Problem 37 - Algebra and Trigonometry

Question

True or False? 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 in which it fails. (a) \(p(n)=n^{2}-n+11\) is prime for all \(n .\) (b) \(n^{2}>n\) for all \(n \geq 2\) . (c) \(n^{2 n+1}+1\) 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) False. d) False. e) True. f) False.

Step 1: Statement Analysis for (a)

The statement is about whether \( p(n)=n^{2}-n+11 \) is prime for all \( n \). To check this, substitute different values of \( n \) and verify the result.For \( n = 0: \)\[ p(0) = 0^2 - 0 + 11 = 11 \] (prime)For \( n = 1: \)\[ p(1) = 1^2 - 1 + 11 = 11 \] (prime)For \( n = 2: \)\[ p(2) = 2^2 - 2 + 11 = 13 \] (prime)Continue until finding a contradiction.For \( n = 11: \)\[ p(11) = 11^2 - 11 + 11 = 121 \] (not prime, because \( 121 = 11 \times 11 \))This counterexample shows that \( p(n) \) is not prime for all \( n \).

Step 2: Statement Analysis for (b)

The statement claims \( n^{2} > n \) for all \( n \geq 2 \). To verify, substitute values starting from \( n = 2 \).For \( n = 2: \)\[ 2^2 = 4 \] and \[ 2 = 2 \], so \( 4 > 2 \),For \( n = 3: \)\[ 3^2 = 9 \] and \[ 3 = 3 \], so \( 9 > 3 \).For any \( n \geq 2 \), \( n^{2} - n = n(n-1) \), which is positive for \( n \geq 2 \) because \( n-1 \geq 1 \). Thus, the statement is true.

Step 3: Statement Analysis for (c)

Check if \( n^{2n+1} + 1 \) is divisible by 3 for \( n \geq 1 \). Test this with small \( n \) values.For \( n = 1: \)\[ 1^{2 \times 1 + 1} + 1 = 2 \] (not divisible by 3)The statement is false, as illustrated by this counterexample.

Step 4: Statement Analysis for (d)

Verify if \( n^{3} \geq(n+1)^{2} \) for \( n \geq 2 \). Simplify and evaluate inequalities.Starting at \( n = 2: \)\[ n^{3} - (n^2 + 2n + 1) = n^{3} - n^2 - 2n - 1 \] should be checked.For \( n = 2: \)\[ 2^3 - (2+1)^2 = 8 - 9 = -1 \], which means \( 8 ot\ge 9 \).For \( n = 3: \)\[ 27 - 16 = 11 \], which is true, but not universally true as shown by \( n = 2 \).Counterexample provides a failure at \( n = 2 \), hence the statement is false.

Step 5: Statement Analysis for (e)

Examine if \( n^3 - n \) is divisible by 3 for \( n \geq 2 \). Simplify as \( n(n^2 - 1) = n(n-1)(n+1) \).Since \( n, n-1, n+1 \) are consecutive integers, one of them must be divisible by 3. Therefore, \( n^3 - n \) is always divisible by 3, making the statement true.

Step 6: Statement Analysis for (f)

Determine whether \( n^{3}-6n^{2}+11n \) is divisible by 6 for all \( n \geq 1 \). Factor to analyze divisibility.\[ f(n) = n(n^2 - 6n + 11) \].Check small values:For \( n = 1: \)\[ 1 imes (1 - 6 + 11) = 6 \] \( (divisible)\)Check other divisible by 2 and 3 scenarios for various \( n \).The function may not always be divisible by 2 and 3 simultaneously showing that the statement could potentially not hold. Inspect for failure, but the assumption covers conditions when divisible by either easily. For each integer \( n \), analyze till failure or satisfaction.

Key Concepts

Prime Numbers

Prime numbers are those natural numbers greater than 1 that have no divisors other than 1 and themselves. This means that a prime number cannot be formed by multiplying two smaller natural numbers.
Understanding prime numbers is vital because they are the building blocks of all whole numbers, much like atoms are for molecules.
To check if a number is prime, you can:

• Examine if it is divisible by any prime number smaller than its square root. If it is, then it's not prime.
• Note that the lowest primes are 2, 3, 5, and 7, and they follow no specific pattern, making prediction without calculation difficult.

Understanding their significance, especially in number theory, helps in solving complex mathematical problems that involve divisibility and factorization.

Divisibility Rules

Divisibility rules are shortcuts you can use to determine whether one number divides evenly into another without having to perform the actual division. Each number has its own rule, making them handy tools for simplifying division tasks in your head.
For example:

• A number is divisible by 3 if the sum of its digits is divisible by 3. This rule helps quickly identify numbers divisible by 3.
• For divisibility by 6, a number must be divisible by both 2 and 3. This combines the rules for both to assure divisibility by 6.

These rules play an instrumental role in number theory, simplifying complex problems involving factorization and greatest common divisors. Mastering them aids in various mathematical contexts, from algebra to competitive math problems.

Polynomial Functions

Polynomial functions are mathematical expressions consisting of variables and coefficients, constructed using only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. The general form of a polynomial in one variable is given by\[ p(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \]where the exponents are non-negative integers, and the coefficients \(a_n, a_{n-1}, \ldots, a_0\) are constants.
Polynomials are crucial because they offer a straightforward way to model relationships in algebra and calculus, from finding the roots of equations to exploring curves and continuity.
Adding or subtracting polynomials involves combining like terms, while multiplying involves distributing each term from one polynomial to the other, often using a method like FOIL for binomials. Understanding how to manipulate polynomials is key in algebraic problem-solving.

Inequalities

Inequalities are mathematical statements used to show the relationship between two values, where one is less than, greater than, less than or equal to, or greater than or equal to another. They are denoted by symbols such as \(<\), \(>\), \(\leq\), and \(\geq\).
Understanding inequalities allows us to solve and graph solutions for ranges rather than specific values, which is especially useful in real-world situations like budgeting or planning.
When solving inequalities:

• Perform the same operation on both sides to maintain the balance, similar to solving equations.
• Be aware that multiplying or dividing both sides by a negative number reverses the inequality sign. This is a critical rule that distinguishes inequality manipulation from equations.

Mastering inequalities enables an accurate assessment of scenarios involving constraints, limits, and conditions.