# one more attempt

• Oct 13th 2009, 10:59 AM
one more attempt
I assume that P(n) is true

p(n): j - i < 2^n => A(i,j) $\displaystyle \leq$ n

for all Z i , j , n with
1 $\displaystyle \leq$ i $\displaystyle \leq$ j $\displaystyle \leq$ N

and n $\displaystyle \leq$ 0

basic step: P(0)

j - i < 1 => A(i,j) $\displaystyle \leq$ 0

now I need to prove P(n+1)
P(n+1) : j - i < 2^(n+1) => A(i,j) $\displaystyle \leq$ n+1

meaning I have to show that A(i,j) $\displaystyle \leq$ n+1

how?
• Oct 13th 2009, 01:54 PM
tonio
Quote:

I assume that P(n) is true

p(n): j - i < 2^n => A(i,j) $\displaystyle \leq$ n

for all Z i , j , n with
1 $\displaystyle \leq$ i $\displaystyle \leq$ j $\displaystyle \leq$ N

and n $\displaystyle \leq$ 0

basic step: P(0)

j - i < 1 => A(i,j) $\displaystyle \leq$ 0

now I need to prove P(n+1)
P(n+1) : j - i < 2^(n+1) => A(i,j) $\displaystyle \leq$ n+1

meaning I have to show that A(i,j) $\displaystyle \leq$ n+1

how?

Do you realize that we have no idea what is P, what is A and what are you talking about at all?

Tonio
• Oct 13th 2009, 02:26 PM