$A=\left\{ 1,2,3,5 \right\}\text{ and }B=\left\{ 4,6,9 \right\}$ Define a Relation R from A to B by $\text{R}=\left\{ \left( \text{x},\text{y} \right):\text{the difference between x and y is odd},x\in A,y\in B \right\}$ Write R in Roster form.

Hint:At first, consider each of the elements of A then compares to that of B by forming ordered pairs in the form (x, y). For example, (1, 4) where 1 is from set A, 4 is from set B, then find their difference and if it is odd then it is included in Relation R otherwise they are excluded.

Complete step by step answer:
In the question, we are given two sets $A=\left\{ 1,2,3,5 \right\}\text{ and }B=\left\{ 4,6,9 \right\}$ and from this we have to define a relation R from A to B such that, difference of x and y is odd. Given that, x comes from set A and y from set B.
So, let's consider element 1 from A, when taking (1, 4) their difference is $\left( \text{4}-\text{1} \right)\Rightarrow \text{3}$ which is odd thus, (1, 4) satisfies to be in Relation R.
When taking (1, 6) their difference is $\left( 6-1 \right)\Rightarrow 5$ which is odd thus, (1, 6) satisfies to be Relation R.
When taking (1, 9) their difference is $\left( 9-1 \right)\Rightarrow 8$ which is not odd, thus, does not satisfies.
So, let's consider element 2 from A, when taking (2, 4) their difference is $\left( 4-2 \right)\Rightarrow 2$ which is not odd, thus, does not satisfy.
When taking (2, 6) their difference is $\left( 6-2 \right)\Rightarrow 4$ which is not odd, thus, does not satisfy.
When taking (2, 9) their difference is $\left( 9-2 \right)\Rightarrow 7$ which is odd, thus, (2, 9) satisfies to be in Relation R.
So, let's consider element 3 from A, when taking (3, 4) their difference is $\left( 4-3 \right)\Rightarrow 1$ which is odd, thus, (3, 4) satisfies to be in Relation R.
When taking (3, 6) their difference is $\left( 6-3 \right)\Rightarrow 3$ which is odd, thus, (3, 6) satisfies to be in Relation R.
When taking (3, 9) their difference is $\left( 9-3 \right)\Rightarrow 6$ which is not odd, thus, does not satisfy.
So, let's consider element 5 from A, when taking (5, 4) their difference is $\left( 5-4 \right)\Rightarrow 1$ which is odd, thus, (5, 4) satisfies to be in Relation R.
When taking (5, 6) their difference is $\left( 6-5 \right)\Rightarrow 1$ which is odd, thus, (5, 6) satisfies to be in Relation R.
When taking (5, 9) their difference is $\left( 9-5 \right)\Rightarrow 4$ which is not odd, thus, does not satisfy.
We have got the details of each pair considered above. Now, compiling the ones resulting in their difference as odd, we will get the Relation R.
Thus, the Relation \[R=\left\{ \left( 1,4 \right),\left( 1,6 \right),\left( 2,9 \right),\left( 3,4 \right),\left( 3,6 \right),\left( 5,4 \right),\left( 5,6 \right) \right\}\]

Note:
We can also solve this and find the pair in another way. At first, we can use the property that, only a difference of odd and even or even and odd is always odd. So, then the pair should be of odd-even or even-odd pair to get the Relation R. So, using this, we can easily avoid pairs like (1,9), (2,4), (2,6), (3,9), (5,9).