Question

What is a strong induction?

Answer 1

Strong induction (also known as complete induction) is a proof technique used in mathematics and computer science to prove statements about integers or other well-ordered sets. It’s a variation of regular mathematical induction, but it provides a slightly stronger approach when proving a statement.

In regular induction, you prove that a statement is true for a base case (usually the smallest integer, often 0 or 1), and then prove that if it’s true for one case, it’s true for the next case.

Strong induction works similarly, but instead of proving that the statement is true for the next integer based on just the previous one, you assume the statement is true for all integers less than or equal to a certain integer \( k \), and use this to prove that it’s true for \( k+1 \).

Here’s how strong induction works:

Steps of Strong Induction:

1. Base Case: Prove the statement is true for the first integer (or integers). This could be \( P(0) \), \( P(1) \), or any starting point, depending on the problem.
   

   

1. Inductive Hypothesis: Assume that the statement is true for all integers from the base case up to some integer \( k \). That is, assume that \( P(0) \), \( P(1) \), ..., \( P(k) \) are true.
   

1. Inductive Step: Prove that under the assumption of the inductive hypothesis, the statement is true for \( k+1 \). This step shows that if the statement holds for all integers up to \( k \), it must also hold for the next integer, \( k+1 \).
   

Why Use Strong Induction?

You use strong induction when the proof requires considering multiple previous cases, not just the immediate previous case. It is particularly useful when the statement at \( k+1 \) depends on several previous values, not just \( k \).

Example:

Let’s say we want to prove that every integer \( n \geq 0 \) can be written as a sum of 4’s and 5’s.

1. Base Case:
   

   - For \( n = 0 \), we can write 0 as the sum of zero 4's and zero 5's, so the statement holds for \( n = 0 \).
   - For \( n = 4 \), we can write 4 as the sum of one 4, so the statement holds for \( n = 4 \).
   - For \( n = 5 \), we can write 5 as the sum of one 5, so the statement holds for \( n = 5 \).

1. Inductive Hypothesis: Assume that for all \( n \leq k \), the statement holds. That is, every integer from 0 to \( k \) can be written as a sum of 4’s and 5’s.
   

1. Inductive Step:
   

   - We want to show that the statement holds for \( k+1 \).
   - Since \( k+1 \geq 4 \), we know that \( k+1-4 = k-3 \) is also greater than or equal to 0, and by the inductive hypothesis, we can express \( k-3 \) as the sum of 4’s and 5’s.
   - Therefore, by adding one more 4 to this sum, we can write \( k+1 \) as the sum of 4’s and 5’s.

Thus, by strong induction, every integer \( n \geq 0 \) can be written as the sum of 4’s and 5’s.

This is the power of strong induction—it lets you assume that a statement is true for a range of previous values, making it easier to prove for larger values.