Prove 11 Divides 10^(2n)-1 For All Natural Numbers Using Induction

Introduction to Mathematical Induction and Divisibility

In the realm of mathematics, particularly in areas like elementary number theory and discrete mathematics, proof by induction stands as a cornerstone technique. It's a powerful method for establishing that a statement holds true for all natural numbers. Our focus here is to demonstrate, using induction, that $11$ divides $10^{2n} - 1$ for every natural number $n$. This involves showing that the expression $10^{2n} - 1$ is always a multiple of $11$. Divisibility, a fundamental concept in number theory, plays a crucial role in various mathematical applications and proofs. Understanding how numbers divide each other forms the basis for many advanced topics, including modular arithmetic and cryptography. The beauty of mathematical induction lies in its ability to tackle problems involving an infinite number of cases with a finite amount of work. By establishing a base case and proving an inductive step, we can confidently assert the truth of a statement for all natural numbers. This approach is particularly useful when dealing with sequences, series, and properties that depend on an integer variable. In our case, we'll see how induction elegantly confirms that $10^{2n} - 1$ is indeed divisible by $11$ for any natural number $n$.

Understanding the Problem: $11$ Divides $10^{2n}-1$

At the heart of our discussion is the assertion that $11$ divides $10^{2n} - 1$ for all natural numbers $n$. To fully grasp this statement, let's break it down. A natural number is a positive integer (1, 2, 3, ...). The expression $10^{2n}$ represents $10$ raised to the power of $2n$, where $n$ is any natural number. When we subtract $1$ from $10^{2n}$, we get a number that we claim is always divisible by $11$. Divisibility, in mathematical terms, means that one number can be divided by another number with no remainder. In this context, it means that when we divide $10^{2n} - 1$ by $11$, we should always get an integer result. To illustrate, let's consider a few examples. When $n = 1$, $10^{2n} - 1 = 10^2 - 1 = 100 - 1 = 99$, which is indeed divisible by $11$ (since $99 = 11 imes 9$). When $n = 2$, $10^{2n} - 1 = 10^4 - 1 = 10000 - 1 = 9999$, which is also divisible by $11$ (since $9999 = 11 imes 909$). These examples provide initial evidence for our claim, but they don't constitute a proof for all natural numbers. This is where mathematical induction comes into play. It provides a rigorous framework to prove statements like this that hold for an infinite number of cases. By following the principle of mathematical induction, we can confidently establish the truth of our assertion for all natural numbers $n$.

The Principle of Mathematical Induction: A Step-by-Step Guide

Mathematical induction is a powerful technique used to prove statements that hold for all natural numbers. The principle revolves around two key steps: the base case and the inductive step. The base case is the foundation of our proof. It involves showing that the statement is true for the smallest natural number, typically $n = 1$. This establishes the starting point for our induction. The inductive step is where the real magic happens. We assume that the statement is true for some arbitrary natural number $k$ (this is called the inductive hypothesis) and then use this assumption to prove that the statement is also true for the next natural number, $k + 1$. This step demonstrates that if the statement holds for a particular number, it must also hold for the next one. Think of it like a chain reaction: if the first domino falls (base case) and each domino knocks over the next one (inductive step), then all the dominoes will fall. Once we've successfully completed both the base case and the inductive step, the principle of mathematical induction allows us to conclude that the statement is true for all natural numbers. This is because the base case establishes the initial truth, and the inductive step ensures that the truth propagates from one number to the next indefinitely. In our specific problem, we'll first show that $11$ divides $10^{2n} - 1$ when $n = 1$ (base case). Then, we'll assume that $11$ divides $10^{2k} - 1$ for some natural number $k$ (inductive hypothesis) and use this assumption to prove that $11$ also divides $10^{2(k+1)} - 1$. By following these steps, we'll rigorously establish that $11$ divides $10^{2n} - 1$ for all natural numbers $n$.

Base Case: Proving the Statement for $n = 1$

The first step in our inductive proof is to establish the base case. This means we need to show that the statement we're trying to prove holds true for the smallest natural number, which is $n = 1$. In our scenario, the statement is that $11$ divides $10^{2n} - 1$. So, for the base case, we need to demonstrate that $11$ divides $10^{2(1)} - 1$. Let's substitute $n = 1$ into the expression $10^{2n} - 1$: $10^{2(1)} - 1 = 10^2 - 1 = 100 - 1 = 99$ Now, we need to check if $11$ divides $99$. This is straightforward: $99 = 11 imes 9$ Since $99$ is a multiple of $11$, we can confidently say that $11$ divides $99$. This confirms that the statement holds true for $n = 1$. Therefore, we have successfully established the base case for our inductive proof. This is a crucial step because it forms the foundation upon which the rest of the proof rests. Without a valid base case, the entire inductive argument would fall apart. With the base case secured, we can now move on to the next step: the inductive step, where we'll assume the statement is true for an arbitrary natural number $k$ and then use this assumption to prove it's also true for $k + 1$.

Inductive Hypothesis: Assuming the Statement for $n = k$

Now, let's move on to the inductive hypothesis, a crucial step in our proof by induction. Here, we assume that the statement we're trying to prove is true for some arbitrary natural number $k$. In our case, this means we assume that $11$ divides $10^{2k} - 1$ for some natural number $k$. In mathematical notation, this assumption can be expressed as: $10^2k} - 1 = 11m$ where $m$ is an integer. This equation states that $10^{2k} - 1$ is a multiple of $11$. This assumption is the cornerstone of the inductive step. We're essentially saying, "Let's suppose that the statement is true for some number $k$." The key is that we don't specify what $k$ is; it could be any natural number. The power of induction lies in the fact that if we can show that the statement is also true for the next number, $k + 1$, then we've effectively created a chain reaction that proves the statement for all natural numbers greater than or equal to the base case. The inductive hypothesis allows us to work with a concrete assumption, which we can then manipulate and use to prove the next step. Without this assumption, we wouldn't have a starting point to bridge the gap between the case $n = k$ and the case $n = k + 1$. So, with the inductive hypothesis firmly in place, we're ready to tackle the final and most challenging part of the proof the inductive step, where we'll use our assumption to demonstrate that $11$ also divides $10^{2(k+1) - 1$.

Inductive Step: Proving the Statement for $n = k + 1$

The heart of the induction proof lies in the inductive step. Here, we leverage our inductive hypothesis – the assumption that $11$ divides $10^{2k} - 1$ for some natural number $k$ – to prove that the statement also holds for the next natural number, $k + 1$. In other words, we aim to show that $11$ divides $10^{2(k+1)} - 1$. Let's start by expanding the expression $10^{2(k+1)} - 1$: $10^2(k+1)} - 1 = 10^{2k + 2} - 1$ Now, we can rewrite this expression using the properties of exponents $10^{2k + 2 - 1 = 10^2k} imes 10^2 - 1 = 10^{2k} imes 100 - 1$ This is where our inductive hypothesis comes into play. We know that $10^{2k} - 1 = 11m$ for some integer $m$. We can rearrange this equation to get $10^{2k = 11m + 1$ Now, we can substitute this expression for $10^{2k}$ back into our equation: $10^2k} imes 100 - 1 = (11m + 1) imes 100 - 1$ Expanding this, we get $(11m + 1) imes 100 - 1 = 1100m + 100 - 1 = 1100m + 99$ Now, we can factor out $11$ from the expression: $1100m + 99 = 11(100m + 9)$ Since $100m + 9$ is an integer (because $m$ is an integer), we've shown that $10^{2(k+1) - 1$ is a multiple of $11$. This means that $11$ divides $10^{2(k+1)} - 1$, which is exactly what we wanted to prove. By successfully completing the inductive step, we've demonstrated that if the statement is true for $n = k$, it's also true for $n = k + 1$. This, combined with our base case, allows us to conclude that the statement holds for all natural numbers.

Conclusion: $11$ Divides $10^{2n}-1$ for All Natural Numbers

In this comprehensive exploration, we have successfully proven, using the principle of mathematical induction, that $11$ divides $10^{2n} - 1$ for all natural numbers $n$. Our journey began with a clear understanding of the problem, where we established the assertion that the expression $10^{2n} - 1$ is always divisible by $11$ for any natural number $n$. We then delved into the core concept of mathematical induction, outlining the two essential steps: the base case and the inductive step. The base case, where we demonstrated that the statement holds true for $n = 1$, laid the foundation for our proof. We showed that $10^{2(1)} - 1 = 99$, which is indeed divisible by $11$. Next, we formulated the inductive hypothesis, assuming that the statement is true for some arbitrary natural number $k$. This assumption, expressed as $10^{2k} - 1 = 11m$ (where $m$ is an integer), became the cornerstone of our inductive step. The inductive step, the heart of our proof, involved using the inductive hypothesis to prove that the statement also holds for the next natural number, $k + 1$. Through a series of algebraic manipulations, we showed that $10^{2(k+1)} - 1$ can be expressed as $11(100m + 9)$, which is clearly a multiple of $11$. By successfully establishing both the base case and the inductive step, we have rigorously proven that $11$ divides $10^{2n} - 1$ for all natural numbers $n$. This demonstration highlights the power and elegance of mathematical induction as a technique for proving statements that hold for an infinite number of cases. The principle of mathematical induction is a fundamental tool in various branches of mathematics, including number theory, discrete mathematics, and computer science. Its ability to provide conclusive proofs for statements involving natural numbers makes it an indispensable technique for mathematicians and researchers alike.