Problem 42 - Algebra and Trigonometry

Question

Some statements are false for the first few positive integers, but true for some positive integer \(m\) on. In these instances, you can prove \(S_{n}\) for \(n \geq m\) by showing that \(S_{m}\) is true and that \(S_{k}\) implies \(S_{k+1}\) when \(k \geq m .\) Prove that \(2^{n}>n^{2}\) for \(n \geq 5 .\) Show that the formula is true for \(n=5\) and then use step 2 of mathematical induction.

Step-by-Step Solution

Answer

The statement that \(2^n > n^2\) for \(n \geq 5\) has been proved using the principle of mathematical induction. The key steps were first establishing the truth of the base case, then verifying that if the statement holds for a particular \(k\) then it also holds for \(k + 1\). This approach confirms that the statement is true for all \(n \geq 5\).

Step 1: Base Case

First, verify that the formula holds for \(n = 5\). That is, confirm that \(2^5 > 5^2\). Calculate each side of the inequality: \(2^5 = 32\) and \(5^2 = 25\). Since \(32 > 25\), the statement is true for \(n = 5\).

Step 2: Inductive Step

Assume that our inequality holds for some arbitrary \(k \geq 5\); that is, assume \(2^k > k^2\). Now we want to prove that if the equation is true when \(n = k\), then it is also true when \(n = k + 1\). That is, if:\(2^k > k^2\), can we show that \(2^{k+1} > (k+1)^2\)?

Step 3: Apply the Inductive Hypothesis

We proceed by multiplying both sides of the assumption \(2^k > k^2\) by 2. Doing so yields \(2^{k+1} > 2k^2\). Note that for \(k \geq 5\), \(2k^2 > (k^2 + 2k + 1)\), which is equivalent to \((k+1)^2\). We use this to rewrite the previous inequality as \(2^{k+1} > (k+1)^2\). Therefore, if our inequality holds for \(k\), then it also holds for \(k + 1\).

Step 4: Conclusion

Thus, by the principle of mathematical induction, we have shown that for all \(n \geq 5\), \(2^n > n^2\) is indeed true.

Key Concepts

Understanding Inequalities

An inequality is a mathematical expression involving two values, showing that one value is smaller or larger than the other. In the context of our problem, we are comparing two expressions: \(2^n\) and \(n^2\). Inequalities are crucial because they help identify the relative sizes of numbers or expressions, leading to understanding how one grows in relation to another.
When solving inequalities, consider:

• When the inequality holds true (for which range or set of numbers).
• The nature of the inequality (whether it's a strict inequality like \(>\) or \(<\), or inclusive like \(\geq\)).
• How operations apply to each part of the inequality, keeping the relationship intact.

By demonstrating that \(2^n > n^2\) for \(n \geq 5\), we show the exponential function \(2^n\) grows faster than the quadratic function \(n^2\) as \(n\) increases.

Exploring Exponentiation

Exponentiation is a mathematical operation that involves raising a base number to the power of an exponent. In our problem, we are particularly interested in the exponential expression \(2^n\). The exponent indicates how many times the base is multiplied by itself.
Key points to consider include:

• As \(n\) increases, \(2^n\) grows rapidly compared to polynomial expressions like \(n^2\).
• Exponentiation can significantly alter the size of a number, making it much larger or smaller depending on the exponent.
• This rapid growth is why for \(n \geq 5\), \(2^n\) outpaces \(n^2\).

The power of exponentiation is at the heart of proving such inequalities, especially in problems involving growth rates.

The Base Case

The base case is a fundamental part of mathematical induction. It serves as the starting point that demonstrates the base condition is true. In our example, the base case confirms that \(2^n > n^2\) is true for \(n=5\). This means checking the inequality for this initial value.
Here's why it's crucial:

• Establishing a true base case is necessary to apply the principle of induction.
• Verifying the base case prevents false assumptions about the inequality's truthfulness for all integers \(n\).
• Without a true base case, the chain of logic in mathematical induction cannot begin.

By showing that \(2^5 = 32\) and \(5^2 = 25\), we verify that the base case holds because \(32 > 25\).

Understanding the Induction Hypothesis

The induction hypothesis is a provisional assumption used in the proof technique of mathematical induction. It involves assuming that a given statement or inequality is true for some arbitrary value, \(k\).
This hypothesis serves as a bridge:

• Assume the inequality \(2^k > k^2\) holds for a specific \(k \geq 5\).
• Use this assumption to prove it holds for \(k + 1\): that is, \(2^{k+1} > (k+1)^2\).
• This step is crucial for proving the generality of the statement, extending its validity from a particular case to all cases \(n \geq 5\).

The induction hypothesis underpins the transition from one value of \(n\) to the next, allowing us to show a pattern holds true according to the logic of mathematical induction.