Every positive integer N can be represented in base 2 by (N)a = b,b,_1 ... by, where N=bg+b -2+ +by_12" 1 +5,2" such that for each i = 0,...,n, b = 0 or b; = 1, and b, # 0. In this problem, we consider writing numbers in an alternative form of base 2 where { N)z't =b,b,_1...bby if N=by+b -2+ 1)+ +by1(2" 2+ 1) + by, - (2" +1) such that for each i =0,...,n, by =0or b; =1 and b, 0. Noting e 20 +1=2 o2l +1=3 e 224+1=5 e 281 1=9 o 241 =17 we get (12)2 = 10100 since 12=9+3=1-940-54+1-3+0-2+0-1, and (29)3" = 110100 since 29 =17+ 94+3=1-17+1-940-54+1-340.240-1. (a) Show that for any positive integer N there are by, . . ., b, such that (N)3" = b,b, 1 ... byby. {b) Using this definition, 3 has two representations in alternative base 2: (3)3" = 11 and (3)3" = 100 However, some numbers, such as 4 and 7, have unique representations in alternative base 2. For example (4)3" = 101 and (7)3" = 1010, and neither of these numbers have a second representation in alternative base 2. Show that there are infinitely many positive integers that have a unigue representation in alternative base 2, and there are infinitely many mumbers that have at least two representations in alternative base 2