You are currently viewing Solving Mathematical Induction: The Basics

Solving Mathematical Induction: The Basics

Mathematical induction is a powerful proof technique used in mathematics to prove statements about natural numbers. It can be used to prove statements about infinite sets of natural numbers, such as the set of all positive integers. The technique is based on the idea that if a statement is true for a specific number, then it is true for all numbers greater than that specific number. In this post, we’ll go over the basics of mathematical induction, why it’s important, and the steps for solving problems using this technique.

Steps for solving mathematical induction

Solving Mathematical Induction

The process of solving mathematical induction can be broken down into two steps: the base case and the inductive step.

  1. Base case: The first step in solving mathematical induction is to prove the statement for the smallest natural number. This is known as the base case. The base case must be proven before the inductive step can be completed.
  2. Inductive step: The second step is to assume that the statement is true for some arbitrary natural number, and then prove that it is also true for the next natural number. This is known as the inductive step. To complete the inductive step, we use the assumption that the statement is true for some natural number to prove that it is true for the next natural number.

Examples of mathematical induction

Example 1: Prove that 1 + 2 + 3 + … + n = n(n+1)/2 for all natural numbers.

Base case: The statement is true for n = 1, as 1 = 1(1+1)/2.

Inductive step: Assume that the statement is true for some natural number k, that is, 1 + 2 + 3 + … + k = k(k+1)/2. We must prove that it is also true for k+1. Using the assumption that the statement is true for k, we can add k+1 to both sides of the equation to get 1 + 2 + 3 + … + k + (k+1) = k(k+1)/2 + (k+1) = (k+1)(k+2)/2. This shows that the statement is true for k+1, which completes the inductive step.

Example 2: Prove that 2^n > n for all natural numbers.

Base case: The statement is true for n = 1, as 2^1 = 2 > 1.

Inductive step: Assume that the statement is true for some natural number k, that is, 2^k > k. We must prove that it is also true for k+1. Using the assumption that the statement is true for k, we can multiply both sides of the equation by 2 to get 2^k * 2 > k * 2. This simplifies to 2^(k+1) > k+1, which shows that the statement is true for k+1, which completes the inductive step.

Related:

Conclusion

In this post, we covered the basics of mathematical induction, the steps for solving problems using this technique, and provided examples to illustrate how it works. Remember that mathematical induction is a powerful proof technique that can be used to prove statements about infinite sets of natural numbers. Keep practicing with mathematical induction and you’ll soon master this technique.