Applications of PMI in Proving Inequalities

Introduction

As your interest in mathematics grows, and you begin to work with material that takes proof seriously, you’ll reach a point where you start proving inequalities. 

Generally, it is easier to prove that something is equal to something else. You can manipulate both sides in the same way to reach the equation in question. There may be a need for some clever substitutions, but you only have some things to do. 

However, due to transitivity, proving inequalities is trickier. Let’s say, for some a and b, you want to show that a < b. It may not look very easy. If there is no obvious solution, you will have to take the help of c to show that a < c < b, thus indicating that a < b. 

The Principle of Mathematical induction (PMI) is a mathematical technique used to prove a variety of mathematical statements. It helps in proving identities, proving inequalities, and proving divisibility rules. 

Proof by Mathematical Induction

Imagine there is an infinite ladder.

1. You can reach the first rung of the ladder. 
2. If you can reach one rung of the ladder, you can also reach the next one. 

The first statement concludes that you can climb the first rung of the ladder. The second statement proves that you can climb to the second rung. Using the second statement again, it is believed that you can climb the third and the next set of rungs. You can apply the second statement as many times as you like to reach any particular rung. 

This particular example highlights proof by mathematical induction. 

The Transitive Property of Inequality

The transitive property of inequality proves several statements:

• If a < b and b < c, then a < c, where a and b are real numbers. 

It is also possible to make the statement by turning around the relationships or by making inclusive statements. 

• If a > b and b > c, then a > c.
• If a ≤ b and b ≤ c, it concludes that a ≤ c.
• If a ≥ b and b ≥ c, it leads to a ≥ c.

Arithmetic Properties of Inequalities

When it comes to other mathematical operations, the transitive property of inequality is as follows:

Addition

• When a < b, then a + c < b + c.
• When a > b, then a + c > b + c.

Subtraction

• When a < b, then a – c < bc.
• When a > b, then a – c > bc.

The properties are also valid for inclusive statements with ≤ and ≥. 

• When a ≥ b, then a + c ≥ b + c.
• When a ≤ b, then a + c ≤ b + c.

Similarly, 

• When a ≥ b, then a – c ≥ bc.
• When a ≤ b, then a – c ≤ bc.

Multiplication and Division

Whenever you multiply or divide an inequality with a negative term, the inequality sign is reversed. 

For instance, 

If 3 > 1 and you are multiplying both sides with (-3), it will be expressed as:

3 (-3) < 1 (-3)

= – 9 < -3

= -3 < -1

For positive numbers,

• When a < b and c > 0, then a/c < b/c and ac < bc. 
• When a > b and c > 0, then a/c > b/c and ac > bc.

Applications of PMI in Proving Inequalities

There are two steps involved in the principles of mathematical induction for proving inequalities. In the first step, you prove that the given statement is true for the initial value. It is known as the base step and is a factual statement. In the next step, you need to prove that the statement is true for the nth value. It is also known as the induction step and leads to proving that the statement holds for (n+1)the iteration. The statement here is a conditional one. 

Let us understand this with the help of various examples:

Example 1

Prove that 3n > n where n is a positive integer.

When n = 1, the statement is expressed as, 

31 > 1, which is true. 

Now, let us express the statement as 3k > k, where the positive integer k has a value higher than 1. To prove this statement, multiply it by 3 on both sides, 

3k X 3 > 3k

3k+1 > 3k.

For this example, we kept k > 1. Therefore, 3k > k + 1 is always true. The above statement can be expressed as,

3k+1 > 3k > k + 1. 

As a result, this particular statement is true when n = k as well as when n = k + 1. The statement is true for all positive values of n. 

Example 2

Prove that when a > 0, the inequality (1 + a)n > 1 + na is true for all positive integers when n ≥ 2.

(1 + a)2 = a2 + 2a + 1 > 1 + 2a

It means that the statement is true when n = 2. 

Take that (1 + a)k > 1 + ka for a positive integer greater than 2. Then multiply both sides of the equation by 1 + a. The statement is now expressed as,

(1 + a)k X (1 + a) = (1 + a)k+1 > (1 + ka)(1 + a)

      = ka2 + a(k + 1) + 1

      > 1 + (k + 1)a. 

As a result, the statement is true for n = k as well as for n = k + 1. It is proved that the inequality is true for all positive integers ≥ 2. 

Example 3

Use mathematical induction to prove n2 > 4n + 1 for n ≥ 6.

Let’s first verify if this statement is true for the base case. 

62 > 4(6) + 1

36 > 25, which is true. 

Now let us look at this statement with an arbitrary integer, k.

k2 > 4k + 1 fork > 6. If this statement is true by assumption, then the next value of k, which is (k + 1) should hold for the principles of mathematical induction to prove that the entire statement is true for all values of n. 

LHS = (k + 1)2

        = k2 + 2k + 1

        > (4k + 1) + 2k + 1 (using assumption k2 > 3k + 1)

        = 6k + 2

        = 3k + 3k + 2

        ≥ 3k + 20 (since k ≥ 6)

        > 4k + 5

        = 4k + 4 + 1

        = 4(k + 1) + 1

        = RHS

It proves LHS > RHS.

As a result, if the statement is true for n = k, it is also true for n = k + 1. And the statement is true for all positive integers n ≥ 6. 

Conclusion

Solving and proving any statement of equality is easy. But when you need to state and prove inequalities, you require the principle of mathematical induction. It is because you need to prove that the statement holds for all the natural numbers contained within. 

For additional support on the principle of mathematical induction, you can refer to the extensive study material by Unacademy. The courses also cover many other subjects, such as Physics, Chemistry, Biology, and more to help you prepare for your IIT entrance exam.