In the previous section, we completed a discussion on composite functions and invertible functions. In this section, we will see binary operations.
Some basics can be written in 3 steps:
1. We are already familiar with the four fundamental operations:
addition, subtraction, multiplication and division.
2. Let us write the three main features of these operations:
(i) Given any two numbers a and b.
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We can find four new numbers:
♦ a+b
♦ a-b
♦ ab
♦ a/b where b ≠ 0.
(ii) Consider any of the four operations. That operation must be carried out between two numbers only.
(iii) If there is a third number, we must follow the procedure given below:
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If three numbers a, b and c are to be added, then we must first add a and b. To the sum of a and b, we can add c.
♦ That means: a+b+c = (a+b) + c.
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If three numbers a, b and c are to be multiplied, then we must first multiply a and b. The product (ab) can be multiplied by c.
♦ That means: abc = (ab)c.
3. The word ‘binary’ means two.
So addition, subtraction, multiplication and division are examples of binary operation.
Now we can write a general definition which is applicable to all binary operations. It can be written in 7 steps:
1. Let us denote any binary operation by one symbol ‘*’
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That is., the symbol ‘*’ can be:
'+' or '-' or '×' or '÷' or any other binary operation.
2. * can be considered as a function.
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This can be explained using an example.
♦ Consider the function f(x) = x2 + 3. This function tells us to square the input x and then add 3.
♦ In a similar way, the function * will tell us what to do when two numbers a and b are given.
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For the function f, the input is a single value x. So we write f(x)
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But for the function *, there will be two input values a and b. So we write *(a,b)
3. For the function f, we specify the domain. Input values for f are taken from the domain.
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In a similar way, we must specify the domain for function * also. This can be done in 4 steps:
(i) Let set A contain all the numbers which can be subjected to the binary operation *.
(ii) But for the function *, we cannot input individual numbers from A. We can input only pairs.
(iii) Recall that the set A × A will contain all the possible pairs that can be formed from the numbers in A.
(iv) So A × A can be taken as the domain.
4. The output of function * is not a pair. It is a single number. So set A can be taken as the codomain.
5. So we have the domain and codomain. We can write four of the many possible ways in which the function * can be defined:
(i) +: A × A →A is defined as +(a,b) = a+b
(ii) -: A × A →A is defined as -(a,b) = a-b
(iii) ×: A × A →A is defined as ×(a,b) = a × b
(iv) ÷: A × A →A is defined as ÷(a,b) = a÷b
6. So it is clear that, * can be described as a function.
♦ It has input and output.
♦ It has domain and codomain.
• However, from now on wards, we will call '*' as a binary operation. We will not call '*' as a function.
7. It is important to note that, only one output must be present for each input.
• That means, in the Venn diagram, only one arrow must diverge from each element of the domain.
• So for every input (a,b), there will be a unique output.
Let us see a solved example:
Solved example 17.29
Show that addition, subtraction and multiplication are binary operations on R, but division is not a binary operation on R. Further, show that division is a binary operation on the set R* of nonzero real numbers.
Solution:
Part (i):
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Binary operation on R means:
Domain is R × R and codomain is R
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Let us check the four binary operations:
1. Is the following operation possible ?
+: R × R →R is defined as +(a,b) = a+b
Answer:
R × R will contain ordered pairs of the form (a,b), where a and b are two real numbers. It is possible to add any two real numbers. The sum will also be a real number. The sum will be a unique number in the codomain R. So this operation is possible.
2. Is the following operation possible ?
-: R × R →R is defined as -(a,b) = a-b
Answer:
R × R will contain ordered pairs of the form (a,b), where a and b are two real numbers. It is possible to find the difference between any two real numbers. The difference will also be a real number. The difference will be a unique number in the codomain R. So this operation is possible.
3. Is the following operation possible ?
×: R × R →R is defined as ×(a,b) = a × b
Answer:
R ×
R will contain ordered pairs of the form (a,b), where a and b are two
real numbers. It is possible to find the product of any two real
numbers. The product will also be a real number. The product will be a unique number in the codomain R. So this operation is
possible.
4. Is the following operation possible ?
÷: R × R →R is defined as ÷(a,b) = a ÷ b
Answer:
R ×
R will contain ordered pairs of the form (a,b), where a and b are two
real numbers. This operation is not defined when b = 0. So this operation is not possible when the domain is R × R.
Part (ii):
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Is the following operation possible ?
÷: R* × R* → R* is defined as ÷(a,b) = a ÷ b
Answer:
R* ×
R* will contain ordered pairs of the form (a,b), where a and b are two
real numbers. But R* is the set of non-zero real numbers. This set will not contain zero. So there will not be any ordered pairs in which b is zero. That means, when this binary operation is carried out, there will not be any division by zero. The quotient will be a unique number in the codomain R*. So this operation is possible.
Solved example 17.30
Show that subtraction and
division are not binary operations on N.
Solution:
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Binary operation on N means:
Domain is N × N and codomain is N.
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Let us check the binary operations of subtraction and division:
1. Is the following operation possible ?
-: N × N → N is defined as -(a,b) = a-b
Answer:
N ×
N will contain ordered pairs of the form (a,b), where a and b are two
natural numbers. It is possible to find the difference between any two natural numbers. But the difference need not be a natural number. For example, if a = 5 and b = 9, then (a-b) will be -4. This -4 is not a natural number. It will not be present in the codomain N. So this operation is
not possible.
2. Is the following operation possible ?
÷: N × N → N is defined as ÷(a,b) = a ÷ b
Answer:
N ×
N will contain ordered pairs of the form (a,b), where a and b are two
natural numbers. It is possible to find the quotient when any natural number is divided by any
natural number. But the quotient need not be a natural number. For
example, if a = 5 and b = 9, then 5/9 is not a
natural number. 5/9 is not present in the codomain N. So this operation is not possible.
Solved example 17.31
Show that *: R × R →R is defined as *(a,b) = a+4b2 is a binary operation.
Solution:
1. R ×
R will contain ordered pairs of the form (a,b), where a and b are two
real numbers.
2. First we find 4b2. Then we add it to a.
• The sum will be a real number. The sum will be a unique number in the codomain R.
3. So for any pair (a,b), the given operation is valid. That means, it is a binary operation.
Solved example 17.32
Let P be the set of all subsets of a given set X.
Show that:
(i) ∪: P × P → P given by (A, B) → A ∪ B
(ii) ∩ : P × P → P given by (A, B) → A ∩ B
are binary operations on the set P.
Solution:
• Note that, in the discussion so far in this section, we denoted the input as (a,b). Here a and b are numbers.
• But for the present problem, input is given as (A,B). This is because, A and B are sets. We are inputting sets.
Part (i):
1. (A,B) is an ordered pair. This pair is taken from the set P × P.
• When we are given the input (A,B), we can find the union of A and B.
• This A∪B will be a set.
• A∪B will be present in P. This is because, P is the set of all subsets.
• Also, A∪B will be unique. We will get only one set when we calculate the union.
2. So we can write:
♦ The given operation,
♦ carries the element (A,B) of the domain P×P,
♦ to an unique element A∪B of the codomain P
• Therefore, the given operation is a binary operation.
Part (ii):
1. (A,B) is an ordered pair. This pair is taken from the set P × P.
• When we are given the input (A,B), we can find the intersection of A and B.
• This A∩B will be a set.
• A∩B will be present in P. This is because, P is the set of all subsets.
• Also, A∩B will be unique. We will get only one set when we calculate the union.
2. So we can write:
♦ The given operation,
♦ carries the element (A,B) of the domain P×P,
♦ to an unique element A∩B of the codomain P
• Therefore, the given operation is a binary operation.
In the next section, we will see a few more solved examples. We will also see some properties of binary operations.
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