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High School Geometry: Help and Review13 chapters | 162 lessons

Instructor:
*Laura Pennington*

Laura has taught collegiate mathematics and holds a master's degree in pure mathematics.

Mathematical induction is a method of proof that is often used in mathematics and logic. We will learn what mathematical induction is and what steps are involved in mathematical induction.

Have you ever met someone who doesn't like Rice Krispies treats? I haven't, and I would be willing to bet that everyone likes them! Suppose we wanted to prove that everyone in the world likes Rice Krispies treats. We might start by saying that I like them, my mom likes them, my friend likes them, and so on. However, unless we asked every single person in the world if they liked these treats (a fairly impossible feat since someone new is born every second or so), it wouldn't actually be proven. Proving that it is true for a select population makes the likelihood that everyone likes them higher, but doesn't truly prove that the whole world likes them.

It's problems similar to these that we can prove using a method called **mathematical induction**. Mathematical induction is a form of proof used to prove a property about all of the elements in an infinite set. When we are dealing with a problem involving an infinite set, it would be impossible to prove something is true about that set by just proving it for individual cases. Therefore, we use mathematical induction.

As mentioned, we use mathematical induction when we want to prove a property for an infinite number of elements. This is the main indicator that mathematical induction is a good method to use. There are a few questions involved in mathematical induction.

1.) Do we want to prove something for a set of elements that is infinite?

2.) Would it be easy to prove the property for the first element in the set?

3.) If we were to assume the property was true for the first *k* elements, can we use that to show that it is also true for the (*k* + 1)st element?

If the answers to these questions are yes, then mathematical induction is definitely the way to go, and the good news is that by answering these questions, you have already basically performed the proof! For example, suppose we wanted to prove that the sum of the first *n* positive integers is equal to (*n*(*n* + 1)) / 2.

The sum of the first *n* positive integers is given by the formula

The set of positive integers is an infinite set, so the answer to question 1 is yes. We could easily add up the first one, two, or three integers and make sure the formula holds, so the answer to question 2 is yes. Lastly, if we assume the property is true for the first *k* cases, can we use that to show it is true for the (*k* + 1)st case? If the answer to this is also yes, then that means our formula is true for all of the members of our infinite set. Why is this? Well, if we know that the formula is true for 3, then it must be true for 3+1 = 4. And if it is true for 4, then it must be true for 4+1 = 5. And so on to infinity. This is the key step in mathematical induction and will be easier to see when we go through the example below.

Mathematical induction is based on the fact that if something is true for the first *k* terms, and we show it is true for the (*k* + 1)st term, then it is true for all terms in an infinite set. This fact leads us to the steps involved in mathematical induction.

1.) Show the property is true for the first element in the set. This is called the **base case**.

2.) Assume the property is true for the first *k* terms and use this to show it is true for the (*k* + 1)st term. This is called the **induction step**.

Do you see now how answering the earlier mentioned questions actually does most of the work for you? Let's illustrate this by proving our example. That is, let's use induction to show that the sum of the first *n* positive integers is equal to (*n*(*n* + 1)) / 2.

The first step is to prove the base case, so we will show our property is true for *n* = 1, *n* = 2, and *n* = 3, even though it is only necessary to prove it for *n* = 1.

We've shown our base case to be true.

The second step is to assume that the property is true for the first *k* positive integers and show it is true for the (*k* + 1)st integer. That is, we will assume the sum of the first *k* positive integers is equal to (*k*(*k* + 1)) / 2, and use this to show that the sum of the first *k* + 1 positive integers is equal to ((*k* + 1)((*k* + 1) + 1)) / 2 or ((*k* + 1)(*k* + 2)) / 2.

We start with our assumption and manipulate it by adding *k* +1 to both sides. This gives us the left hand side of what we want to show. By simplifying the right hand side, we have shown that if the property is true for *k*, then the property is true for *k* + 1. Therefore, by mathematical induction, the sum of the first *n* positive integers is equal to (*n*(*n* + 1)) / 2.

**Mathematical induction** is a method of proof used when we want to prove a property for all the of elements in an infinite set. To perform mathematical induction, we first show that the property is true for the first element in the set. This is called the **base case**. Next, we assume the property is true for the first *k* elements and then show that it is true for the (*k* + 1)st element. This is called the **induction step**. The more you practice mathematical induction, the easier it becomes. This method of proof is extremely useful and is used quite often in mathematics and logic.

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High School Geometry: Help and Review13 chapters | 162 lessons

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