Question

Using mathematical induction, prove that \(\mathrm{x}^{2 \mathrm{n}}-\mathrm{y}^{2 \mathrm{n}}\) is divisible by \(\mathrm{x}+\mathrm{y}\)

Short answer

Using mathematical induction, we prove that \(x^{2n} - y^{2n}\) is divisible by \(x+y\) for all positive integers \(n\). We first establish the base case for \(n=1\) as \(x^2 - y^2 = (x+y)(x-y)\), showing that it is divisible by \(x+y\). Next, we assume the statement true for some \(n=k\), then prove it for \(n=k+1\) using the inductive hypothesis. Finally, we demonstrate that both terms are divisible by \(x+y\) and thus prove the statement.

Step-by-step solution

Base Case

We start by verifying the base case when \(n = 1\). This means that we want to check if \(x^{2\cdot 1} - y^{2\cdot 1}\) is divisible by \(x+y\). We have: \[x^2 - y^2 = (x+y)(x-y).\] Here, we can clearly see that \(x^2-y^2\) is divisible by \(x+y\). Thus, the statement holds for \(n=1\).

Inductive Hypothesis

Now, let's assume that the statement is true for some positive integer \(n=k\). This means that \(x^{2k} - y^{2k}\) is divisible by \(x + y\).

Inductive Step

We want to show that the statement holds for \(n = k + 1\). That is, we want to prove that \(x^{2(k+1)} - y^{2(k+1)}\) is divisible by \(x + y\). We rewrite the expression and use the inductive hypothesis as follows: \begin{align*} x^{2(k+1)} - y^{2(k+1)} &= x^{2k+2} - y^{2k+2} \\ &= x^2x^{2k} - y^2y^{2k} \\ &= x^2(x^{2k} - y^{2k}) + (x^2y^{2k} - y^2y^{2k}) \\ &= x^2(x^{2k} - y^{2k}) + y^{2k}(x^2 - y^2). \end{align*} By the inductive hypothesis, \(x^{2k} - y^{2k}\) is divisible by \(x + y\), so \(x^2(x^{2k} - y^{2k})\) is also divisible by \(x + y\). From the base case, we know that \(x^2 - y^2\) is divisible by \(x + y\), so \(y^{2k}(x^2 - y^2)\) is also divisible by \(x + y\). Since both terms are divisible by \(x+y\), their sum, \(x^{2(k+1)} - y^{2(k+1)}\), is also divisible by \(x+y\).

Conclusion

By proving the base case and the inductive step, we have shown that the statement holds for all positive integers \(n\), which means that \(x^{2n} - y^{2n}\) is divisible by \(x+y\) using mathematical induction.

Key concepts

Divisibility

Understanding divisibility in mathematics is crucial. It tells us whether one number can be "divided" by another without leaving a remainder.
For example, if we say a number A is divisible by B, it means when you divide A by B, the remainder is zero.
In the exercise, we are investigating if the polynomial expression \(x^{2n} - y^{2n}\) can be divided by \(x + y\) without leaving a remainder.

This is essential in mathematical induction problems where proving divisibility often confirms a general rule. By breaking down the expression and checking for conditions of divisibility, we can ascertain the truth of statements over integers.

Polynomial Expressions

Polynomials are algebraic expressions that consist of variables and coefficients. They are written in the form of \(ax^n + bx^{n-1} + ... + z\).
In this problem, we are dealing with a specific polynomial expression \(x^{2n} - y^{2n}\).
This format represents two terms raised to a power, which can be expanded and factored using algebraic identities.

For instance, \(x^2 - y^2\) is a polynomial that factors neatly into \((x+y)(x-y)\). Understanding how polynomials work and how they can be factored is key in problems involving divisibility, as these skills facilitate identifying factors like \(x + y\) within the expression.

Inductive Hypothesis

The inductive hypothesis is a core part of proving statements by induction. It involves assuming that the statement is true for a particular case, often \(n = k\).
For this problem, we assume that the expression \(x^{2k} - y^{2k}\) is divisible by \(x + y\) for some integer \(k\). This is our hypothesis.

The strength of this assumption allows us to build upon it, showing that the next case, \(n = k + 1\), is also true given the hypothesis holds true. It provides a foothold or stepping stone that we extend using algebraic manipulation to demonstrate its truth for the next integer.

Base Case

The base case in mathematical induction serves as the starting point of the proof. It's like the first domino in a line; once you prove it, the rest follow.
For our exercise, the base case involves proving the statement is true when \(n = 1\). We check if the simpler polynomial \(x^2 - y^2\) is divisible by \(x + y\).

By factoring, it becomes clear that \(x^2 - y^2 = (x+y)(x-y)\), confirming divisibility for \(n=1\). Verifying this base case is a crucial first step; it grounds the subsequent inductive steps ensuring the statement holds for all positive integers.