Question

State the reason why the relation $S=\left( a,b \right)\in R\times R:a\le {{b}^{3}}$ on the set R of the real numbers is not transitive.

Answer 1

To solve the above question, we will assume some values of ‘a’ and ‘b’ which will satisfy $a\le {{b}^{3}}.$ Then, we will consider another number c such that b and c will satisfy the relation $b\le {{c}^{3}}.$ Then we will check the same relations for a and c. If ‘a’ will always be less than ${{c}^{3}}$ then the relation will always be transitive, otherwise, it will not be.

Complete step by step answer:
Before we solve this question, we must know what a relation and transitive relation is. A relation between two sets is a collection of ordered pairs containing one object from each set. If the object x is from the first set and the object y is from the second set, then the objects are said to be related if the ordered pair (x, y) is in the relation. Now, we will see what a transitive relation is. If A be any set then a relation R on A will be transitive only if $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ $\Rightarrow \left( a,c \right)\in R$ for all $a,b,c\in A.$ Thus, if ‘a’ and ‘b’ satisfy some relation and (b, c) also satisfies the same relation then the relation will be transitive only if a and c will also satisfy the same relation. In this question, we have to prove why the relation given in the question is not transitive. For this, we will assume some values of a, b and c and then we will try to prove it is not a transitive relation. Let a = 7 and b = 2.
Let us check whether $\left( a,b \right)$ satisfies the relation $a\le {{b}^{3}}$.
$\Rightarrow 7\le {{\left( 2 \right)}^{3}}$.
$\Rightarrow 7\le 2\times 2\times 2$.
$\Rightarrow 7\le 8$.
We can see that $\left( a,b \right)$ satisfied the given relation.
Now, we will consider c = 1.5. Let us check whether $\left( b,c \right)$ satisfies the relation $b\le {{c}^{3}}$.
$\Rightarrow 2\le {{\left( 1.5 \right)}^{3}}$.
$\Rightarrow 2\le 1.5\times 1.5\times 1.5$.
$\Rightarrow 2\le 3.375$.
We can see that $\left( b,c \right)$ satisfied the given relation.
Now, we will check whether (a, c) satisfies the relation.
$\Rightarrow 7\le {{\left( 1.5 \right)}^{3}}$
$\Rightarrow 7\le 1.5\times 1.5\times 1.5$
$\Rightarrow 7\le 3.375$
Thus, we can say that the values (a, c) do not satisfy the same relation.

Hence, we can say that the relation $S=\left( a,b \right)\in R\times R:a\le {{b}^{3}}$ on the set of R is not transitive because (a, b) and (b, c) are satisfying the relation but (a, c) is not satisfying it. For, the relation to be transitive (a, c) should also satisfy the condition.

Note: If we put a = 0, b = 1 and c = 2, then the relation (a, b), (b, c) and (a, c) will satisfy the relation $a\le {{b}^{3}},b\le {{c}^{3}}$ and $a\le {{c}^{3}}$ respectively but we cannot say that the relation will always be transitive on this basis alone because there may be some values of a, b and c which will not satisfy it. For the relation to be transitive, each value of a, b, and c in the real number must satisfy the relation.