1. Understand the two relations

Given $A=\{1,2,3,\dots,20\}$.

• $R_1=\{(a,b): b \text{ is divisible by } a\}$

  This means: $a\mid b$

• $R_2=\{(a,b): a \text{ is an integral multiple of } b\}$

  This means: $b\mid a$

We need the number of elements in: $R_1-R_2$ which means ordered pairs $(a,b)$ such that: $a\mid b \quad \text{but} \quad b\nmid a$

2. Interpretation

If $a\mid b$, then $b=ka$ for some integer $k\ge 1$.

For both $a\mid b$ and $b\mid a$ to hold, we must have $a=b$. So the pairs in $R_1\cap R_2$ are exactly: $(1,1),(2,2),\dots,(20,20)$ Thus, $|R_1\cap R_2|=20$

Hence, $|R_1-R_2|=|R_1|-|R_1\cap R_2|=|R_1|-20$

So we only need to compute $|R_1|$.

3. Count $|R_1|$

For each $a\in A$, count the number of $b\in A$ such that $a\mid b$. That number is: $\left\lfloor \frac{20}{a} \right\rfloor$

Therefore, $|R_1|=\sum_{a=1}^{20}\left\lfloor \frac{20}{a} \right\rfloor$

Now compute:

$\left\lfloor \frac{20}{1} \right\rfloor=20$ $\left\lfloor \frac{20}{2} \right\rfloor=10$ $\left\lfloor \frac{20}{3} \right\rfloor=6$ $\left\lfloor \frac{20}{4} \right\rfloor=5$ $\left\lfloor \frac{20}{5} \right\rfloor=4$ $\left\lfloor \frac{20}{6} \right\rfloor=3$ $\left\lfloor \frac{20}{7} \right\rfloor=2$ $\left\lfloor \frac{20}{8} \right\rfloor=2$ $\left\lfloor \frac{20}{9} \right\rfloor=2$ $\left\lfloor \frac{20}{10} \right\rfloor=2$ $\left\lfloor \frac{20}{11} \right\rfloor=1$ $\left\lfloor \frac{20}{12} \right\rfloor=1$ $\left\lfloor \frac{20}{13} \right\rfloor=1$ $\left\lfloor \frac{20}{14} \right\rfloor=1$ $\left\lfloor \frac{20}{15} \right\rfloor=1$ $\left\lfloor \frac{20}{16} \right\rfloor=1$ $\left\lfloor \frac{20}{17} \right\rfloor=1$ $\left\lfloor \frac{20}{18} \right\rfloor=1$ $\left\lfloor \frac{20}{19} \right\rfloor=1$ $\left\lfloor \frac{20}{20} \right\rfloor=1$

Adding: $|R_1|=20+10+6+5+4+3+2+2+2+2+1+1+1+1+1+1+1+1+1+1$ $|R_1|=67$

4. Now compute $|R_1-R_2|$

$|R_1-R_2|=67-20=47$

5. Final answer

The number of elements in $R_1-R_2$ is: $\boxed{47}$

6. Compare with stored answer

Stored correct answer is $46$, but the correct count is $47$.

A quick check:

• Total pairs with $a\mid b$ = $67$
• Pairs where also $b\mid a$ are only $(a,a)$ for $a=1$ to $20$, so $20$ pairs
• Hence difference = $67-20=47$

So the stored answer appears to be incorrect.