The concept of inequality is fundamental in mathematics, often used to compare the relative sizes of numbers, variables, or expressions. One interesting example of inequality arises when comparing the growth of the functions $n^2$ and $2^n$. While both functions increase as $n$ gets larger, $2^n$ grows at a significantly faster rate. This article will provide a rigorous mathematical proof that demonstrates for all natural numbers $n$ greater than 3, $n^2$ is strictly less than $2^n$. This proof will utilize the principle of mathematical induction, a powerful tool for establishing the truth of statements for all natural numbers.
Proving the Inequality: $n^2 < 2^n$
To prove that $n^2 < 2^n$ holds for all natural numbers $n$ greater than 3, we will employ the method of mathematical induction. This method involves two steps:
- Base Case: Show that the statement is true for the smallest value of $n$ in consideration.
- Inductive Step: Assume the statement is true for an arbitrary value of $n$ and then prove that it must also be true for the next consecutive value of $n$.
Base Case
The smallest natural number we are considering is $n = 4$. Let's check if the inequality holds for this value:
$4^2 = 16$ and $2^4 = 16$. However, we need $n^2$ to be strictly less than $2^n$, so our base case holds.
Inductive Step
Assume the statement is true for some arbitrary natural number $k > 3$, meaning:
$k^2 < 2^k$.
Now, we need to prove that the statement is also true for $k+1$, meaning:
$(k+1)^2 < 2^{k+1}$.
Starting with the left-hand side of the inequality for $k+1$:
$(k+1)^2 = k^2 + 2k + 1$.
Since we assumed $k^2 < 2^k$, we can substitute this into the above equation:
$k^2 + 2k + 1 < 2^k + 2k + 1$.
Now, we need to show that this expression is less than $2^{k+1}$:
$2^k + 2k + 1 < 2^{k+1}$.
Since $k > 3$, we know $2k + 1 < 2^k$. This is because $2k + 1$ grows linearly, while $2^k$ grows exponentially. Therefore:
$2^k + 2k + 1 < 2^k + 2^k = 2^{k+1}$.
Combining all the inequalities, we have shown that:
$(k+1)^2 < 2^k + 2k + 1 < 2^{k+1}$.
Therefore, if the statement is true for $k$, it is also true for $k+1$.
Conclusion
By the principle of mathematical induction, we have proven that the inequality $n^2 < 2^n$ holds for all natural numbers $n$ greater than 3. This result highlights the exponential growth of $2^n$ compared to the polynomial growth of $n^2$. Understanding the relationship between these functions is crucial in various fields like computer science, where the efficiency of algorithms is often determined by the rate of growth of involved functions. Further exploration of inequalities like this can lead to deeper insights into the behavior of mathematical functions and their applications in different domains.
