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Chapter 1: Problem 3

Prove that for all \(n \in \mathbb{N}, 7^{n}-1\) is divisible by 3.

Short Answer

Expert verified
By mathematical induction, \( 7^{n} - 1 \) is divisible by 3 for all \( n \in \mathbb{N} \).

Step by step solution

01

Base case for induction

We start by testing the base case, where \( n = 1 \). We calculate \( 7^{1} - 1 = 6 \), which is divisible by 3. Thus, the base case holds true.
02

Induction hypothesis

Assume that for some positive integer \( k \), the expression \( 7^{k} - 1 \) is divisible by 3. This means \( 7^{k} - 1 = 3m \) for some integer \( m \).
03

Induction step

We need to prove that \( 7^{k+1} - 1 \) is divisible by 3 using our hypothesis. We start with the expression \( 7^{k+1} - 1 \) and rewrite it as \( 7 imes 7^{k} - 1 \).
04

Simplifying the induction step

Using the induction hypothesis, substitute \( 7^{k} = 3m + 1 \) into \( 7^{k+1} - 1 = 7 \times 7^{k} - 1 \). This gives us \( 7 \times (3m + 1) - 1 = 21m + 7 - 1 = 21m + 6 \).
05

Concluding the proof with divisibility

Since \( 21m + 6 = 3(7m + 2) \), it is clearly divisible by 3. Hence, \( 7^{k+1} - 1 \) is divisible by 3. The proof by induction is complete.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Divisibility
Divisibility is a fundamental concept in mathematics that helps us determine if one number can be divided by another without leaving a remainder. In this exercise, the goal is to show that \( 7^n - 1 \) is divisible by 3 for all natural numbers \( n \).
When we say a number like \( 6 \) is divisible by \( 3 \), it means that when \( 6 \) is divided by \( 3 \), the remainder is \( 0 \). This concept is crucial for validating whether mathematical statements hold true across various integers.
In our solution, the expression \( 7^n - 1 \) being divisible by 3 tells us that, theoretically, if you were to divide \( 7^n - 1 \) by 3, you should expect a whole number result. This forms the crux of verifying mathematical claims using divisibility rules.
Base Case
The base case is our starting point in mathematical induction. It helps establish the first step in the chain of reasoning that confirms the statement we want to prove.
In this problem, the base case considers \( n = 1 \). By calculating \( 7^1 - 1 \), we find it equals \( 6 \). Since \( 6 \) is easily divisible by \( 3 \) (as \( 6 \div 3 = 2 \)), the base case holds true.
Establishing a solid base case is essential because a weak or incorrect base can invalidate the entire induction process. Without the base case, the domino effect required in induction fails to start.
Inductive Hypothesis
The inductive hypothesis is the assumption we make for a particular step in order to prove the next one. It's like assuming one domino will fall to ensure others do too.
This hypothesis involves presuming the statement holds for some integer \( k \), that is, \( 7^k - 1 \) is divisible by 3. Formally, this means we assume \( 7^k - 1 = 3m \) for some integer \( m \), implying it's divisible by 3.
This assumption provides a foundation to show that if the hypothesis is true for \( k \), then it should also be true for \( k+1 \), which is the next logical step in the sequence.
Inductive Step
The inductive step is where we prove the proposition holds for \( k+1 \), given the assumption in the inductive hypothesis. This is akin to proving the next domino falls.
To do so, we begin with the expression \( 7^{k+1} - 1 \). Rewriting it as \( 7 \times 7^k - 1 \) allows us to substitute \( 7^k \) from our hypothesis: \( 7^k = 3m + 1 \).
Substituting this into the expression, we get \( 7 \times (3m + 1) - 1 \), which simplifies to \( 21m + 6 \), clearly divisible by 3, as \( 21m + 6 = 3(7m + 2) \).
Thus, by confirming the proposition for \( k+1 \) with direct calculation and simplification, the inductive step completes, confirming the statement for all \( n \) due to the domino effect triggered by a true base case.

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Most popular questions from this chapter

Given two real numbers \(x\) and \(y\), prove that \(\max \\{x, y\\}=\frac{x+y+|x-y|}{2}\) and \(\min \\{x, y\\}=\frac{x+y-|x-y|}{2}\).

Let \(f: X \rightarrow Y\) be a function. Prove that: (a) If \(f\) is one-to-one, then \(A=f^{-1}(f(A))\) for every subset \(A\) of \(X\). (b) If \(f\) is onto, then \(f\left(f^{-1}(B)\right)=B\) for every subset \(B\) of \(Y\).

Let \(a \geq-1\). Prove by induction that $$ (1+a)^{n} \geq 1+n a \text { for all } n \in \mathbb{N}. $$

Suppose \(A\) and \(B\) are nonempty subsets of \(\mathbb{R}\) that are bounded above. Define $$A+B=\\{a+b: a \in A \text { and } b \in B\\}$$ Prove that \(A+B\) is bounded above and $$\sup (A+B)=\sup A+\sup B.$$

Let \(A\) be a nonempty subset of \(\mathbb{R}\) and \(\alpha \in \mathbb{R}\). Define \(\alpha A=\\{\alpha a: a \in A\\}\). Prove the following statements: (a) If \(\alpha>0\) and \(A\) is bounded above, then \(\alpha A\) is bounded above and \(\sup \alpha A=\alpha \sup A\). (b) If \(\alpha<0\) and \(A\) is bounded above, then \(\alpha A\) is bounded below and inf \(\alpha A=\alpha \sup A\).
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