By de nition of oddness, there must exist some integer k such that n = 2 k +1. Of even, a = 2m for integer m) = (2m) + (2n + 1) (defn. Web a + b = (2m) + b (defn. Of odd, b = 2n + 1 for integer n) = 2(m + n) + 1 (taking 2 as common factor) = 2p + 1 (p = m + n and addition is closed on integers) = odd (defn. Web contents tableofcontentsii listoffiguresxvii listoftablesxix listofalgorithmsxx prefacexxi. A b = ( 2 k + 1) ( 2 m + 1) = 4 k m + 2 k + 2 m + 1 = 2 ( 2 k m + k + m) + 1. Web iproof:assume n is odd. Then, n2= 4 k +4 k +1 = 2(2 k2+2 k)+1 , which is odd. Thus, if n is odd, n2is also odd. Therefore ab a b is odd.
A b = ( 2 k + 1) ( 2 m + 1) = 4 k m + 2 k + 2 m + 1 = 2 ( 2 k m + k + m) + 1. Of odd, b = 2n + 1 for integer n) = 2(m + n) + 1 (taking 2 as common factor) = 2p + 1 (p = m + n and addition is closed on integers) = odd (defn. By de nition of oddness, there must exist some integer k such that n = 2 k +1. Web contents tableofcontentsii listoffiguresxvii listoftablesxix listofalgorithmsxx prefacexxi. Web a + b = (2m) + b (defn. Of even, a = 2m for integer m) = (2m) + (2n + 1) (defn. A b = ( 2 k + 1) ( 2 m + 1) = 4 k m + 2 k + 2 m + 1 = 2 ( 2 k m + k + m) + 1. Web ab = (2k+1)(2m+1) = 4km+2k+2m+1 = 2(2km+k+m)+1. Web iproof:assume n is odd. Thus, if n is odd, n2is also odd. Therefore ab a b is odd.
