Problem 31 - 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 \(j\). $$ 2 n+2 \leq 2^{n} $$

Step-by-Step Solution

Answer

The smallest positive integer \( j \) is 3.

Step 1: Understanding the Problem

We need to find the smallest positive integer \( j \) such that for all integers\( n \geq j \), the inequality \( 2n + 2 \leq 2^n \) holds true. We will use the principle of mathematical induction to confirm that this inequality is valid for all integers greater than \( j \).

Step 2: Finding the Base Case

To find the smallest such \( j \), we start by testing small values of \( n \) in the inequality. For \( n = 1 \):\( 2(1) + 2 = 4 \leq 2^1 = 2 \), which is false. Next, for \( n = 2 \): \( 2(2) + 2 = 6 \leq 2^2 = 4 \), also false. For \( n = 3 \): \( 2(3) + 2 = 8 \leq 2^3 = 8 \), which is true. This suggests checking further.

Step 3: Confirming the Validity of j

For \( n = 4 \), check if\( 2(4) + 2 = 10 \leq 2^4 = 16 \), which is true. This suggests \( j = 3 \) may be the correct minimum integer. We should perform mathematical induction from \( n = 3 \).

Step 4: Inductive Step Setup

Assume the statement is true for some arbitrary integer \( k \geq 3 \): \( 2k + 2 \leq 2^k \). We must prove that the statement holds for \( k + 1 \), i.e., \( 2(k+1) + 2 \leq 2^{k+1} \).

Step 5: Inductive Step Verification

Starting from the assumption: \( 2k + 2 \leq 2^k \), add 2 to both sides: \( 2k + 4 \leq 2^k + 2 \). To show \( 2(k+1) + 2 = 2k + 4 \leq 2^{k+1} = 2 \cdot 2^k \), note \( 2^{k} \geq 4 \) for \( k \geq 3 \). Therefore, \( 2^{k} + 2 \leq 2 \cdot 2^{k} \), proving \( 2k + 4 \leq 2^{k+1} \). This completes the induction.

Step 6: Conclusion

We have shown that for \( n = 3 \), the inequality holds true and have proven it by induction for all \( n \geq 3 \). Therefore, the smallest positive integer \( j \) is 3.

Key Concepts

Base Case

In mathematical induction, the base case is the starting point to show that a mathematical statement holds true for a particular initial value. In our problem, we needed to find the smallest positive integer \( j \) where the inequality \( 2n + 2 \leq 2^n \) is true for all integers \( n \geq j \).

We started by testing small values of \( n \) to determine this base case:

• For \( n = 1 \), \( 4 \leq 2 \) is false.
• For \( n = 2 \), \( 6 \leq 4 \) is also false.
• For \( n = 3 \), \( 8 \leq 8 \) is true.

Continuing to \( n = 4 \) showed \( 10 \leq 16 \), which is true. The base case, then, appears to be \( n = 3 \), as the statement holds from this point. This is crucial because it sets the foundation for subsequent steps in the proof process by proving initial applicability.

Inductive Step

The inductive step is vital in mathematical induction, as it establishes the truth of a statement for all integers beyond the base case. Here’s how this works: assume the statement is true for some arbitrary positive integer \( k \geq 3 \) (from the base case): \( 2k + 2 \leq 2^k \).

Now, demonstrate it holds for \( k + 1 \). Transform it into \( 2(k+1) + 2 \leq 2^{k+1} \):

• Using the assumption \( 2k + 2 \leq 2^k \), add 2 to both sides gives \( 2k + 4 \leq 2^k + 2 \).
• We must establish \( 2^k + 2 \leq 2 \cdot 2^k \).
• Knowing \( 2^k \geq 4 \) for \( k \geq 3 \) ensures \( 2^k + 2 \leq 2^{k+1} \).

The inductive step confirms the inequality for all \( n \geq 3 \), proving the statement universally beyond \( j \) using logical progression from the assumption.

Inequality

Inequalities deal with the order between expressions. In this exercise, the inequality \( 2n + 2 \leq 2^n \) was used. This signifies that for the appropriate values of \( n \), the expression on the left is either less than or equal to the power of two on the right. Working with such inequalities often involves:

• Determining specific values for which the inequality holds true to establish a starting point.
• A methodical approach (like induction) to prove it holds for a sequence of integers.

In our example, ensuring the inequality holds required initial tests and the power of mathematical induction to provide validity beyond the smallest true case.

Positive Integer

Positive integers are all the whole numbers greater than zero. They are vital in mathematics, especially in topics like induction, where propositions must often hold for all positive integers starting from a specific point.

In this problem, finding the smallest positive integer \( j \) was crucial, as it served as a threshold from which the inequality \( 2n + 2 \leq 2^n \) held true. Here's how positive integers were part of our solution:

• Determined \( j \), a positive integer, as 3, from where the inequality holds.
• Carried the induction proof from this integer forward to all successive integers, affirming the continuous truth of the statement.

Positive integers provided the domain in which we needed to verify the statement's validity owing to their nature of laying down an enumerable infinite set to work upon rationale like induction.