proof the mathematical induction
To prove a statement using mathematical induction, we must first show that it holds for a base case, usually when n=1 or n=0. Then, we assume that the statement holds for some arbitrary positive integer k, and we use this assumption to prove that it must also hold for the next integer k+1. The steps to prove a statement using mathematical induction are as follows:
Step 1: Base Case
Show that the statement holds for the smallest possible value of n. That is, show that the statement is true when n=1 or n=0 (depending on the problem). This step is important because it provides a starting point for the induction process.
Step 2: Inductive Hypothesis
Assume that the statement is true for some arbitrary positive integer k. This assumption is called the inductive hypothesis.
Step 3: Inductive Step
Using the inductive hypothesis, prove that the statement must also be true for the next integer, k+1. This step involves showing that if the statement holds for k, then it must also hold for k+1.
Step 4: Conclusion
Conclude that the statement is true for all positive integers n, using the principle of mathematical induction.
Here is an example of how to use mathematical induction to prove that the sum of the first n positive integers is n(n+1)/2:
Step 1: Base Case
When n=1, the sum of the first n positive integers is simply 1, which is equal to 1(1+1)/2. Therefore, the statement is true when n=1.
Step 2: Inductive Hypothesis
Assume that the statement is true for some arbitrary positive integer k. That is, assume that the sum of the first k positive integers is k(k+1)/2.
Step 3: Inductive Step
Using the inductive hypothesis, we must show that the statement is also true for k+1. The sum of the first k+1 positive integers is:
1 + 2 + 3 + ... + k + (k+1)
By the inductive hypothesis, the sum of the first k positive integers is k(k+1)/2, so we can substitute this expression to get:
1 + 2 + 3 + ... + k + (k+1) = k(k+1)/2 + (k+1)
We can simplify this expression by finding a common denominator:
1 + 2 + 3 + ... + k + (k+1) = (k(k+1) + 2(k+1))/2
1 + 2 + 3 + ... + k + (k+1) = (k+1)(k+2)/2
Therefore, the statement is true for k+1 as well.
Step 4: Conclusion
By the principle of mathematical induction, the statement is true for all positive integers n. Therefore, we have proven that the sum of the first n positive integers is n(n+1)/2.