Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

We know that: Prime number has only two factors that is 1 and the number itself whereas composite number has factors other than 1 and itself. It can be observed that, 7 × 11 × 13 + 13 = 13 × ( 7 × 11 + 1 ) = 13 × ( 77 + 1 ) = 13 × 78 = 13 × 13 × 2 × 3 From the above obtained values, we can conclude that 7 × 11 × 13 + 13 is a composite number since it has more than 2 factors. We can write, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × ( 7 × 6 × 4 × 3 × 2 × 1 + 1 ) = 5 × ( 1008 + 1 ) = 5 × 1009 ∵ 1009 cannot be factorised further. The given expression has 5 and 1009 as its factors. Hence, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.

If x = 3 + 1 3 - 1 and y = 3 - 1 3 + 1 then prove that x 2 + y 2 x 2 - y 2 = 7 3 12 .

Given x = 3 + 1 3 - 1 and y = 3 - 1 3 + 1 To prove x 2 + y 2 x 2 - y 2 = 7 3 12 . x 2 = 3 + 1 3 - 1 2 = 3 2 + 1 + 2 3 3 2 + 1 - 2 3 = 3 + 1 + 2 3 3 + 1 - 2 3 = 4 + 2 3 4 - 2 3 = 2 + 3 2 - 3 Similarly, y 2 = 2 - 3 2 + 3 LHS = x 2 + y 2 x 2 - y 2 = 2 + 3 2 - 3 + 2 - 3 2 + 3 2 + 3 2 - 3 - 2 - 3 2 + 3 = 2 + 3 2 + 2 - 3 2 2 - 3 2 + 3 2 + 3 2 - 2 - 3 2 2 - 3 2 + 3 = 4 + 3 + 4 3 + 4 + 3 - 4 3 4 + 3 + 4 3 - 4 + 3 - 4 3 = 14 7 + 4 3 - 7 + 4 3 = 14 8 3 × 3 3 = 7 3 4 × 3 = 7 3 12 = RHS

If the prime factorization of a natural number n is 2 3 × 3 2 × 5 2 × 7 , write the number of consecutive zeroes in n .

It is given that, n = 2 3 × 3 2 × 5 2 × 7 To find the number of zeroes in the above, we need to identify the equal powers of 2 and 5 . ⇒ n = ( 7 × 9 × 2 ) × ( 2 2 × 5 2 ) ⇒ n = ( 7 × 9 × 2 ) × 100 ⇒ n = 12600 Hence, the number of consecutive zeroes are 2 .

Prove that p + q is an irrational, where p , q are primes.

Let us assume that p + q is a rational number. Then, there exist co-prime positive integers a and b such that, p + q = a b ; b ≠ 0 ⇒ a b - p = q Squaring both sides, ⇒ a b - p 2 = ( q ) 2 ⇒ a 2 b 2 - 2 a b p + p = q ⇒ a 2 b 2 + p - q = 2 a b p ⇒ a 2 + p - q b 2 b 2 = 2 a b p ⇒ a 2 - p - q b 2 2 a b = p Here, L.H.S. is a rational number while p is irrational because p is prime. Hence, our assumption is wrong. Thus, p + q is an irrational number.

Prove that there is no positive integer n for which n - 1 + n + 1 is rational.

Suppose there exists a positive integer n for which n - 1 + n + 1 is a rational number. Let p q = n - 1 + n + 1 , where p and q are positive integers and q ≠ 0 ⇒ q p = 1 n - 1 + n + 1 ⇒ q p = 1 n - 1 + n + 1 × n - 1 - n + 1 n - 1 - n + 1 = n - 1 - n + 1 ( n - 1 ) - ( n + 1 ) = n - 1 - n + 1 - 2 ⇒ 2 q p = n + 1 - n - 1 Thus, we have, ( n - 1 + n + 1 ) + ( n + 1 - n - 1 ) = p q + 2 q p ⇒ 2 n + 1 = p 2 + 2 q 2 p q ⇒ n + 1 = p 2 + 2 q 2 2 p q - - - - i Also, we have, ( n - 1 + n + 1 ) - ( n + 1 - n - 1 ) = p q - 2 q p ⇒ 2 n - 1 = p 2 - 2 q 2 p q ⇒ n - 1 = p 2 - 2 q 2 2 p q - - - - i i From i and i i , we get, n + 1 and n - 1 are rational. [Since p , q are integers, p 2 + 2 q 2 2 p q and p 2 - 2 q 2 2 p q are rationals]. From i and i i , we also get, n + 1 and n - 1 are perfect square of positive integers. Now, ( n + 1 ) - ( n - 1 ) = 2 , which is not possible since any two perfect squares differ by at least 3 . Hence, there is no positive integer n for which n - 1 + n + 1 is a rational number. [proved]