0.5 Discrete structures recursion  (Page 7/8)

    F<- 2 * 3

    i<- 3 + 1

producing F = 6 and i = 4.

Since i = 4, the while loop is not entered any longer, F = 6 is returned and the algorithm is terminated.

To prove that the algorithm is correct, let us first note that the algorithm stops after a finite number of steps. For i increases one by one from 1 and n is a positive integer. Thus i eventually becomes equal to n.

Next, to prove that it computes n!, we show that after going through the loop k times, F = k ! and i = k + 1 hold. This is a loop invariant and again we are going to use mathematical induction to prove it.

Proof by induction.

Basis Step: k = 1. When k = 1, that is when the loop is entered the first time, F = 1 * 1 = 1 and i = 1 + 1 = 2. Since 1! = 1, F = k! and i = k + 1 hold.

Induction Hypothesis: For an arbitrary value m of k, F = m! and i = m + 1 hold after going through the loop m times.

Inductive Step: When the loop is entered (m + 1)-st time, F = m! and i = (m+1) at the beginning of the loop. Inside the loop,

    F<- m!* (m + 1)

    i<- (m + 1) + 1

producing F = (m + 1)! and i = (m + 1) + 1.

Thus F = k! and i = k + 1 hold for any positive integer k.

Now, when the algorithm stops, i = n + 1. Hence the loop will have been entered n times. Thus F = n! is returned. Hence the algorithm is correct.

Mathematical induction -- second principle

There is another form of induction over the natural numbers based on the second principle of induction to prove assertions of the form ∀x P(x). This form of induction does not require the basis step, and in the inductive step P(n) is proved assuming P(k)   holds for all k<n . Certain problems can be proven more easily by using the second principle than the first principle because P(k) for all k<n can be used rather than just P(n - 1) to prove P(n).

Formally the second principle of induction states that

      if ∀n [ ∀k [ k<n $\to$ P(k) ] $\to$ P(n) ] , then ∀n P(n) can be concluded.

Here ∀k [ k<n $\to$ P(k) ] is the induction hypothesis.

The reason that this principle holds is going to be explained later after a few examples of proof. Example 1: Let us prove the following equality using the second principle:

For any natural number n , 1 + 3 + ... + ( 2n + 1 ) = ( n + 1 )2.

Proof: Assume that 1 + 3 + ... + ( 2k + 1 ) = ( k + 1 )2   holds for all k,   k<n.

Then 1 + 3 + ... + ( 2n + 1 ) = ( 1 + 3 + ... + ( 2n - 1 ) ) + ( 2n + 1 )

= n2 + ( 2n + 1 ) = ( n + 1 )2 by the induction hypothesis.

Hence by the second principle of induction 1 + 3 + ... + ( 2n + 1 ) = ( n + 1 )2   holds for all natural numbers.

Example 2: Prove that for all positive integer n, ${\sum }_{i=1}^n$ i ( i! ) = ( n + 1 )! - 1

Proof: Assume that

1 * 1! + 2 * 2! + ... + k * k! = ( k + 1 )! - 1   for all k,   k<n.

Then 1 * 1! + 2 * 2! + ... + ( n - 1 ) * ( n - 1 )! + n * n!

= n! - 1 + n * n!    by the induction hypothesis.

= ( n + 1 )n! - 1

Hence by the second principle of induction ${\sum }_{i=1}^n$ i ( i! ) = ( n + 1 )! - 1   holds for all positive integers.

Example 3: Prove that any positive integer n, n>1, can be written as the product of prime numbers.