Question

Tems.] Use the second principle of induction to prove that every positive integer n has a factorization of the form 2m, where m is odd. (Hint: For n > 1, n is either odd or is divisible by 2.)

Answer 1

I used the hint and the second principle of mathematical induction to prove the statement.

also known as strong induction. Want to use second principle to prove that every (positive ) integer n has factorisation of the form atm. miodd. (base case) for n=1. n=2%) n=2 nail hence, it can be written in the form 2km where mis odd. base case is clear. (Induction hypothesis) for some, noll, for all Isern, p can be written in the form ak. m. with modd (Roduction step) To show that nt I can be written in that form. case 1. ntl is odd. then n+1=20 (n+1) hence, this case is clear case 2 nt is even htl is divisble by a. let ht=q. then lsqsn = q can be written as akim (mi odd] 4 using induchine hypothesis. 7 h = q => ht) = atm = Intl = Q k t'mk hence, it can be written in this from i from second principle of mathematical incluction it follows that for every positive integer can be written as 2km. with mi odd. a o