Let \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) be relations on the set \(\{1,2, \ldots, 50\}\) such that \(\mathrm{R}_{1}=\left\{\left(\mathrm{p}, \mathrm{p}^{\mathrm{n}}\right): \mathrm{p}\right.\) is a prime and \(\mathrm{n} \geq 0\) is an integer \(\}\) and \(\mathrm{R}_{2}=\left\{\left(\mathrm{p}, \mathrm{p}^{\mathrm{n}}\right)\right.\) : p is a prime and \(\mathrm{n}=0\) or 1\(\}\). Then, the number of elements in \(R_{1}-R_{2}\) is \_\_\_\_ .

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Sets Relations and Functions

Let

R_{1}

and

R_{2}

be relations on the set

{1, 2, \dots, 50}

such that

R_{1} = {(p, p^{n}) : p

is a prime and

n \geq 0

is an integer

}

and

R_{2} = {(p, p^{n})

: p is a prime and

n = 0

or 1

}

.
Then, the number of elements in

R_{1} - R_{2}

is ____ .

Answer: 8

There are fifteen primes in

{1, 2, \dots, 50}

, namely

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

. The relation

R_{1}

pairs each prime

p

with every power

p^{n} \leq 50

for integer

n \geq 0

, whereas

R_{2}

keeps only the powers with

n = 0

n = 1

. Since every pair of

R_{2}

already lies in

R_{1}

, the difference

R_{1} - R_{2}

consists exactly of the pairs

(p, p^{n})

with

n \geq 2

and

p^{n} \leq 50

. For

p = 2

these are

4, 8, 16, 32

; for

p = 3

they are

9, 27

; for

p = 5

only

25

; and for

p = 7

only

49

; no larger prime has even its square within

50

. Counting gives

4 + 2 + 1 + 1 = 8

, so

R_{1} - R_{2}

has

8

elements.

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Integer answer