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