03/06/2011, 09:41 PM

6. (a + 1)[n]b > a[n]b

Proof:

For n = 1 & 2 lemma 6 is evident.

Let's assume lemma 6 to be true for a given n > 1.

b = 1

=====

(a + 1)[n+1] 1 = a + 1 > a = a [n+1] 1

Assume (a + 1) [n+1] b > a [n+1] b for some b > 0

Then we wish to prove that (a + 1) [n+1] (b + 1) > a [n+1] (b + 1)

This is proved by:

(a + 1) [n+1] (b + 1)

= (a + 1) [n] ((a + 1) [n+1] b)

> (a + 1) [n] (a [n+1] b)

> a [n] (a [n+1] b)

= a [n+1] (b + 1)

This proves lemma 6 for n + 1 by induction applied to b.

So lemma 6 also has been proved for all n by induction applied to n.

Proof:

For n = 1 & 2 lemma 6 is evident.

Let's assume lemma 6 to be true for a given n > 1.

b = 1

=====

(a + 1)[n+1] 1 = a + 1 > a = a [n+1] 1

Assume (a + 1) [n+1] b > a [n+1] b for some b > 0

Then we wish to prove that (a + 1) [n+1] (b + 1) > a [n+1] (b + 1)

This is proved by:

(a + 1) [n+1] (b + 1)

= (a + 1) [n] ((a + 1) [n+1] b)

> (a + 1) [n] (a [n+1] b)

> a [n] (a [n+1] b)

= a [n+1] (b + 1)

This proves lemma 6 for n + 1 by induction applied to b.

So lemma 6 also has been proved for all n by induction applied to n.