously satisfy the two conditions.
• So there is no identity element for subtraction on R.
2. Next we want to prove that, there is no identity element for division on R.
• Let us check whether the element ‘x’ satisfies the two conditions:
(i) a ÷ x = a, ∀ a ∈ R.
(ii) x ÷ a = a, ∀ a ∈ R.
• No real number x can simultaneously satisfy the two conditions.
• So there is no identity element for division on R.
We saw that, zero is the identity for the addition operation on R. But instead of R, if the set is N, then we cannot consider zero as the identity. This is because, set N does not contain element zero. In fact, the addition operation on N does not have any identity.
Inverse element
This can be explained in 6 steps:
1.
We know that, any number 'a' can be added to the number (-a) in any
order. We can write (a+ -a) or (-a+a). Both will give the same answer zero.
• For example: (8+ -8) = (-8+8) = 0
♦ Zero is the identity for addition.
2. But for subtraction, 'any order' is not possible. (a- -a) need not be equal to (-a-a).
• For example: (8- -8) ≠ (-8-8)
•
Recall that, subtraction does not have identity element.
•
Since there is no identity element, we can exclude subtraction from our present discussion on inverse element.
3.
We know that, any number 'a' can be multiplied by the number '1/a' in any
order. We can write (a × 1/a) or (1/a × a). Both will give the same answer 1.
• For example: (8 × 1/8) = (1/8 × 8) = 1
♦ 1 is the identity for multiplication.
4. But for division, 'any order' is not possible. (a ÷ 1/a) need not be equal to (1/a ÷ a).
• For example: (8 ÷ 1/8) ≠ (1/8 ÷ 8)
•
Recall that, division does not have identity element.
•
Since there is no identity element, we can exclude division from our present discussion on inverse element.
5. We can write a summary:
• For binary addition involving any two numbers 'a' and (-a), order is not important. The result will be the addition identity zero.
• We exclude binary subtraction from our present discussion. This is because, there is no subtraction identity.
• For binary multiplication involving any two numbers 'a' and '1/a', order is not important. The result will be the multiplication identity 1.
• We exclude binary division from our present discussion. This is because, there is no division identity.
6. Based on the above steps, we can write:
•
Given a binary operation ∗: A × A → A, with the identity element e in A.
•
An element a is said to be invertible with respect to the operation ∗, if there exists an element b in a such that a∗b = e = b∗a.
•
b is called the inverse of a and is denoted by a-1.
Let us see a solved example:
Solved example 17.39
Show that –a is the inverse of a for the addition operation ‘+’ on R and 1/a is the inverse of a ≠ 0 for the multiplication operation ‘×’ on R.
Solution:
1. If a number b has to qualify as the inverse element of a, it must satisfy two conditions:
(i) a ∗ b = e, ∀ a,b ∈ R.
(ii) b ∗ a = e, ∀ a,b ∈ R.
•
In the above two conditions, we specify set R because, it is given that: the operation is on R.
2. We want to prove that '-a' is the inverse of 'a' for addition on R.
•
We see that, '-a' satisfies both conditions because:
(i) a + -a = 0, ∀ a, -a ∈ R.
(ii) -a + a = 0, ∀ a, -a ∈ R.
(0 is the e for addition)
•
So '-a' is the inverse of 'a' for addition on R.
3. Next we want to prove that '1/a' is the inverse of 'a' for multiplication on R.
•
We see that, '1/a' satisfies both conditions because:
(i) a × 1/a = 1, ∀ a, 1/a ∈ R.
(ii) 1/a × a = 1, ∀ a, 1/a ∈ R.
(1 is the e for multiplication)
•
So '1/a' is the inverse of 'a' for multiplication on R.
Solved example 17.40
Show that '–a' is not the inverse of a ∈ N for the addition operation + on N and '1/a' is not the inverse of a ∈ N for multiplication operation × on N, for a ≠ 1.
Solution:
1. If a number b has to qualify as the inverse element of a, it must satisfy two conditions:
(i) a ∗ b = e, ∀ a,b ∈ N.
(ii) b ∗ a = e, ∀ a,b ∈ N.
•
In the above two conditions, we specify set N because, it is given that: the operation is on N.
2. Consider addition.
•
We see that, '-a' can be substituted in the place of b. The two equations will be satisfied.
•
But '-a' is not an element of N. So we cannot use '-a'.
3. Consider multiplication.
•
We see that, '1/a' can be substituted in the place of b. The two equations will be satisfied.
•
But '1/a' is not an element of N. So we cannot use '1/a'
The link below gives a few more solved examples
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