Today our topic of discussion is Ordered Pair.
Ordered Pair
Ordered Pair
Amena and Sumena of class VIII occupied 1st and 2nd position repectively in the annual examination. According to merit this can be expressed as (Amena, Sumena) pair. Similar definite pairs are called ordered pairs.
So, for a pair of elements if the positions are given and expressed, then the pair is called Ordered Pair.
If the first element of an ordered pair is x_{1} and the second element is y, then the ordered pair is denoted by (z, y). For the ordered pairs (z, y) and (a,b) will the equal or (x, y) = (a, b) if and only if x = a and y = b
Example 11. If (2x + y, 3) = (6, x – y) then, determine (x,y).
Solution: Given, (2x+y, 3) = (6, x-y)
As per conditions of ordered pairs,
2x + y =6*****(1)
x – y =3*….. (2)
By adding equations (1) and (2) we get 3x = 9 or x = 3 Putting value of r in equation (1) we get, 6 + y = 6 or y = 0
therefore (x, y) = (3, 0)
(Cartesian Product)
Mr Karim decided to colour interior wall of his rooms using white or blue colour, and outside wall using red, yellow or green. The set of colours of interior wall is A= (white, blue) and that of outside wall is B = {red, yellow and green} Mr Karim can colour walls of his rooms by (white, red), (while, yellow), (white, green), (blue, red), (blue, yellow), (blue, green) ordered pairs. The above ordered pairs are writtena s below:
AB = {(white, red), (while, yellow), (white, green), (blue, red), (blue, yellow), (blue, green)}
The above set of ordered pairs is called Cartesian Product set. In set building method, A* B={ (x,y):x ∈ A and y ∈ B}
AB is read as A cross B.
Example 12. If P = {1, 2, 3} , Q = {3, 4} , R =P ∩ Q ,then determine PR and RQ
Solution: Given, P = {1, 2, 3}, Q = {3, 4}
and R=P ∩ Q={ 1,2,3} ∩ {3, 4} = {3} .
PR = {1, 2, 3}.{3} = {(1, 3), (2, 3), (3, 3)} and RQ = {3}.{3, 4} ={(3, 3), (3, 4)}
Example 13. Determine the set of natural numbers that have the same residue 23 when they divide 311 and 419.
Solution: The natural numbers that leave 23 as residue when they divide 311 and 419 will be bigger than 23 and will be common factors of both 311-23 = 288 and 419-23396.
Let the set of factors of 288 are larger than 23 be A.
Here 288 = 1 x 2882 x 144 = 3 x 964 x 726 x 488 x 369 x 32 = 12 * 24 = 16 * 18
∴ A={24, 32, 36, 48, 72 ,96,144,288}
Let the set of factors of 396 greater than 23 be B.
Here 396 1 x 3962 x 1983 x 132 = 4 x 996 x 669 x 4411 x 36= 12 * 33 = 18 * 22
∴ B={33, 36, 44, 66 ,99,132,198,396}
∴ A cap B={24, 32, 36, 48, 72 ,96,144,288}∩{33,36,44,66,99,132,198,396}
∴A cap B = {36}
∴ The required set is (36)
Example 14. Of 100 students 88 passed in Bangla, 80 in Mathematics and 70 in both subjects. Express the information in Venn diagram, and find how many students failed in both subjects.
Solution: In the Venn diagram rectangular area represents U the set of 100 students. The set of students who passed in Bangla and Mathematics are denoted respectively by B and M. As a result the Venn diagram is divided into 4 disjoint sets denoted by P, Q, R and F.
Here, the set of students passed in both subjects is Q =B cap M, cardinality of which is 70.
P is the set of students passed in Bangla only. Its cardinality is |P| = 88 – 70 = 18 R is the set of students who passed only in Mathmatics… |R| = 80 – 70 = 10
Students who passed in at least one subject is corresponding cardinality is 18 + 10 + 70 = 98 P cup Q cup R=B cup M, and
F is the set of students who failed in both the subjects, and |F| = 100 – 98 = 2
∴ number of students failed in both the subjects is 2.
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