Discrete Mathematics..

Hints and Solutions to Selected Exercises 6375. m is a factor of n ⇒ m = k 1 n where k 1 is a positive integer.n is a factor of m ⇒ n = k 2 m where k 2 is a positive integer.Therefore⇒⇒⇒n = k 2 m= k 1 k 2 nk 1 k 2 = 1 (sincen ̸= 0)k 1 = k 2 = 1 (sincek 1 and k 2 are positive integers)m = n.6. The contrapositive is ‘If n is divisible by 5 then n 2 is divisible by 5’. If nis divisible by 5 then⇒⇒n = 5kwhere k is an integern 2 = 25k 2= 5(5k 2 ) where 5k 2 is an integern 2 is divisible by 5.9. If n − 2 is divisible by 4 then⇒Thenn − 2 = 4kn + 2 = 4k + 4= 4(k + 1).where k is an integer⇒n 2 − 4 = (n − 2)(n + 2)= 4k × 4(k + 1)= 16k(k + 1)n 2 − 4 is divisible by 16.10. Assume that an integer n has a smallest factor greater than 1, which is notprime, and show that this leads to a contradiction.Exercises 2.32. If n = 1, 2 n = 2 > 1, so that the proposition holds for n = 1.