Introduction for math term relation:
When comparing (relate) the objects (human beings) the concept of relation becomes very important. In a similar fashion we connect two sets (set of objects) by means of relation.
Let A = {2, 5, 7} and B = {4, 6, 8}. Consider a subset of the Cartesian product. A × B = {(2,4), (2,6, (2,8), (5,6), (5,8, (7,8)} If R denotes the relation “is less than”, then 2R4 (since 2 < 4), 2R6, 2R8, 5R6, 5R8, 7R8. This gives R = {(2,4), (2,6, (2,8), (5,6), (5,8, (7,8)}. The relation is R from A to B. The math term relation example problems and practice problems are given below.
I like to share this Graphing Ordered Pairs with you all through my article.
Example problems for math term relation:
Example problem 1:
Give examples of relations of the following type
(i) Reflexive and transitive but not symmetric
(ii) Neither reflexive nor symmetric but transitive
(iii) Nither reflexive nor transitive but symmetric.
Solution:
(i) The given relation “a divisor of ” is reflexive and transitive but not symmetric on the set of M natural numbers.
(ii) Consider the given relation “is greater than” on N. Since a > a is not true, the relation is not reflexive. If a > b, then b R a is not true. The above example relation is transitive.
(iii) Consider the relation R = {(2,3), (3,2),(3,4),(4,3)} on the set A = {2,3,4}. Evidently R is not reflexive as (2,2), (3,3), (4,4) are not in R. Since (2,3) and (3,2) are in R, it is symmetric. It is not transitive as (2,3), (3,2) ? R, but (2,2) ? R.
Example problem 2:
Let M represents the set of integers. S on I is defined as a S b if a – b is an even integer. Prove that S is an equivalence relation.
Solution:
By definition a S b if a – b is an even integer. Since a – a = 0 is even, we have a S a. Hence S is reflexive. Let a S b so that a – b is an even integer.
But then b – a = – (a – b) is also even and so b R a. Hence S is symmetric. Let a S b, and b S c so that a – b and b – c are even integers. Then (a – c) = (a – b) + (b – c) is also an even integer. Hence S is transitive so that S is an equivalence relation.
These example problems are very helpful to study of math term relation.
Having problem with college level math help keep reading my upcoming posts, i will try to help you.
Practice problems for math term relation:
Practice problem 1:
Given that A = {6, 7, 8, 9, 10} and B = {2, 3, 4, 5}. Write down all ordered pairs (a, b) such that a is divisible by b and hence write down the set of ordered pairs given the relation ‘is a multiple of’ from A to B.
Answer: (6,2), (6,3), (8,2), (8,4), (9,3), (10,2), (10,5)
Practice problem 2:
In the set N of natural numbers, examine the kinds of relation given below
(i) If a^2 + b^2 is a perfect square, a R b
(ii) if a^2 = b, a R b.
Answer: (i) symmetric (ii) not a relation
When comparing (relate) the objects (human beings) the concept of relation becomes very important. In a similar fashion we connect two sets (set of objects) by means of relation.
Let A = {2, 5, 7} and B = {4, 6, 8}. Consider a subset of the Cartesian product. A × B = {(2,4), (2,6, (2,8), (5,6), (5,8, (7,8)} If R denotes the relation “is less than”, then 2R4 (since 2 < 4), 2R6, 2R8, 5R6, 5R8, 7R8. This gives R = {(2,4), (2,6, (2,8), (5,6), (5,8, (7,8)}. The relation is R from A to B. The math term relation example problems and practice problems are given below.
I like to share this Graphing Ordered Pairs with you all through my article.
Example problems for math term relation:
Example problem 1:
Give examples of relations of the following type
(i) Reflexive and transitive but not symmetric
(ii) Neither reflexive nor symmetric but transitive
(iii) Nither reflexive nor transitive but symmetric.
Solution:
(i) The given relation “a divisor of ” is reflexive and transitive but not symmetric on the set of M natural numbers.
(ii) Consider the given relation “is greater than” on N. Since a > a is not true, the relation is not reflexive. If a > b, then b R a is not true. The above example relation is transitive.
(iii) Consider the relation R = {(2,3), (3,2),(3,4),(4,3)} on the set A = {2,3,4}. Evidently R is not reflexive as (2,2), (3,3), (4,4) are not in R. Since (2,3) and (3,2) are in R, it is symmetric. It is not transitive as (2,3), (3,2) ? R, but (2,2) ? R.
Example problem 2:
Let M represents the set of integers. S on I is defined as a S b if a – b is an even integer. Prove that S is an equivalence relation.
Solution:
By definition a S b if a – b is an even integer. Since a – a = 0 is even, we have a S a. Hence S is reflexive. Let a S b so that a – b is an even integer.
But then b – a = – (a – b) is also even and so b R a. Hence S is symmetric. Let a S b, and b S c so that a – b and b – c are even integers. Then (a – c) = (a – b) + (b – c) is also an even integer. Hence S is transitive so that S is an equivalence relation.
These example problems are very helpful to study of math term relation.
Having problem with college level math help keep reading my upcoming posts, i will try to help you.
Practice problems for math term relation:
Practice problem 1:
Given that A = {6, 7, 8, 9, 10} and B = {2, 3, 4, 5}. Write down all ordered pairs (a, b) such that a is divisible by b and hence write down the set of ordered pairs given the relation ‘is a multiple of’ from A to B.
Answer: (6,2), (6,3), (8,2), (8,4), (9,3), (10,2), (10,5)
Practice problem 2:
In the set N of natural numbers, examine the kinds of relation given below
(i) If a^2 + b^2 is a perfect square, a R b
(ii) if a^2 = b, a R b.
Answer: (i) symmetric (ii) not a relation
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