Induction and Recursion

The following are some notes taken during a lecture.

Mathematical Induction

Suppose we have an infinite ladder:

We can reach the first rung of the ladder.
If we can reach a particular rung of the ladder, then we can reach the next rung.

From (1), we can reach the first rung. Then by applying (2), we can reach the second rung. Applying (2) again, the third rung. And so on. We can apply (2) any number of times to reach any particular rung, no matter how high up.

This example motivates proof by mathematical induction.

Principle of Mathematical Induction

Principle of Mathematical Induction: To prove that P(n) is true for all positive integers n, we complete these steps:

Basis Step: Show that P(1) is true.
Inductive Step: Show that P(k) → P(k + 1) is true for all positive integers k.

To complete the inductive step, assuming the inductive hypothesis that P(k) holds for an arbitrary integer k, show that must P(k + 1) be true.

Climbing an Infinite Ladder Example:

BASIS STEP: By (1), we can reach rung 1.
INDUCTIVE STEP: Assume the inductive hypothesis that we can reach rung k. Then by (2), we can reach rung k + 1.

Hence, P(k) → P(k + 1) is true for all positive integers k. We can reach every rung on the ladder.

Mathematical induction can be expressed as the rule of inference

where the domain is the set of positive integers.

In a proof by mathematical induction, we don’t assume that P(k) is true for all positive integers! We show that if we assume that P(k) is true, then P(k + 1) must also be true.
Proofs by mathematical induction do not always start at the integer 1. In such a case, the basis step begins at a starting point b where b is an integer. We will see examples of this soon.

Validity of Mathematical Induction ???

Mathematical induction is valid because of the well ordering property, which states that every nonempty subset of the set of positive integers has a least element. Here is the proof using Contradiction

Suppose that P(1) holds and P(k) → P(k + 1) is true for all positive integers k.
Assume there is at least one positive integer n for which P(n) is false. Then the set S of positive integers for which P(n) is false is nonempty.
By the well-ordering property, S has a least element, say m.
We know that m can not be 1 since P(1) holds.
Since m is positive and greater than 1, m − 1 must be a positive integer. Since m − 1 < m, it is not in S, so P(m − 1) must be true.
But then, since the conditional P(k) → P(k + 1) for every positive integer k holds, P(m) must also be true. This contradicts P(m) being false.
Hence, P(n) must be true for every positive integer n.

Mathematical Induction: Example

Sum of all Integers where i=1 and goes to n. Proof using Induction

Conjecturing and Proving a Formula

Strong Induction

Strong Induction: To prove that P(n) is true for all positive integers n, where P(n) is a propositional function, complete two steps:

Basis Step: Verify that the proposition P(1) is true.
Inductive Step: Show the conditional statement [P(1) ∧ P(2) ∧∙∙∙ ∧ P(k)] → P(k + 1) holds for all positive integers k.

Strong induction is sometimes called the second principle of mathematical induction or complete induction

Strong induction tells us that we can reach all rungs if:

We can reach the first rung of the ladder.
For every integer k, if we can reach the first k rungs, then we can reach the (k + 1)st rung.

To conclude that we can reach every rung by strong induction:

BASIS STEP: P(1) holds
INDUCTIVE STEP: Assume P(1) ∧ P(2) ∧∙∙∙ ∧ P(k) holds for an arbitrary integer k, and show that P(k + 1) must also hold.

We will have then shown by strong induction that for every positive integer n, P(n) holds, i.e., we can reach the nth rung of the ladder.

Proof Using Strong Induction ???

Example: Show that if n is an integer greater than 1, then n can be written as the product of primes.

Solution: Let P(n) be the proposition that n can be written as a product of primes.

BASIS STEP: P(2) is true since 2 itself is prime.
INDUCTIVE STEP: The inductive hypothesis is P(j) is true for all integers j with 2 ≤ j ≤ k. To show that P(k + 1) must be true under this assumption, two cases need to be considered:
- If k + 1 is prime, then P(k + 1) is true.
- Otherwise, k + 1 is composite and can be written as the product of two positive integers a and b with 2 ≤ a ≤ b < k + 1. By the inductive hypothesis a and b can be written as the product of primes and therefore k + 1 can also be written as the product of those primes.

Hence, it has been shown that every integer greater than 1 can be written as the product of primes.

Example: Prove that every amount of postage of 12 cents or more can be formed using just 4-cent and 5-cent stamps.

Solution: Let P(n) be the proposition that postage of n cents can be formed using 4-cent and 5-cent stamps.

BASIS STEP: P(12), P(13), P(14), and P(15) hold.
P(12) uses three 4-cent stamps.
P(13) uses two 4-cent stamps and one 5-cent stamp.
P(14) uses one 4-cent stamp and two 5-cent stamps.
P(15) uses three 5-cent stamps.

INDUCTIVE STEP: The inductive hypothesis states that P(j) holds for 12 ≤ j ≤ k, where k ≥ 15. Assuming the inductive hypothesis, it can be shown that P(k + 1) holds.

Using the inductive hypothesis, P(k − 3) holds since k − 3 ≥ 12. To form postage of k + 1 cents, add a 4-cent stamp to the postage for k − 3 cents.

Hence, P(n) holds for all n ≥ 12.

Well-Ordering Property

Well-ordering property: Every nonempty set of nonnegative integers has a least element.
The well-ordering property is one of the axioms of the positive integers.
The well-ordering property can be used directly in proofs.
The well-ordering property can be generalized.

Definition: A set is well ordered if every subset has a least element.

N is well ordered under ≤.
The set of finite strings over an alphabet using lexicographic ordering is well ordered.

Example: Use the well-ordering property to prove the division algorithm, which states that if a is an integer and d is a positive integer, then there are unique integers q and r with 0 ≤ r < d, such that a = dq + r.

Solution: Let S be the set of nonnegative integers of the form a − dq, where q is an integer. The set is nonempty since −dq can be made as large as needed.

By the well-ordering property, S has a least element r = a − dq0. The integer r is nonnegative. It also must be the case that r < d. If it were not, then there would be a smaller nonnegative element in S, namely, a − d(q0 + 1) = a − dq0 − d = r − d > 0.
Therefore, there are integers q and r with 0 ≤ r < d.

Which form of Induction Should I Choose?

We can always use strong induction instead of mathematical induction. But there is no reason to use it if it is simpler to use mathematical induction.
In fact, the principles of mathematical induction, strong induction, and the well-ordering property are all equivalent.
Sometimes it is clear how to proceed using one of the three methods, but not the other two.

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Goal of strong induction is to prove P(n), for n >= a

Step 1: (Basis) P(a), P(a+1), … P(b)
Step 2: (Induction) Assume P(i) for a <= i <= k, then prove P(k+1)
Example:

GIVEN: a[1] = 1, a[2] = 3, a[k] = a[k – 2] + 2a[k-1] (recursive definition, to get to ak you must know previous values)
EX: a[3] = a[1] + 2a[2] (a[1] = a[k – 2] and 2a[2] = 2a[k – 1]); if a1 = 1 and a2 = 3, then we get a3 = 1 + 2(3) = 7
1, 3, 7
CLAIM: a[k] is odd always
Proof by Induction
- Step 1: a[1] = 1 and a[2] = 3 are odd, was GIVEN
- Step 2: Assume that a[i] is odd, for all 1 <= i <= k
  - Then lets look at a[k+1] which is the term after a[k], we get: a[k+1] = a[k-1] + 2a[k]
    - We assumed a[i] is odd, therefore
    - a[k+1] = 2p+1 + 2(2q + 1) where p and q are some integers
    - a[k+1] = 2(p + 2q + 1) + 1
    - a[k+1]

Example

Prove that every amount of postage of $0.08 or more can be formed using $0.03 stamps and $0.05 stamps. Let this be P(n).

Think it through first… 8,9,10,11,12,13

Basis: (initial values)

P(8) is true using .03×1 + .05×1
P(9) is true .03×3
P(10) is true .05×2

Induction:

P(j) is true for 8<=j<=k and k>=10
We want k+1 cents of postage, since k>=10, we know P(k-2) is true by adding .03×1 stamp, which is k+1 of postage
Therefore P(k-2) + .03 –> P(k+1)
Examples:
- k=11; P(11-2) + .03 –> P(11+1)
  - P(9) + .03 –> P(12)
  - .09 + .03 –> .12 TRUE
- K(8) + .03 –> P(11)
  - .08 + .03 –> .11

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