Problem 35 - Algebra and Trigonometry

Question

Determine whether each statement is true or false. If you think the statement is true, prove it. If you think it is false, give an example where it fails. (a) \(p(n)=n^{2}-n+11\) is prime for all \(n\) (b) \(n^{2}>n\) for all \(n \geq 2\) (c) \(2^{2 n+1}+1\) is divisible by 3 for all \(n \geq 1\) (d) \(n^{3} \geq(n+1)^{2}\) for all \(n \geq 2\) (e) \(n^{3}-n\) is divisible by 3 for all \(n \geq 2\) (f) \(n^{3}-6 n^{2}+11 n\) is divisible by 6 for all \(n \geq 1\)

Step-by-Step Solution

Answer

(a) False, (b) True, (c) True, (d) False, (e) True, (f) True.

Step 1: Analyze Statement (a)

The statement asks if \( p(n)=n^{2}-n+11 \) is prime for all \( n \). Let's test the formula with different natural numbers. For \( n = 0 \), \( p(0) = 0^2 - 0 + 11 = 11 \), which is prime. For \( n = 1 \), \( p(1) = 1^2 - 1 + 11 = 11 \), which is also prime.Check for \( n = 2 \): \( p(2) = 2^2 - 2 + 11 = 13 \), prime. However, try \( n = 3 \): \( p(3) = 3^2 - 3 + 11 = 17 \), again prime.Now, let's try \( n = 5 \): \( p(5) = 5^2 - 5 + 11 = 31 \), prime. Things change at \( n = 11 \): \( p(11) = 11^2 - 11 + 11 = 121 \), which is not prime because \( 121 = 11 \times 11 \). Therefore, the statement is false.

Step 2: Evaluate Statement (b)

The statement is \( n^{2} > n \) for all \( n \geq 2 \). Testing small values of \( n \):- For \( n = 2 \), \( 2^2 = 4 > 2 \).- For \( n = 3 \), \( 3^2 = 9 > 3 \).These calculations show that the inequality holds.In general, \( n^2 - n = n(n-1) \). When \( n \geq 2 \), both factors \( n \) and \( n-1 \) are positive integers, making their product larger than zero; hence the inequality is true. Thus, the statement is true.

Step 3: Analyze Statement (c)

The statement is \( 2^{2n+1} + 1 \) is divisible by 3 for all \( n \geq 1 \). We'll check some initial values of \( n \):- For \( n = 1 \), \( 2^{2 \times 1 + 1} + 1 = 2^3 + 1 = 9 \), divisible by 3.- For \( n = 2 \), \( 2^{2 \times 2 + 1} + 1 = 2^5 + 1 = 33 \), divisible by 3.Continuing this reasoning, observe that powers of 2 alternate modulo 3: \( 2^1 \equiv 2 \mod 3 \), \( 2^2 \equiv 1 \mod 3 \), \( 2^3 \equiv 2 \mod 3 \), and so forth.Thus, \( 2^{2n+1} \equiv 2 \mod 3 \) implying \( 2^{2n+1} + 1 \equiv 2 + 1 = 3 \equiv 0 \mod 3 \).Therefore, the statement is true.

Step 4: Evaluate Statement (d)

The statement is \( n^{3} \geq (n+1)^{2} \) for all \( n \geq 2 \). Let's test for smaller values:- For \( n = 2 \), \( 2^{3} = 8 \) and \((2+1)^{2} = 9 \). The inequality does not hold.To prove or disprove generally, rewrite: \( n^3 - (n+1)^2 = n^3 - (n^2 + 2n + 1) = n^3 - n^2 - 2n - 1 = n(n-1)(n-2) - 1 \)For \( n = 2 \), it equals \(-1 \). This inequality is false as proven with that example.

Step 5: Analyze Statement (e)

The statement is \( n^{3} - n \) is divisible by 3 for all \( n \geq 2 \). Consider the expression: \( n^3 - n = n(n^2 - 1) = n(n-1)(n+1) \).This is a product of three consecutive integers, one of which is necessarily divisible by 3. Thus, \( n^3 - n \) is divisible by 3, making the statement true.

Step 6: Evaluate Statement (f)

The statement is \( n^{3} - 6 n^{2} + 11 n \) is divisible by 6 for \( n \geq 1 \). Simplify: \[ n(n^2 - 6n + 11) \]Verify for consecutive small values of \( n \):- For \( n = 1, n^3 - 6n^2 + 11n = 1 - 6 + 11 = 6 \), divisible by 6.- For \( n = 2, 8 - 24 + 22 = 6\), divisible by 6.The polynomial factors into \( n(n-2)(n-1) \), ensuring divisibility by 2 and proves divisibility by 3 due to \( n-1, n, n+1 \) structure, hence divisible by 6. Therefore, the statement is true.

Key Concepts

Prime Numbers

Prime numbers are fascinating numbers greater than 1 that have no divisors other than 1 and themselves. They are fundamental in mathematics because they serve as the "building blocks" of whole numbers. When analyzing whether a formula like \( p(n) = n^2 - n + 11 \) produces prime numbers, we need to test it for different values of \( n \). This involves substituting values of \( n \) in the formula and checking if the result is a prime number. For small values like \( n = 0, 1, 2, 3 \), the formula seems to yield prime numbers. However, when you reach \( n = 11 \), the result \( p(11) = 121 \), which is not prime because it is divisible by 11. This example highlights how functions can sometimes produce primes but not consistently for all inputs.

Divisibility

Divisibility means that one number can be divided by another without leaving a remainder. A number \( a \) is divisible by a number \( b \) if \( a \mod b = 0 \). For instance, to determine the divisibility of \( 2^{2n+1} + 1 \) by 3, we start by calculating values:

• For \( n = 1 \), the result is 9, divisible by 3
• For \( n = 2 \), the result is 33, also divisible by 3

To generalize this, notice the pattern of base 2: \( 2^1 \equiv 2 \mod 3 \), \( 2^2 \equiv 1 \mod 3 \), so powers alternate. Especially interesting is that \( 2^{2n+1} \equiv 2 \mod 3 \), ensuring \( 2^{2n+1} + 1 \equiv 0 \mod 3 \). This systematic analysis confirms the statement's divisibility for all \( n \geq 1 \).

Inequalities

An inequality compares two quantities, showing if one is less than, greater than, or not equal to the other. In the statement \( n^2 > n \) for \( n \geq 2 \), testing simple values like 2 or 3 shows it's true because squaring a number makes it larger unless the number is 0 or 1.

• For \( n = 2 \), \( 2^2 = 4 \) which is greater than 2.
• For \( n = 3 \), \( 3^2 = 9 \) which is greater than 3.

This inequality indicates growth outpaces linearly increasing numbers. Also, given \( n(n-1) > 0 \) when \( n \geq 2 \), it becomes evident that the inequality holds true under these constraints.

Polynomial Functions

Polynomial functions like \( n(n^2 - 6n + 11) \) are expressions consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents. The function \( n^3 - 6n^2 + 11n \) simplifies beautifully to \( n(n-2)(n-1) \), revealing interesting properties, specifically divisibility.

• It shows that for \( n \geq 1 \),\( n(n-2)(n-1) \) captures multiples of 2 and 3 inherently.
• By expanding the expression, you can verify it covers all factors needed for divisibility by 6.

This factorization approach demonstrates why the expression \( n^3 - 6n^2 + 11n \) is divisible by 6 for natural numbers greater than or equal to 1. Such insights into polynomial structures help break down complex algebraic setups into simpler, understandable parts.