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Let us consider the set A as given below.

When take off is transitive it means to remove something (that something is an object). When take off doesn’t have an object after it, it is intransitive and can mean to leave the ground. Also note that the same phrasal verb, for example take off, can have more than one meaning (yes, even 7 or 8 different meanings).

A = {a, b, c}

• Is It Transitive Calculator Worksheet There is another way two relations can be combined that is analogous to the composition of functions. Composition – Let be a relation from to and be a relation from to, then the composite of and, denoted by, is the relation consisting of ordered pairs where and for which there exists an element such.
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• Transitive closure of above graphs is 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 Recommended: Please solve it on “ PRACTICE ” first, before moving on to the solution. The graph is given in the form of adjacency matrix say ‘graphVV’ where graphij is 1 if there is an edge from vertex i to vertex j or i is equal to j, otherwise graphij is 0.

Let R be a transitive relation defined on the set A.

Then,

R = { (a, b), (b, c), (a, c)}

That is,

If 'a' is related to 'b' and 'b' is related to 'c', then 'a' has to be related to 'c'.

In simple terms,

a R b, b R c -----> a R c

Example :

Let A = { 1, 2, 3 } and R be a relation defined on set A as 'is less than' and R = {(1, 2), (2, 3), (1, 3)} Verify R is transitive.

Solution :

From the given set A, let

a = 1

b = 2

c = 3

Then, we have

(a, b) = (1, 2) -----> 1 is less than 2

(b, c) = (2, 3) -----> 2 is less than 3

(a, c) = (1, 3) -----> 1 is less than 3

That is, if 1 is less than 2 and 2 is less than 3, then 1 is less than 3.

More clearly,

1R2, 2R3 -----> 1R3

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Clearly, the above points prove that R is transitive.

Important Note :

For a particular ordered pair in R, if we have (a, b) and we don't have (b, c), then we don't have to check transitive for that ordered pair.

So, we have to check transitive, only if we find both (a, b) and (b, c) in R.

Practice Problems

Problem 1 :

Let A = {1, 2, 3} and R be a relation defined on set A as

R = {(1, 1), (2, 2), (3, 3), (1, 2)}

Verify R is transitive.

Solution :

To verify whether R is transitive, we have to check the condition given below for each ordered pair in R.

That is,

(a, b), (b, c) -----> (a, c)

Let's check the above condition for each ordered pair in R.

From the table above, it is clear that R is transitive.

Note :

For the two ordered pairs (2, 2) and (3, 3), we don't find the pair (b, c). So, we don't have to check the condition for those ordered pairs.

Problem 2 :

Is It Transitive Calculator Math

Let A = {1, 2, 3} and R be a relation defined on set A as

R = {(1, 1), (2, 2), (1, 2), (2, 1)}

Verify R is transitive.

Solution :

To verify whether R is transitive, we have to check the condition given below for each ordered pair in R.

That is,

(a, b), (b, c) -----> (a, c)

Let's check the above condition for each ordered pair in R.

From the table above, it is clear that R is transitive.

Problem 3 :

Let A = {1, 2, 3} and R be a relation defined on set A as

R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}

Verify R is transitive.

Solution :

To verify whether R is transitive, we have to check the condition given below for each ordered pair in R.

That is,

(a, b), (b, c) -----> (a, c)

Let's check the above condition for each ordered pair in R.

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In the table above, for the ordered pair (1, 2), we have both (a, b) and (b, c). But, we don't find (a, c).

That is, we have the ordered pairs (1, 2) and (2, 3) in R. But, we don't have the ordered pair (1, 3) in R.

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So, we stop the process and conclude that R is not transitive.

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