Problem 27 - Precalculus Mathematics for Calculus

Question

Show that \(x-y\) is a factor of \(x^{n}-y^{n}\) for all natural numbers \(n\) \(\left[\text {Hint: } x^{k+1}-y^{k+1}=x^{k}(x-y)+\left(x^{k}-y^{k}\right) y .\right]\)

Step-by-Step Solution

Answer

By induction, \(x-y\) is a factor of \(x^n-y^n\) for all natural \(n\).

Step 1: Understanding the Problem

We are asked to show that \(x-y\) is a factor of \(x^n-y^n\) for any natural number \(n\). The problem provides a hint formula \(x^{k+1}-y^{k+1}=x^k(x-y)+(x^k-y^k)y\) which will be useful in our proof.

Step 2: Base Case: When n=1

For \(n=1\), this becomes \(x^1-y^1 = x-y\). Clearly, \(x-y\) is a factor of itself. This establishes the base case for our proof by induction.

Step 3: Inductive Hypothesis

Assume that for some natural number \(k\), \(x-y\) is a factor of \(x^k-y^k\). This means there exists some polynomial \(Q(x,y)\) such that \(x^k-y^k = (x-y)Q(x,y)\).

Step 4: Applying the Inductive Step

Using the given hint, express \(x^{k+1}-y^{k+1}\) as \(x^k(x-y)+(x^k-y^k)y\). This can be rewritten, using our hypothesis, to \((x-y)(x^k + yQ(x,y))\). This shows \(x-y\) is a factor of \(x^{k+1}-y^{k+1}\).

Step 5: Conclusion by Induction

Since \(x-y\) being a factor holds for \(n=1\) and assuming it for \(n=k\) leads to it being true for \(n=k+1\), by mathematical induction, \(x-y\) is a factor of \(x^n-y^n\) for all natural numbers \(n\).

Key Concepts

Mathematical Induction

Mathematical induction is a powerful mathematical technique used to prove statements about natural numbers. It is much like how falling dominoes create a sequential cascade; if the first domino falls, and each domino can knock over its successor, then all the dominoes will end up falling. This same principle applies to mathematical induction.

The process consists of two main parts:

• Base Case: You start by proving that the statement holds true for the first natural number, often zero or one. For example, for our problem, we show that for \(n=1\), \(x-y\) is undoubtedly a factor of \(x^1 - y^1 = x-y\).
• Inductive Step: You then assume the statement is true for some arbitrary natural number \(k\). This is known as the inductive hypothesis. You must demonstrate that if the statement holds for \(n=k\), then it must also hold for \(n=k+1\). In our case, we assume that \(x-y\) is a factor of \(x^k - y^k\), and show how it must also be a factor of \(x^{k+1} - y^{k+1}\).

Together, these steps confirm the truth of a statement for all natural numbers.

Natural Numbers

Natural numbers are the set of positive integers beginning from 1, which we denote as \(\mathbb{N}\). These numbers are foundational in mathematics and are used to count and order objects. They do not include zero, negative numbers, fractions, or decimals.

Natural numbers have several key properties that make them essential for various branches of mathematics:

• Well-Ordering Principle: This means every non-empty set of natural numbers has a smallest element. This principle plays a crucial role in proofs by induction.
• Closure: Natural numbers are closed under addition and multiplication, but not under subtraction or division. This means that adding or multiplying two natural numbers yields another natural number.

The concept of natural numbers is myopic yet powerful, providing the backbone to more complex numbers seen in mathematics.

Polynomials

Polynomials are algebraic expressions that include variables raised to whole number exponents. They are comprised of terms that can be constants, variables, or products of constants and variables. A typical polynomial might look something like \(a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0\). Here, \(a_n, a_{n-1}, \ldots, a_0\) are constants known as coefficients, and \(n\) is a non-negative integer.

Key aspects of polynomials are:

• Degree: This is the highest power of the variable in the polynomial. The degree signifies the polynomial's most significant term.
• Roots or Zeros: The values that make the polynomial zero. For example, if \(x-y\) is a factor of a polynomial, then \(y\) is a root.
• Factorization: Factorizing a polynomial means expressing it as a product of its factors. This is crucial for simplifying expressions and solving equations.

Understanding polynomials is critical for solving algebraic equations, modeling real-world scenarios, and analyzing functions.