Strong Form Induction

P(0): Show that some property P is true for 0
Assume P(0),P(1),...,P(k) for a particular k
P(0) and P(1) and ... and P(k) => P(k+1): Show that if P is true for 
   all integers <= k, then P is true for k+1
Conclude that P(n) holds for all integers.

The gist:
  Check if a property P works for a base case
  Assume P is true for all integers from the base case value to k
  Check if P(k+1) (the next case) produces a true result.
  If so, since k is arbitrary, all values of k will work.

Example:
          
  P: Every integer n >=2 is divisible by a prime number 
  Base case: P(2) is true because 2/2 is legal and 2 is prime

  Inductive hypothesis: Assume P is true for 2 <= i < k.
                        So, P(2), P(3), ..., P(k) is true.
                        (use k-1 to make things easier)

  Inductive step: Need to show P works for *next* integer, k

        Note: k is either prime or not prime

        1) k is prime: k is divisible by itself, which is prime. OK.

        2) k is not prime: k = a*b, where 2<=a<k and 2<=b<k
  
           Since a and b are within 2<=i<k and we've assumed 
           P works for that range, a&b are divisible by prime number(s)

           So, k is divisible by at least one prime number.
           
  Conclusion: Therefore, all integers n >= 2 are divisible by a prime number.