Proving Inequality by Induction: $n! > 2^n$
Mathematical induction is a powerful tool for proving statements about natural numbers. It involves establishing a base case and an inductive step. In this article, we will use mathematical induction to prove the inequality $n! > 2^n$ for all natural numbers $n \geq 4$.
Base Case:
Let's start with the base case, $n = 4$. We need to show that $4! > 2^4$.
- $4! = 4 \times 3 \times 2 \times 1 = 24$
- $2^4 = 2 \times 2 \times 2 \times 2 = 16$
Since $24 > 16$, the inequality holds for the base case.
Inductive Step:
Now, assume that the inequality holds for some arbitrary natural number $k \geq 4$. This means we are assuming that $k! > 2^k$. Our goal is to prove that the inequality also holds for the next natural number, $k+1$. In other words, we need to show that $(k+1)! > 2^{k+1}$.
Starting with the left-hand side of the inequality we want to prove:
$(k+1)! = (k+1) \times k!$
Since we assumed $k! > 2^k$, we can substitute this into the equation:
$(k+1)! = (k+1) \times k! > (k+1) \times 2^k$
Now, we need to show that $(k+1) \times 2^k$ is greater than $2^{k+1}$.
Notice that for $k \geq 4$, $(k+1) > 2$. Therefore,
$(k+1) \times 2^k > 2 \times 2^k = 2^{k+1}$
This shows that $(k+1)! > 2^{k+1}$.
Conclusion:
We have shown that the inequality holds for the base case $n=4$, and we have proven that if the inequality holds for an arbitrary natural number $k \geq 4$, then it also holds for $k+1$. Therefore, by the principle of mathematical induction, the inequality $n! > 2^n$ holds for all natural numbers $n \geq 4$.
This proof demonstrates the power of mathematical induction in establishing the truth of statements about natural numbers. It provides a rigorous and systematic way to prove inequalities, and it can be applied to a wide range of mathematical problems. The understanding of induction helps us to build a deeper understanding of how mathematical proofs work and to develop new and creative ways to solve problems.
