NUMBER THEORY
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Prove that for every positive integer n, there is a nonnegative integer i and an odd integer j so that n = 2^i * j

hope you mean …
N = [2^I] *[J] …ASSUMING SO …
ANY POSITIVE INTEGER N CAN BE WRITTEN AS
N = 2M ….OR …..N=2M+1…….WHERE M IS AN INTEGER
CASE 1…N=2M
AS PER PRIME FACTOR THEOREM ..
M=[{P1^A1}{P2^A2}{P3^A3}……ETC]…..WHERE P1,P2,…ETC ARE PRIME NUMBERS
SAY …P1=2 , P2=3 , P3=5….ETC...AND A1 , A2 …ETC ARE NON NEGATIVE INTEGERS .
SO WE GET …
N=2M = 2*[{P1^A1}{P2^A2}{P3^A3}……ETC]…..
THAT IS …
N = [2^(A1+1)] * [{P2^A2}{P3^A3}………….] = [2^I] * [J ] ….…PROVED
CASE 1…N=2M+1
AS PER PRIME FACTOR THEOREM ..
N=[{P2^A2}{P3^A3}……ETC]…..WHERE P1,P2,…ETC ARE PRIME NUMBERS
SAY …P1=2 , P2=3 ,P3=5... ETC...AND A1 , A2 …ETC ARE NON NEGATIVE INTEGERS .
SO WE GET …
N= [2^0]*[{P2^A2}{P3^A3}……ETC] = [ 2^0] * J ……PROVED ..