# How to Use Mathematical Induction in Discrete Math

April 28, 2023
Selena Mendoza
Discrete Mathematics
Dr. Selena has a background in discrete mathematics, specializing in combinatorics, and graph theory. After receiving her Ph.D. in mathematics from the University of California, Berkeley, she went on to work as a lecturer at various community colleges. Her areas of interest in the study are computational complexity, combinatorial optimization, and algorithm analysis.

In discrete mathematics, mathematical induction is a crucial tool for proving claims involving integers or other discrete structures. Several different types of propositions, including those involving inequalities, divisibility, and combinatorics, can be proven using the potent method of induction. We will give an outline of mathematical induction in this blog article, including its principles to discrete mathematics.

## The Principle of Mathematical Induction

We can prove statements about numbers using the notion of mathematical induction, a potent proof method. According to the principle,

we can show two things if we want to prove a statement P(n) for all positive integers n:

• In the basic situation, which is often n=1, the claim is accurate.
• If the assertion is accurate for every given positive integer k, we can demonstrate that it is likewise accurate for k+1.

To illustrate how this functions, let's have a look at an example.

Let's say that we want to establish the following truth for all positive integers n:

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

First, we demonstrate that the statement is accurate for the simplest case, n=1:

1 = 1(1+1)/2

The assertion is accurate for n=1 because the left and right sides of the equation are both equal to 1.

The next step is to assume that the assertion is accurate for any positive integer k, and then attempt to demonstrate that it is accurate for k+1. Thus.

We presume:

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

And we aim to demonstrate that:

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

To accomplish this, we start with the left side of the equation and multiply both sides by (k+1):

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

This expression can be made simpler by factoring out (k+1)/2:

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

Now we can simplify the expression on the right-hand side using the formula for the sum of the first k integers:

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

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

The assertion is accurate for all positive integers n because this is precisely what we set out to demonstrate.

## Induction Proofs for Inequalities

Inequalities can also be demonstrated through mathematical induction.

We need to demonstrate two things to use mathematical induction to prove an inequality:

• In the base situation, often n=1, the inequality is true.
• If the inequality holds for every given positive integer k, we may demonstrate that it also holds for k+1.

Let's examine a case in point. Consider that we wish to demonstrate the following inequality for all positive integers n:

1 + 2 + 3 + ... + n < (n+1)^2

First, we demonstrate that the inequality holds true in the simplest case, where n equals 1, using the formula 1 (1+1)2.

The inequality is valid for n=1 since its left and right sides are both equal to 1.

Then, assuming that the inequality holds for every given positive integer k,

We attempt to demonstrate that it also holds for k+1. Thus, we presume:

1 + 2 + 3 + ... + k < (k+1)^2

And we aim to demonstrate that:

1 + 2 + 3 + ... + k + (k+1) < (k+2)^2

1 + 2 + 3 + ... + k + (k+1) < (k+1)^2 + (k+1)

This expression can be made simpler by factoring out (k+1):

1 + 2 + 3 + ... + k + (k+1) < (k+1)(k+2)

Now we can simplify the statement on the left-hand side using the formula for the sum of the first k integers:

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

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

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

Nearly what we intended to demonstrate, but we need to demonstrate that (k+2)2/2 is greater than (k+1)2. To accomplish this,

We demonstrate that (k+2)2/2 - (k+1)2 > 0:

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

2k+3/2 is greater than 0 since k is a positive integer. As a result, we have demonstrated:

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

Therefore, for any positive integer n, the inequality holds true.

## Strong Induction

Only regular induction has been discussed thus far; it starts by assuming that the proposition is true for any k and then shows that it is also true for k+1. A different type of induction, known as strong induction, presupposes that the statement is accurate for all positive integers less than k and then establishes that it is accurate for k.

• We must demonstrate two things to apply strong induction.
• In the basic situation, which is often n=1, the claim is accurate.
• If the premise holds for all positive integers less than or equal to k, we may demonstrate that the premise holds for k+1.

Let's examine an illustration to learn how these functions work.

Let's say that we want to establish the following truth for all positive integers n:

A sum of different powers of 2 can be used to represent any positive integer n.

First, we demonstrate that the statement is accurate for the n=1 base case. Since 1 equals 20, this is a trivial situation.

After that, we presumptively wish to show that the proposition holds true for all positive integers less than or equal to k+1.

As a sum of powers of 2, we can express k+1 as follows:

Where m1 > m2 >... > mk >= 0, k+1 = 2m1 + 2m2 +... + 2mk. Since m1 is assumed to be the highest power of 2 that is less than or equal to k+1, it follows that m1 = log2(k+1).

Think about the number k+1 - 2m1, now. We know that k+1 - 2m1 is less than 2m1 because 2m1 is the highest power of 2 that is less than or equal to k+1. Therefore, since any power of two that appears in k+1 - 2m1 cannot also appear in 2m1, we can write k+1 - 2m1 as a sum of different powers of two.

Consequently, we have k+1 = 2m1 + (k+1 - 2m1).

The induction hypothesis informs us that both m1 and k+1 - 2m1 can be expressed as a sum of different powers of 2 because they are both less than or equal to k. As a result, k+1 can be expressed as the sum of several powers of 2.

This concludes the strong induction proof. We have demonstrated the validity of the assertion for the case where n=1 and, assuming the validity of the statement for all positive integers less than or equal to k, we have demonstrated the validity of the statement for k+1. Therefore, for all positive integers n, the proposition is true.

## Tips and Tricks for Using Mathematical Induction

1. Specify the claim that has to be proven: It's crucial to grasp exactly what you are attempting to show before beginning the proof. Find the statement and make sure it is precise and well-defined.
2. Confirm the default case: Show that the assertion is accurate for the least value of n, often 1. Although it is frequently the simplest stage, it is crucial to make sure the proof is accurate.
3. Assume the assertion is true for k: Assume the statement is true for some positive integer k for purposes of regular induction. Assume that the assertion is accurate for all positive integers less than or equal to k to perform strong induction.
4. Establish that the claim is accurate for k+1: Using the presumption from step 3, establish that the claim is accurate for k+1 (for weak induction) or for all positive integers less than or equal to k+1 (for regular induction).
5. Use simple language: Induction proofs can be complicated, thus it's crucial to explain each stage of the proof using simple language. To make the evidence easier to understand, use logical and mathematical symbols as needed.
6. Practice: The more you use mathematical induction, just like any other ability, the more proficient you'll become. To improve your comprehension and abilities, try tackling various proofs and challenges.

## Mathematical Induction Examples

Here are some illustrations of discrete mathematics applications of mathematical induction:

Example 1: Prove that the sum of the first n positive integers is n(n+1)/2.

Base Case: Assume that the statement is accurate for some positive integer k, i.e., that the product of the first k positive numbers is equal to k(k+1)/2.

Inductive hypothesis: To prove that the statement is accurate for k+1, we must first demonstrate that the sum of the first k+1 positive integers is equal to (k+1)(k+2)/2.

Inductive Step: The formula is as follows: 1 + 2 +... + k + (k+1) = (k(k+1)/2) + (k+1) = (k+1)(k+2)/2

The claim is therefore accurate for k+1.

The assertion holds for all positive numbers n, according to mathematical induction.

Example 2: Prove that every integer greater than 1 can be expressed as a product of prime numbers.

Base Case: n=2. Since 2 is a prime number by itself, it is a prime number. So, for n=2, the assertion is accurate.

Inductive Hypotheses: Every positive integer higher than 1 and less than or equal to k can be written as a product of prime numbers, according to the inductive hypothesis. Assume that this is true for some positive integer k.

Inductive Step: to demonstrate that the statement—according to which every integer bigger than one and less than or equal to k+1 can be written as a product of prime numbers—is true for k+1.

K+1 is a prime number product if k+1 is a prime number. Otherwise, if a and b are positive integers and 1 a = b k+1, respectively, then k+1 is composite and can be expressed as k+1 = ab. According to the induction hypothesis, both a and b can be written as prime number products. Consequently, k+1 can also be written as a prime number product.

By mathematical induction, the assertion is true for all positive numbers greater than 1 because it holds true for n=2 and is true for k+1 whenever it holds true for k.

Example 3: Prove that every positive integer can be expressed as a sum of distinct powers of 2.

Base Case: n=1; 1 can be stated as the sum of several powers of 2 since 1 is a power of 2. The claim is therefore accurate when n=1.

Inductive Hypothesis: Assume that every positive integer less than or equal to k can be stated as the sum of several powers of 2 and that the statement is true for some positive integer k.

Inductive Step: The statement that every positive number less than or equal to k+1 may be written as the sum of different powers of 2 must be proven to be true for k+1.

If k+1 is a power of two, it can be written as the sum of several different powers of two. In the absence of such, let m be the biggest power of 2 that is less than or equal to k+1. The induction hypothesis states that k+1-m is a positive integer smaller than m and may be written as the sum of different powers of 2. Therefore, by adding m to the sum that expresses k+1-m, k+1 can be written as a sum of several powers of 2.

By mathematical induction, the assertion must be true for all positive integers since it is true for n=1 and is true for k+1 whenever it is true for k.

## Conclusion

A potent method for demonstrating statements about positive integers is mathematical induction. Regular induction presupposes that the assertion is accurate for k and establishes that it is accurate for k+1. To prove that a proposition is true for k, strong induction first assumes that it is true for all positive integers less than or equal to k. Two steps must be taken to employ mathematical induction. For the base case, which is often n=1, you must first demonstrate that the statement is true. To use regular induction or strong induction, you must first assume that the assertion is true for some value of k and then demonstrate that it is also true for k+1 or all positive integers less than or equal to k. Discrete mathematics uses mathematical induction extensively to prove numerous significant theorems and statements. Any math or computer science student would benefit from mastering this strategy.