Question

Use mathematical induction to prove that \(n^{2}

Short answer

In summary, using mathematical induction, we have proved that the inequality \(n^2 < n!\) is true for all integers \(n \geq 4\). We demonstrated this by showing that the statement is true for the base case \(n=4\), assuming it is true for an arbitrary \(k \geq 4\), and then proving it to be true for \(k+1\). This demonstrates that the inequality holds for all values of \(n\) greater than or equal to 4.

Step-by-step solution

Base case

We start by checking if the statement is true for \(n=4\): \(4^2 = 16\) and \(4!=4\times 3\times 2\times1=24\). Since \(16 < 24\), the statement is true for \(n=4\).

Inductive hypothesis

Assume that the statement is true for an arbitrary \(k \geq 4\), which means \(k^2 < k!\).

Inductive step

We want to prove that the statement is also true for \(k+1\). We have to show that \((k+1)^2 < (k+1)!\). Using the inductive hypothesis, we know \(k^2 < k!\). Now, multiply both sides by \(k+1\): \(k^2(k+1) < k!(k+1)\). We need to show that the left-hand side is less than the right-hand side. Observe that: \((k+1)^2 = k^2 + 2k + 1\) and \(k^2(k+1) = k^3 + k^2\). Since \(k\geq 4\), it is clear that \(k^3>k^2\), so: \(k^2+2k+1 < k^3+k^2 < k!(k+1)\). Therefore, we have shown that \((k+1)^2 < (k+1)!\) when \(k^2 < k!\).

Conclusion

By mathematical induction, we have proved that the statement \(n^2<n!\) holds for all \(n \geq 4\).

Key concepts

Base Case

The first step in mathematical induction is to establish the 'base case'. This is like the foundation of a building. We must prove that the statement is true for the initial value. In our scenario, the starting point is when \( n = 4 \). We evaluate both sides of the inequality: \( 4^2 = 16 \) and \( 4! = 4 \times 3 \times 2 \times 1 = 24 \).
Since \( 16 < 24 \), the base case holds true. This step guarantees that the statement works at least for one beginning instance, which is crucial to proceed to the next steps.

Inductive Hypothesis

The inductive hypothesis is where we make an educated assumption. Here, we assume that our statement is correct for an arbitrary integer \( k \), equal to or greater than 4. This assumption is our stepping stone. Imagine it like saying, "Let's pretend this bridge holds" before confidently crossing it.
So, we assume \( k^2 < k! \) to be true. This assumption will lead us to the next phase, where we aim to show that if the statement holds for \( k \), it must also hold for \( k+1 \).

Inductive Step

The inductive step is where we take on the task of proving the hypothesis for \( k+1 \). This follows the logic of dominoes: if one falls, the next one will fall too. Our goal is to demonstrate that if \( k^2 < k! \) is true, then \( (k+1)^2 < (k+1)! \) must also be true. We start with the expression for \( (k+1)^2 = k^2 + 2k + 1 \) and relate it to \( (k+1)! \).
By combining the hypothesis \( k^2 < k! \) with algebra, we multiply both sides by \( (k+1) \) and rearrange to confirm \( (k+1)^2 < (k+1)! \), completing the inductive step. Checking these inequalities ensures every domino in the sequence falls, proving the statement true indefinitely for all \( n \geq 4 \).

Factorial Inequality

Factorial inequalities involve comparing a factorial with another mathematical expression. In this case, the inequality we're interested in is \( n^2 < n! \). Factorials grow rapidly because each term is a multiplication of all positive integers up to \( n \). Hence, for \( n \geq 4 \), it's expected that \( n! \) quickly surpasses \( n^2 \).
Understanding this innate growth behavior of factorials helps compare them efficiently in inequalities. As \( n \) increases, \( n! \) expands faster than \( n^2 \), readily confirming the inequality for larger \( n \). This growth aspect is strategic in the proof by induction and illustrates how factorials can dominate quadratics like \( n^2 \) for sufficiently large numbers.