Problem 30 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 27-32: Find the smallest positive integer \(j\) for which the statement is true. Use the extended principle of mathematical induction to prove that the formula is true for every integer greater than \(j\). $$ n^{2} \leq 2^{n} $$

Step-by-Step Solution

Answer

The smallest positive integer \( j \) is 4.

Step 1: Initial Examination

Firstly, examine the given inequality \( n^2 \leq 2^n \) for small values of \( n \). Start testing with \( n = 1 \): \( 1^2 = 1 \leq 2^1 = 2 \). Hence, the inequality holds true for \( n = 1 \).

Step 2: Testing for Higher Values of n

Continue testing the inequality for \( n = 2 \): \( 2^2 = 4 \leq 2^2 = 4 \). Still holds true. For \( n = 3 \): \( 3^2 = 9 \leq 2^3 = 8 \), which does not hold. Now, check \( n = 4 \): \( 4^2 = 16 \leq 2^4 = 16 \). The inequality holds again. Check \( n = 5 \): \( 5^2 = 25 \leq 2^5 = 32 \). It holds true. Thus, for \( n = 3 \), the inequality doesn’t hold, and for \( n = 4 \) upwards, it holds.

Step 3: Identify the Smallest Positive Integer j

Since for \( n = 3 \), the statement is false and for \( n = 4 \) and greater, it’s true; the smallest integer \( j = 4 \) for which the inequality holds for all \( n \geq j \).

Step 4: Inductive Hypothesis

Assume that the inequality is true for all integers from 4 up to \( k \) (i.e., \( k^2 \leq 2^k \)). Our goal is to prove it for \( n = k + 1 \).

Step 5: Induction Step

For \( n = k + 1 \), prove that \( (k+1)^2 \leq 2^{k+1} \). \((k+1)^2 = k^2 + 2k + 1\). By the inductive hypothesis, \( k^2 \leq 2^k \). Thus, \( k^2 + 2k + 1 \leq 2^k + 2k + 1 \). We need to show \( 2^k + 2k + 1 \leq 2^k \cdot 2 = 2^{k+1} \). Since \( 2k + 1 \leq 2^k \) for \( k \geq 4 \), the inequality holds.

Step 6: Conclusion

Thus, by the principle of mathematical induction, \( n^2 \leq 2^n \) is verified as true for all integers \( n \geq 4 \).

Key Concepts

Inequality

An inequality is a statement that compares two values, expressions, or quantities. It implies that one side of the statement is less than or equal to, greater than or not equal to the other side.
In algebra, inequalities are as crucial as equations. They allow us to express a range of values rather than a single number. For example, in the inequality \( n^2 \leq 2^n \), the value of \( n^2 \) is bounded above by the value of \( 2^n \).
Studying inequalities helps in understanding relationships among numbers and changes brought by operations. You have to manipulate these inequalities carefully by performing operations equally on both sides without changing their signs.

• "Less than" is denoted by \(<\)
• "Greater than" is denoted by \(>\)
• "Less than or equal to" is denoted by \(\leq\)
• "Greater than or equal to" is denoted by \(\geq\)

Integer

Integers are the set of whole numbers and their opposites, including zero. This means they are the numbers ..., -3, -2, -1, 0, 1, 2, 3, ... and so on.
Unlike decimals or fractions, integers do not have fractional or decimal parts. They are used in a wide variety of contexts, including counting, ordering, and in arithmetic operations.
In the context of our problem, integers are those input values of \( n \) we are interested in. Specifically, we focus on positive integers, from a certain point \( j \) and beyond, since integers are discrete, ensuring we find the smallest one satisfying the condition. This leads us through the determination of \( j \) where \( n^2 \leq 2^n \) is true for all \( n \geq j \).

• Whole, non-fractional numbers
• Include negatives, zero, and positives
• Vital for counting and ordering

Power Comparison

When comparing powers of numbers, specifically looking at expressions like \( n^2 \) against \( 2^n \), it means observing how quickly these quantities grow as \( n \) increases. Functions involving exponents can increase very rapidly compared to polynomial functions.
For our inequality problem, when comparing \( n^2 \) and \( 2^n \), it is essential to realize that, generally, exponential functions will eventually outpace polynomial functions. This is manifestly evident as you compute higher and higher powers of two, observing where it overcomes each square of the same integer.
The dynamics between how \( n^2 \) increases (a quadratic growth) and \( 2^n \) (an exponential growth) reveal why the inequality consistently holds past a certain point. Understanding this concept helps predict the behavior and the magnitude of these expressions as \( n \) grows.

Inductive Hypothesis

The inductive hypothesis is a cornerstone of mathematical induction, allowing you to prove statements for all integers sequentially. It acts as your assumptive step, where you believe that your inequality or statement is true for some integer \( k \).
In our case, the inductive hypothesis assumes that \( n^2 \leq 2^n \) is valid for some specific value \( n=k \). From this assumption, we prove that if it holds for \( n=k \), it must also be true for \( n=k+1 \). This chain reaction leads to the conclusion that the inequality is valid for all integers starting from the base integer \( j=4 \).
This method is built upon two key steps:

• Base Case: Show the inequality holds for the starting integer \( j \).
• Inductive Step: Assume true for \( n=k \), and prove for \( n=k+1 \).

When both steps are proven, you achieve a solid logical bridge allowing extension of validity to all specified integers.