If P = {1, 2}, form the set P × P × P.

P× P× P = {(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1), (2,2,2)}.

If set A has 3 elements and set B = {3, 4, 5}, then find the number of elements in (A × B).

Given, set A has 3 elements, and the elements of set B are {3, 4, and 5}. So, the number of elements in set B = 3 Then, the number of elements in (A × B) = (Number of elements in A) × (Number of elements in B) = 3 × 3 = 9 Therefore, the number of elements in (A × B) will be 9.

If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.

Given, G = {7, 8} and H = {5, 4, 2}, G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)} H × G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}

If A = {–1, 1}, find A × A × A.

A × A × A = {(–1, –1, –1), (–1, –1, 1), (–1, 1, –1), (–1, 1, 1), (1, –1, –1), (1, –1, 1), (1, 1, –1), (1, 1, 1)}

If A × B = {( a , x ), ( a , y ), ( b , x ), ( b , y )}. Find A and B.

A is the set of all first elements, and B is the set of all second elements. Therefore, A = {a, b} and B = {x, y}

Let A and B be two sets such that n (A) = 3 and n (B) = 2. If ( x , 1), ( y , 2), ( z , 1) are in A × B, find A and B, where x , y and z are distinct elements.

Given, n(A) = 3 and n(B) = 2; and (x, 1), (y, 2), (z, 1) are in A × B. We know that, A = Set of first elements of the ordered pair elements of A × B B = Set of second elements of the ordered pair elements of A × B So, clearly, x, y, and z are the elements of A; and 1 and 2 are the elements of B. As n(A) = 3 and n(B) = 2, it is clear that set A = {x, y, z} and set B = {1, 2}

The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0, 1). Find the set A and the remaining elements of A × A.

n(A) = 3, A = {–1, 0, 1} Hence, the remaining elements of set A × A are as follows: (–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), and (1, 1)

A relation R from a non-empty set A to a non-empty set B is a subset of the cartesian product A × B. The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in A × B. The second element is called the image of the first element.

The set of all first elements of the ordered pairs in a relation R from a set A to a set B is called the domain of the relation R.

The set of all second elements in a relation R from a set A to a set B is called the range of the relation R. The whole set B is called the codomain of the relation R. Note that range ⊂ codomain.

(i) A relation may be represented algebraically either by the Roster method or by the Set-builder method. (ii) An arrow diagram is a visual representation of a relation.

Let A = {1, 2, 3, 4, 5, 6}. Define a relation R from A to A by R = {(x, y) : y = x + 1 } (i) Depict this relation using an arrow diagram. (ii) Write down the domain, codomain and range of R.

(i) By the definition of the relation, R = {(1,2), (2,3), (3,4), (4,5), (5,6)}. (ii) We can see that the domain ={1, 2, 3, 4, 5,} Similarly, the range = {2, 3, 4, 5, 6} and the codomain = {1, 2, 3, 4, 5, 6}.

The Fig 2.6 shows a relation between the sets P and Q. Write this relation (i) in set-builder form, (ii) in roster form. What is its domain and range?

(i) In set-builder form, R = {(x, y): x is the square of y, x ∈ P, y ∈ Q} (ii) In roster form, R = {(9, 3), (9, –3), (4, 2), (4, –2), (25, 5), (25, –5)} The domain of this relation is {4, 9, 25}. The range of this relation is {– 2, 2, –3, 3, –5, 5}.

Let A = {1, 2} and B = {3, 4}. Find the number of relations from A to B.

We have, A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}. Since n (A×B ) = 4, the number of subsets of A×B is 24 . Therefore, the number of relations from A into B will be 24 .

A relation R from A to A is also stated as a relation on A.

Let A = {1, 2, 3, … , 14}. Define a relation R from A to A by R = {( x , y ): 3 x – y = 0, where x , y ∈ A}. Write down its domain, codomain and range.

The relation R from A to A is given as: R = {(x, y): 3x – y = 0, where x, y ∈ A} = {(x, y): 3x = y, where x, y ∈ A} So, R = {(1, 3), (2, 6), (3, 9), (4, 12)} Now, The domain of R is the set of all first elements of the ordered pairs in the relation. Hence, Domain of R = {1, 2, 3, 4} The whole set A is the codomain of the relation R. Hence, Codomain of R = A = {1, 2, 3, …, 14} The range of R is the set of all second elements of the ordered pairs in the relation. Hence, Range of R = {3, 6, 9, 12}

Define a relation R on the set N of natural numbers by R = {( x , y ): y = x + 5, x is a natural number less than 4; x , y ∈ N}. Depict this relationship using roster form. Write down the domain and the range.

The relation R is given by: R = {(x, y): y = x + 5, x is a natural number less than 4, x, y ∈ N} The natural numbers less than 4 are 1, 2, and 3. So, R = {(1, 6), (2, 7), (3, 8)} Now, The domain of R is the set of all first elements of the ordered pairs in the relation. Hence, Domain of R = {1, 2, 3} The range of R is the set of all second elements of the ordered pairs in the relation. Hence, Range of R = {6, 7, 8}

A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {( x , y ): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.

The relation from A to B is given as R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B} Thus, R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}

The figure shows a relationship between the sets P and Q. Write this relation (i) in set-builder form (ii) in roster form What is its domain and range?

From the given figure, it's seen that P = {5, 6, 7}, Q = {3, 4, 5} The relation between P and Q: Set-builder form (i) R = {(x, y): y = x – 2; x ∈ P} or R = {(x, y): y = x – 2 for x = 5, 6, 7} Roster form (ii) R = {(5, 3), (6, 4), (7, 5)} Domain of R = {5, 6, 7} Range of R = {3, 4, 5}

Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by {( a , b ): a , b ∈ A, b is exactly divisible by a }. (i) Write R in roster form (ii) Find the domain of R (iii) Find the range of R

Given, A = {1, 2, 3, 4, 6} and relation R = {(a, b): a, b ∈ A, b is exactly divisible by a} Hence, (i) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)} (ii) Domain of R = {1, 2, 3, 4, 6} (iii) Range of R = {1, 2, 3, 4, 6}

Determine the domain and range of the relation R defined by R = {( x , x + 5): x ∈ {0, 1, 2, 3, 4, 5}

Given, Relation R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}} Thus, R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)} So, Domain of R = {0, 1, 2, 3, 4, 5} and, Range of R = {5, 6, 7, 8, 9, 10}

Write the relation R = {( x , x 3 ): x is a prime number less than 10} in roster form.

Given, Relation R = {(x, x3): x is a prime number less than 10} The prime numbers less than 10 are 2, 3, 5, and 7. Therefore, R = {(2, 8), (3, 27), (5, 125), (7, 343)}

Let A = { x , y , z} and B = {1, 2}. Find the number of relations from A to B.

Given, A = {x, y, z} and B = {1, 2} Now, A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)} As n(A × B) = 6, the number of subsets of A × B will be 26. Thus, the number of relations from A to B is 26.

Let R be the relation on Z defined by R = {( a , b ): a , b ∈ Z, a – b is an integer}. Find the domain and range of R.

Given, Relation R = {(a, b): a, b ∈ Z, a – b is an integer} We know that the difference between any two integers is always an integer. Therefore, Domain of R = Z and Range of R = Z

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45 sec • 2 pts

In our daily life, we come across many patterns that characterise relations such as brother and sister, father and son, teacher and student. In mathematics also, we come across many relations such as number m is less than number n, line l is parallel to line m, set A is a subset of set B. In all these, What do we notice ?

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Why is the concept of function very important in mathematics ?

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