91Ó°ÊÓ

An inequality expresses a relationship between two expressions that may not necessarily be equal but denote a certain order.
In mathematical terms, inequalities use symbols like > (greater than), < (less than), \geq (greater than or equal to), and \leq (less than or equal to).
In the exercise, we need to show that the factorial of a number is always greater than an exponential function for all values greater than 3.

We start by understanding what an inequality signifies in this context.
We want to prove: \[n! > 2^n\] for all \( n > 3\).
This means the number of ways to arrange \(n\) items in a sequence (which is \(n!\)) grows faster than doubling \(n\) over and over (which is what \(2^n\) does).
By showing that the base case holds and then proving that if it holds for one value it must hold for the next, we form a chain of successes that proves the inequality for all acceptable values of \(n\).

A factorial, denoted as \(n!\), is the product of all positive integers up to \(n\).
For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).
Factorials grow extremely quickly as \(n\) increases. This rapid growth is crucial in our exercise.
Starting with a small value to understand how factorials work:

• \(3! = 1 \times 2 \times 3 = 6\)
• \(4! = 1 \times 2 \times 3 \times 4 = 24\)
• \(5! = 1 \times 2 \times 3 \times 4 \times 5 = 120\)

To show \( n! > 2^n \) for all \( n > 3\), we start with the base case, \( n = 4 \).
We see that \( 4! = 24 \) and \( 2^4 = 16 \). Clearly, \( 24 > 16 \).

The next step is to assume it works for some \( n = k \) and prove it must work for \( n = k + 1 \).
By the definition of factorials and performing simple arithmetic, it becomes evident that factorials will always outgrow exponential functions as \( n \) increases.

Exponential functions denote a constant base raised to a variable exponent.
In the problem, \(2^n\) represents an exponential function where the base 2 is raised to the power of \(n\).

Exponential functions grow rapidly, but not as quickly as factorials for large \(n\).
For example:

• \(2^3 = 8\)
• \(2^4 = 16\)
• \(2^5 = 32\)

As seen, even though \(2^n\) grows by doubling each time, factorial growth is much faster.
For instance:

• \(5! = 120\) while \(2^5 = 32\)
• \(6! = 720\) whereas \(2^6 = 64\)

To finalize the proof by induction:
To show \( (k+1)! > 2^{k+1} \) we assume \( k! > 2^k \) and then show \( (k+1)! = (k+1) \times k! > (k+1) \times 2^k\).

This simplifies to validating \( (k+1) > 2 \), a condition that holds true for \(k > 2\).
This concludes that our initial inequality \( n! > 2^n \) stands correct for all integers greater than 3.