Question

Let R be a relation on the set N of all natural numbers, defined by a R b $\Leftrightarrow $ a is a factor of $b^2$. Then R, is
(a) Reflexive but not transitive and symmetric
(b) Reflexive and transitive but not symmetric
(c) Symmetric and transitive but not reflexive
(d) An equivalence relation

Answer 1

Hint: Here, we will use the definitions of reflexive, symmetric and transitive relations to check whether the given relations are reflexive, symmetric or transitive.

Complete step by step answer:
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.
A relation R is reflexive if each element is related itself, i.e. (a, a) $\in $ R, where a is an element of the domain.
A relation R is symmetric in case if any one element is related to any other element, then the second element is related to the first, i.e. if (x, y) $\in $ R then (y, x) $\in $ R, where x and y are the elements of domain and range respectively.
A relation R is transitive in case if any one element is related to a second and that second element is related to a third, then the first element is related to the third, i.e. if (x, y) $\in $ R and (y, z) $\in $ R then (x, z) $\in $ R.
Here, the given relation is:
a R b $\Leftrightarrow $ a is a factor of $b^2$.
We know that ‘a’ is itself a factor of $a^2$. So, the ordered pair (a, a) exists in R.
So, R is reflexive.
Now, we know that if a is a factor of $b^2$, then b is not a factor of $a^2$. So, the ordered pair (b, a) doesn’t exist in R.
This means that R is not symmetric.
Now, let us suppose that (a, b) $\in $ R and (b, c) $\in $ R, then R would be transitive if (a, c) $\in $ R.
Since, (a, b) $\in $ R, this means that a is a factor of $b^2$.
We can write:
$b^2=pa......\left( 1 \right)$, where p is an integer.
Also, (b, c) $\in $ R, this means that b is a factor of $c^2$.
So, we can write:
$c^2=bq............\left( 2 \right)$, where q is an integer.
On putting the value of b from equation (1) in equation (2), we get:
$\begin{align}
  & c^4=paq^2 \\
\end{align}$
Therefore, a is not a factor of c because we can’t say that square root of any integer is integer.
This implies that R is not transitive
Therefore, R is reflexive but not transitive and symmetric.

So, the correct answer is “Option a”.

Note: Students should note here that for a relation to be a particular type of relation, i.e. reflexive, symmetric or transitive, all its elements must satisfy the required conditions. If we are able to find even a single counter example, i.e. if any of the elements doesn’t satisfy the criteria, then we can’t proceed further.