Let A=1,2,3, , 20.

Question

Let $A=\{1,2,3, \ldots, 20\}$. Let $R_1$ and $R_2$ two relation on $A$ such that 
$R_1=\{(a, b): b$ is divisible by $a\}$
$R_2=\{(a, b): a$ is an integral multiple of $b\}$.
Then, number of elements in $R_1-R_2$ is equal to \_\_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 46

Accepted Answer

To determine the number of elements in ${R}_{1}-{R}_{2}$, let's first articulate the meaning of both relations on set $A={1,2,3,\ldots ,20}$:
${R}_{1}$ includes pairs $(a,b)$ where $b$ is divisible by $a$. This includes pairs like $(1,1),(1,2),\ldots ,(1,20)$ for $1$; similar series for $2$ up to $(2,20)$ (excluding odd numbers); for $3$ up to $(3,18)$; and so on, reflecting the divisibility condition.

$\begin{gathered} {R}_{1}:\begin{gathered} (1,1),(1,2)\ldots ,(1,20), \ (2,2),(2,4)\ldots ,(2,20), \ (3,3),(3,6)\ldots ,(3,18), \ (4,4),(4,8)\ldots ,(4,20), \ (5,5)(5,10)\ldots ,(5,20), \ (6,6),(6,12),(6,18),(7,7),(7,14), \ (8,8),(8,16),(9,9),(9,18)(10,10), \ (10,20),(11,11),(12,12)\ldots ,(20,20) \end{gathered} \ n({R}_{1})=66 \end{gathered}$

${R}_{2}$ consists of pairs $(a,b)$ where $a$ is an integral multiple of $b$. Essentially, this relationship is the reverse of ${R}_{1}$. However, for the essence of ${R}_{1}-{R}_{2}$, the key overlap comes with pairs where $a=b$, since those are the pairs that clearly reflect both conditions identically (i.e., a number is always divisible by itself, and it is always a multiple of itself). There are 20 such pairs corresponding to each integer in $A$ from 1 to 20, resulting in the common elements between ${R}_{1}$ and ${R}_{2}$ (the intersection ${R}_{1}\cap {R}_{2}$) being 20 pairs.

$\begin{gathered} {R}_{1}\cap {R}_{2}={(1,1),(2,2),\ldots (20,20)} \ n({R}_{1}\cap {R}_{2})=20 \end{gathered}$
The difference ${R}_{1}-{R}_{2}$ seeks elements present in ${R}_{1}$ but not in ${R}_{2}$. Given that ${R}_{1}$ and ${R}_{2}$ share 20 elements that are identical, to find ${R}_{1}-{R}_{2}$, we subtract these 20 common elements from the total in ${R}_{1}$, resulting in $66-20=46$ pairs.

$\begin{gathered} n({R}_{1}-{R}_{2})=n({R}_{1})-n({R}_{1}\cap {R}_{2}) \ =n({R}_{1})-20 \ =66-20 \ {R}_{1}-{R}_{2}=46 \end{gathered}$

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