03/06/2011, 08:56 PM

5a. a > 1 -> a[n]b > b

5b. a > 1 -> a[n](b + 1) > a[n]b

Proof:

For n = 1 & 2 5a and 5b are evident.

Let's assume 5a and 5b to be true for a given n > 1 ==> (i).

Proof of 5a for n + 1:

a [n+1] 1 = a and so a [n+1] 1 > 1

Now assume a [n+1] b > b for some b

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

(Since (i) says: x > y -> a [n] x > a [n] y)

Furthermore a [n] b >= b + 1, so:

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

So 5a has been proven for n + 1 by induction applied to b

Proof of 5b for n + 1:

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

(Since (i) says: a [n] x > x)

So 5a and 5b have been proven for any n by induction applied to n.

5b. a > 1 -> a[n](b + 1) > a[n]b

Proof:

For n = 1 & 2 5a and 5b are evident.

Let's assume 5a and 5b to be true for a given n > 1 ==> (i).

Proof of 5a for n + 1:

a [n+1] 1 = a and so a [n+1] 1 > 1

Now assume a [n+1] b > b for some b

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

(Since (i) says: x > y -> a [n] x > a [n] y)

Furthermore a [n] b >= b + 1, so:

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

So 5a has been proven for n + 1 by induction applied to b

Proof of 5b for n + 1:

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

(Since (i) says: a [n] x > x)

So 5a and 5b have been proven for any n by induction applied to n.