03/06/2011, 03:32 PM
4. n > 2 -> 1[n]b = 1
Proof:
n = 3 gives 1 ^ b = 1 which is indeed true for any b
Now suppose that for a given n > 2, it is proven that
1 [n] b = 1 (for any b)
Then we wish to prove that
1 [n+1] b = 1 (for any b)
b = 1 :
1 [n+1] 1 = 1 by definition
b > 1:
1 [n+1] b = 1 [n] (1 [n+1] (b - 1)) = 1 because of the induction hypothesis
Proof:
n = 3 gives 1 ^ b = 1 which is indeed true for any b
Now suppose that for a given n > 2, it is proven that
1 [n] b = 1 (for any b)
Then we wish to prove that
1 [n+1] b = 1 (for any b)
b = 1 :
1 [n+1] 1 = 1 by definition
b > 1:
1 [n+1] b = 1 [n] (1 [n+1] (b - 1)) = 1 because of the induction hypothesis