Question

The inequality \(\mathrm{n} !>2^{\mathrm{n}-1}\) is true for (a) \(n>2, n \in N\) (b) \(n<2\) (c) \(\mathrm{n} \in \mathrm{N}\) (d) None of these

Short answer

The inequality holds true for \(n > 2\), \(n \in \mathbb{N}\), and also for \(n = 0\). So, the correct option is (a) \(n > 2\), \(n \in \mathbb{N}\), but it must be noted that \(n = 0\) also satisfies the inequality.

Step-by-step solution

Case 1: n = 1

When n = 1, the inequality becomes: (1)! > 2^(1-1) 1 > 1 Since the inequality is false, this case does not hold true.

Case 2: n > 2

Let's check for the values of n in natural numbers when n is greater than 2. For n = 3: (3)! > 2^(3-1) 6 > 4 The inequality holds true for n = 3. For n = 4: (4)! > 2^(4-1) 24 > 8 The inequality holds true for n = 4. We can now try to prove that the inequality holds true for n > 4 using the induction method. Let the inequality be true for n = k, i.e., (k)! > 2^(k-1) To prove: The inequality is true for n = k+1, i.e., (k+1)! > 2^k Now, (k+1)! = (k+1) * k! Since, k! > 2^(k-1), by inductive hypothesis: (k+1)! > (k+1) * 2^(k-1) To prove the inequality (k+1)! > 2^k is true, we need to show that: (k+1) * 2^(k-1) > 2^k Dividing both sides by 2^(k-1), we have: k+1 > 2 Since we assumed k > 2, k+1 > 2 is true. Hence, the inequality n! > 2^(n-1) holds true for n > 2 or n = 3, 4, and so on, for all natural numbers.

Case 3: n < 2

This case suggests that if n = 0 (since natural numbers start from 0), the inequality is: (0)! > 2^(-1) 1 > 0.5 The inequality holds true for n = 0. Thus, for n > 2 and n < 2, the inequality holds true. However, only for n = 1, the inequality does not hold. So, the correct option is: (a) n > 2, n ∈ N, keeping in mind that n = 0 also satisfies the inequality.

Key concepts

Mathematical Induction

Mathematical induction is a powerful proof technique used in mathematics, particularly useful when proving statements about natural numbers. It consists of two main steps. First, we verify the statement for the initial case—this is known as the base case. In the context of factorial inequalities, such as proving !! > 2^{n-1} > 2n−1, we would first check if the statement holds for the smallest value of , typically 1 or 2.

After the base case, comes the induction step. We assume that the statement holds for some arbitrary natural number = ! =k=k—this assumption is the inductive hypothesis. Then, using this assumption, we need to prove that the statement also holds for = ! k+1=k+1. If both these steps are successful, we can conclude that the statement is true for all natural numbers greater than or equal to our base case, due to the principle of induction.

To ensure clarity and understanding while solving factorial inequalities using mathematical induction, one must clearly state the base case and walk through the induction step meticulously. Highlighting the inductive hypothesis and how it's used in the induction step can singularly emphasize the logic behind the proof technique.

Natural Numbers

Natural numbers are the set of positive integers, starting from 1: ! 1, 2, 3, ...=1,2,3,…!. They are the numbers we naturally use for counting objects. In the context of our inequality !! > 2^{n-1} > 2n−1, it’s important to note that the existence of a factorial for a natural number implies the process of multiplying a series of descending natural numbers down to 1. For example, ! 5!=5! is 5 × 4 × 3 × 2 × 1 = 1205×4×3×2×1=120.

When analyzing inequalities involving natural numbers and factorials, one should remember that factorials grow incredibly fast—much faster than exponential functions like ! 2^n2n!. This is a key insight that generally guides the proof when comparing factorials with exponential functions using inequalities. It's also worth mentioning that 0 is sometimes included in the set of natural numbers, especially in the context of combinatorics, which affects how we approach problems like the given exercise.

Inequality Solving

Inequality solving is a branch of mathematics that deals with finding the range of values that satisfy a given inequality. It's an essential skill for analyzing and understanding mathematical relationships and constraints. In the provided exercise with the inequality !! > 2^{n-1} > 2n−1, the task is to determine for what values of the inequality is true.

The process of solving inequalities often involves similar techniques used in solving equations, such as adding, subtracting, multiplying, or dividing both sides without changing the direction of the inequality, provided we multiply or divide by a positive number. However, multiplying or dividing by a negative number requires us to reverse the direction of the inequality sign.

Crucial strategies for dealing with inequalities involving factorials and exponents include finding base cases, considering growth rates, and employing logical reasoning or proof methods like mathematical induction. When aiming to simplify the students' understanding, presenting clear examples that contrast cases like = ! 1=1! (where inequality doesn't hold) with > ! 2>2! offers insight into why certain numbers satisfy the inequality and others do not.