Assume that the case n = k is true, so therefore the case n = k + 1 is also true. How can n=k and n=k+1?

Assume that the case n = k is true, so therefore the case n = k + 1 is also true. How can n=k and n=k+1?

The Principle of Mathematical Induction: Proving n=k and n=k+1

Mathematical induction is a fundamental technique in mathematics that allows us to prove statements that are true for all natural numbers (1, 2, 3, and so on). The basic idea behind this method is to start with a base case, where we show that the statement is true for a specific value of n, and then use that information to prove that the statement is also true for the next value of n.

The principle of mathematical induction can be stated as follows:

  1. Base Case: Prove that the statement is true for a specific value of n, usually n=1 or n=0.
  2. Inductive Step: Assume that the statement is true for a particular value of n, say n=k. Then, prove that the statement is also true for the next value of n, which is n=k+1.

Once we have established the base case and the inductive step, we can conclude that the statement is true for all natural numbers.

Let's consider an example to illustrate this concept:

Suppose we want to prove the statement "the sum of the first n positive integers is equal to n(n+1)/2." In other words, we want to show that the following equation is true for all natural numbers n:

1 + 2 + 3 + ... + n = n(n+1)/2

Step 1: Base Case (n=1) For n=1, the left-hand side of the equation becomes 1, and the right-hand side becomes 1(1+1)/2 = 1. Therefore, the statement is true for n=1.

Step 2: Inductive Step (n=k to n=k+1) Assume that the statement is true for n=k. This means that:

1 + 2 + 3 + ... + k = k(k+1)/2

Now, we need to prove that the statement is also true for n=k+1. To do this, we can add (k+1) to both sides of the equation:

1 + 2 + 3 + ... + k + (k+1) = k(k+1)/2 + (k+1)

Simplifying the right-hand side, we get:

1 + 2 + 3 + ... + k + (k+1) = (k(k+1) + 2(k+1))/2
                           = (k^2 + 3k + 2)/2
                           = (k+1)(k+2)/2

This last expression is exactly the formula we wanted to prove for n=k+1. Therefore, we have shown that if the statement is true for n=k, then it is also true for n=k+1.

By combining the base case (n=1) and the inductive step (n=k to n=k+1), we can conclude that the statement is true for all natural numbers n.

The principle of mathematical induction is a powerful tool that can be used to prove a wide range of mathematical statements. It allows us to establish the truth of a statement for all natural numbers, even if we cannot easily verify it for every individual case. This technique is widely used in various branches of mathematics, computer science, and other fields where mathematical reasoning is essential.

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