Question

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

Answer 1

Correct Answer: 46

\[ n(R_1) = 20 + 10 + 6 + 5 + 4 + 3 + 3 + 2 + 2 + 2 + 1 + \cdots + 1 \quad \text{(10 times)} \]

\[ n(R_1) = 66 \]

\[ R_1 \cap R_2 = \{(1, 1), (2, 2), \ldots, (20, 20)\} \]

\[ n(R_1 \cap R_2) = 20 \]

\[ n(R_1 - R_2) = n(R_1) - n(R_1 \cap R_2) \]

\[ = 66 - 20 \]

\[ R_1 - R_2 = 46 \text{ pairs} \]

Answer 2

Let \( A = \{ 1, 2, 3, \dots, 20 \} \). Define two relations:
\(R_1 = \{(a, b) : b \text{ is divisible by } a\}\)
\(R_2 = \{(a, b) : a \text{ is an integral multiple of } b\}\)
We need the number of elements in \( R_1 - R_2 \).

Concept Used:

Interpret the definitions:
\(R_1\): \( (a,b) \in R_1 \) if \( a \mid b \) (a divides b).
\(R_2\): \( (a,b) \in R_2 \) if \( b \mid a \) (b divides a).
So \(R_1 - R_2\) means all pairs \((a,b)\) such that \(a \mid b\) but \(b \nmid a\).
If \(a \mid b\) and \(b \mid a\), then \(a = b\).
Thus \(R_1 - R_2 = \{(a,b) \in R_1 : a \ne b\}\).

Step-by-Step Solution:

Step 1: Understand \(R_1\) and \(R_2\).

\(R_1 = \{(a,b) : a \mid b, a,b \in A\}\)
\(R_2 = \{(a,b) : b \mid a, a,b \in A\}\)
So \(R_1 \cap R_2 = \{(a,b) : a \mid b \text{ and } b \mid a\} = \{(a,a) : a \in A\}\).

Step 2: Describe \(R_1 - R_2\).

\(R_1 - R_2 = \{(a,b) \in R_1 : (a,b) \notin R_2\} = \{(a,b) : a \mid b \text{ and } a \ne b\}\).

So we need ordered pairs \((a,b)\) with \(a \ne b\) and \(a \mid b\).

Step 3: Count elements of \(R_1 - R_2\).

For each \(a \in A\), count \(b \in A\) such that \(b\) is a multiple of \(a\) and \(b \ne a\).

For \(a = 1\): Multiples of 1 in A = all 20 numbers. Exclude \(b = a = 1\) ⇒ count = 19.

For \(a = 2\): Multiples of 2 in A: 2,4,6,...,20 ⇒ 10 numbers. Exclude \(b = 2\) ⇒ count = 9.

For \(a = 3\): Multiples: 3,6,9,12,15,18 ⇒ 6 numbers. Exclude 3 ⇒ count = 5.

For \(a = 4\): Multiples: 4,8,12,16,20 ⇒ 5 numbers. Exclude 4 ⇒ count = 4.

For \(a = 5\): Multiples: 5,10,15,20 ⇒ 4 numbers. Exclude 5 ⇒ count = 3.

For \(a = 6\): Multiples: 6,12,18 ⇒ 3 numbers. Exclude 6 ⇒ count = 2.

For \(a = 7\): Multiples: 7,14 ⇒ 2 numbers. Exclude 7 ⇒ count = 1.

For \(a = 8\): Multiples: 8,16 ⇒ 2 numbers. Exclude 8 ⇒ count = 1.

For \(a = 9\): Multiples: 9,18 ⇒ 2 numbers. Exclude 9 ⇒ count = 1.

For \(a = 10\): Multiples: 10,20 ⇒ 2 numbers. Exclude 10 ⇒ count = 1.

For \(a = 11\): Multiples: 11 ⇒ 1 number. Exclude 11 ⇒ count = 0.

Similarly for \(a = 12,13,14,15,16,17,18,19,20\), multiples in A are only themselves ⇒ count = 0.

Step 4: Sum the counts.

Sum = 19 + 9 + 5 + 4 + 3 + 2 + 1 + 1 + 1 + 1 + 0 + ... + 0

Let's list clearly:
a=1: 19
a=2: 9
a=3: 5
a=4: 4
a=5: 3
a=6: 2
a=7: 1
a=8: 1
a=9: 1
a=10: 1
a=11..20: 0

Sum = 19 + 9 = 28,
28 + 5 = 33,
33 + 4 = 37,
37 + 3 = 40,
40 + 2 = 42,
42 + 1 = 43,
43 + 1 = 44,
44 + 1 = 45,
45 + 1 = 46.

Step 5: Verify with possible double-check.

We can also compute total \(R_1\) and subtract \(|R_1 \cap R_2| = 20\).
Total \(R_1\) = for each a, number of multiples of a in A =
a=1: 20, a=2: 10, a=3: 6, a=4: 5, a=5: 4, a=6: 3, a=7: 2, a=8: 2, a=9: 2, a=10: 2, a=11: 1, a=12: 1, a=13: 1, a=14: 1, a=15: 1, a=16: 1, a=17: 1, a=18: 1, a=19: 1, a=20: 1.
Sum = 20+10+6+5+4+3+2+2+2+2+1×10 = 56+10=66.
Then \(R_1 - R_2 = 66 - 20 = 46\). Matches.

Hence, the number of elements in \( R_1 - R_2 \) is 46.