An `color{blue}ul("ordered pair")` of elements taken from any two sets `P` and `Q` is a pair of elements written in small brackets and grouped together in a particular order, i.e., `(p,q), p ∈ P` and `q ∈ Q` .
`color{purple}ul(✓✓) color{purple} mathbf(" DEFINITION ALERT")`
Given two non-empty sets `P` and `Q.` The cartesian product `P × Q` is the set of all ordered pairs of elements from `P` and `Q,` i.e.,
` \ \ \ \ \ \ color{green} (P × Q = { (p,q) : p ∈ P, q ∈ Q })`
If either `P` or `Q` is the null set, then `P × Q` will also be empty set, i.e., `P × Q = φ`
`color{red}☛ color{red}(" Example")` : Let `P={1} , Q={2} qquad qquad P xx Q = { (1,2)}`
`P={2}, Q= {1} qquad qquad Q xx P={(2,1)}`
`P xx Q ne Q xx P`
In most of the cases `P xx Q ne Q xx P`
`color{green} ✍️ ` Suppose `A` is a set of `2` colours and `B` is a set of `3` objects, i.e.,
` \ \ \ \ \ \ \ \ \ color{blue } (A = {"red, blue"})` and `color{blue } (B = {b, c, s},)`
where `b, c` and `s` represent a particular bag, coat and shirt, respectively.
`color{blue}( "How many pairs of coloured objects can be made from these two sets ?")`
Proceeding in a very orderly manner, we can see that there will be `6` distinct pairs as given below:
` \ \ \ \ \ \ color{green} "(red, b), (red, c), (red, s), (blue, b), (blue, c), (blue, s)."`
Thus, we get `6` distinct objects (Fig 1).
From the illustration given above we note that
` \ \ \ \ \ \ \ \ \ color{blue } (A × B = {"(red,b), (red,c), (red,s), (blue,b), (blue,c), (blue,s)"}.)`
` "Again, consider the two sets :"`
`color{purple}( ✍️ A = {DL, MP, KA},) ` where `DL, MP, KA` represent Delhi, Madhya Pradesh and Karnataka, respectively and `color{purple} (B = {01,02, 03}) ` representing codes for the licence plates of vehicles issued by `DL, MP` and `KA .`
● If the three states, Delhi, Madhya Pradesh and Karnataka were making codes for the licence plates of vehicles, with the restriction that the code begins with an element from set `A,`
● which are the pairs available from these sets and how many such pairs will there be (Fig 2)?
● The available pairs are:
`(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03), (KA,01), (KA,02), (KA,03)`
and the product of set A and set B is given by
` color{purple}(A × B = {(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03), (KA,01), (KA,02), (KA,03)}.)`
● It can easily be seen that there will be `9` such pairs in the Cartesian product, since there are 3 elements in each of the sets A and B. This gives us 9 possible codes.
`\color{red} \ox \color{red} \mathbf(COMMON \ CONFUSION) ` Also note that the order in which these elements are paired is crucial.
For example, the code `(DL, 01)` will not be the same as the code `(01, DL).`
`color{green} ✍️ color{green} mathbf("KEY POINTS")`
`(i)` Two ordered pairs are equal, if and only if the corresponding first elements are equal and the second elements are also equal.
`(ii)` If there are `p` elements in `A` and `q` elements in `B`, then there will be `pq` elements in `A × B,` i.e., if `color{fuchsia} [n(A) = p]` and `color{fuchsia} [n(B) = q,]` then `color{fuchsia} [n(A × B) = pq.]`
`(iii)` If `A` and `B` are non-empty sets and either `A` or `B` is an infinite set, then so is `A × B.`
`(iv) \ \ A × A × A = {(a, b, c) : a, b, c ∈ A}.` Here `(a, b, c)` is called an ordered triplet.
An `color{blue}ul("ordered pair")` of elements taken from any two sets `P` and `Q` is a pair of elements written in small brackets and grouped together in a particular order, i.e., `(p,q), p ∈ P` and `q ∈ Q` .
`color{purple}ul(✓✓) color{purple} mathbf(" DEFINITION ALERT")`
Given two non-empty sets `P` and `Q.` The cartesian product `P × Q` is the set of all ordered pairs of elements from `P` and `Q,` i.e.,
` \ \ \ \ \ \ color{green} (P × Q = { (p,q) : p ∈ P, q ∈ Q })`
If either `P` or `Q` is the null set, then `P × Q` will also be empty set, i.e., `P × Q = φ`
`color{red}☛ color{red}(" Example")` : Let `P={1} , Q={2} qquad qquad P xx Q = { (1,2)}`
`P={2}, Q= {1} qquad qquad Q xx P={(2,1)}`
`P xx Q ne Q xx P`
In most of the cases `P xx Q ne Q xx P`
`color{green} ✍️ ` Suppose `A` is a set of `2` colours and `B` is a set of `3` objects, i.e.,
` \ \ \ \ \ \ \ \ \ color{blue } (A = {"red, blue"})` and `color{blue } (B = {b, c, s},)`
where `b, c` and `s` represent a particular bag, coat and shirt, respectively.
`color{blue}( "How many pairs of coloured objects can be made from these two sets ?")`
Proceeding in a very orderly manner, we can see that there will be `6` distinct pairs as given below:
` \ \ \ \ \ \ color{green} "(red, b), (red, c), (red, s), (blue, b), (blue, c), (blue, s)."`
Thus, we get `6` distinct objects (Fig 1).
From the illustration given above we note that
` \ \ \ \ \ \ \ \ \ color{blue } (A × B = {"(red,b), (red,c), (red,s), (blue,b), (blue,c), (blue,s)"}.)`
` "Again, consider the two sets :"`
`color{purple}( ✍️ A = {DL, MP, KA},) ` where `DL, MP, KA` represent Delhi, Madhya Pradesh and Karnataka, respectively and `color{purple} (B = {01,02, 03}) ` representing codes for the licence plates of vehicles issued by `DL, MP` and `KA .`
● If the three states, Delhi, Madhya Pradesh and Karnataka were making codes for the licence plates of vehicles, with the restriction that the code begins with an element from set `A,`
● which are the pairs available from these sets and how many such pairs will there be (Fig 2)?
● The available pairs are:
`(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03), (KA,01), (KA,02), (KA,03)`
and the product of set A and set B is given by
` color{purple}(A × B = {(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03), (KA,01), (KA,02), (KA,03)}.)`
● It can easily be seen that there will be `9` such pairs in the Cartesian product, since there are 3 elements in each of the sets A and B. This gives us 9 possible codes.
`\color{red} \ox \color{red} \mathbf(COMMON \ CONFUSION) ` Also note that the order in which these elements are paired is crucial.
For example, the code `(DL, 01)` will not be the same as the code `(01, DL).`
`color{green} ✍️ color{green} mathbf("KEY POINTS")`
`(i)` Two ordered pairs are equal, if and only if the corresponding first elements are equal and the second elements are also equal.
`(ii)` If there are `p` elements in `A` and `q` elements in `B`, then there will be `pq` elements in `A × B,` i.e., if `color{fuchsia} [n(A) = p]` and `color{fuchsia} [n(B) = q,]` then `color{fuchsia} [n(A × B) = pq.]`
`(iii)` If `A` and `B` are non-empty sets and either `A` or `B` is an infinite set, then so is `A × B.`
`(iv) \ \ A × A × A = {(a, b, c) : a, b, c ∈ A}.` Here `(a, b, c)` is called an ordered triplet.
