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Chapter 2: Problem 22

Use induction to prove the statement. \(7^{n}-1\) is divisible by \(6,\) for all \(n \geq 1\)

Short Answer

Expert verified
The statement \(7^n - 1\) is divisible by 6, for all \(n \geq 1\) is true, as proven by mathematical induction. This was demonstrated by verifying the base case \(n = 1\) and showing that, if the statement is valid for \(n = k\), it also holds true for \(n = k + 1\).

Step by step solution

01

Base Case

Firstly, we need to test the base case. The smallest eligible value for \(n\) as specified in the given problem is \(1\). Substituting \(n = 1\) into the expression, we get \(7^1 - 1 = 6\), which is indeed divisible by \(6\). Thus, the statement holds true for \(n = 1\).
02

Inductive Hypothesis

Next step is to make our inductive hypothesis. That is, we assume the statement is true for some \(n = k\), where \(k \geq 1\). In other words, we assume that \(7^k - 1\) is divisible by \(6\).
03

Inductive Step

Then, we need to prove that the statement is also true for \(n = k + 1\), assuming it is true for \(n = k\). So, we apply the inductive hypothesis we've made to show \(7^{k+1} - 1\) is also divisible by 6. We begin by adding and subtracting 7 inside the equation to get, \(7^{k+1} - 1 = 7 * 7^k - 7 + 6 = 6 * 7^k + 7^k - 7\). By our inductive hypothesis, \(7^{k} = 6*j + 1\) for some integer \(j\). Substituting into the equation, it simplifies to \(6 * 7^k + 6*j + 6\). We can rewrite it as \(6 * (7^k + j + 1)\), which is divisible by \(6\), where \(j + 1\) is an integer. As a result, it's proven that if the statement is true for \(n = k\), it's also true for \(n = k + 1\).

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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 describes when one integer is evenly divisible by another. In our case, the statement to prove is that \(7^n - 1\) is divisible by 6 for all \(n \geq 1\). Let's break this down:
  • When a number \(x\) is divisible by another number \(y\), it means that there is an integer \(z\) such that \(x = y \times z\).
  • In our problem, we are focused on the expression \(7^n - 1\). We need to find if this expression can be expressed as \(6 \times z\), meaning that it is divisible by 6 and leaves no remainder.
Understanding divisibility is crucial when working with proofs, especially when using mathematical induction to generalize a statement for all possible integers \(n\). By showing \(7^n - 1\) fits this mold, we effectively demonstrate divisibility by 6.
Inductive Hypothesis
The inductive hypothesis is a critical step in the process of mathematical induction. It's where we assume that the statement is true for a particular instance, \(n = k\), to help us in proving it for the next stage. Here's a closer look at how it works:
  • The inductive hypothesis involves assuming our statement, \(7^k - 1\) is divisible by 6 for some integer \(k \geq 1\).
  • This assumption is not a random guess; it's based on faith in the pattern suggested by the base case and any logical deductions made from the exercise. This assumption is key as it acts as a bridge to prove the statement for \(n = k + 1\).
Using this assumption responsibly ensures the transition from one stage (\(n = k\)) to the next (\(n = k+1\)) is smooth, allowing us to cover all values of \(n\) as desired.
Base Case
In any proof by induction, the base case is your starting point. It's essential because it shows that the statement holds true for the initial value(s) before proceeding to apply induction:
  • In our given statement, the base case begins with checking \(n = 1\). We substitute \(n = 1\) into the expression \(7^n - 1\), resulting in \(7^1 - 1 = 6\).
  • It's clear that 6 is divisible by 6, as 6 \(\div\) 6 = 1, an integer with no remainder.
  • Proving the base case confirms that the statement starts correctly and provides a solid foundation for the inductive step.
Without verifying the base case, advancing the proof to any higher numbers would be resting on incomplete evidence, emphasizing the importance of this step.
Inductive Step
The inductive step is where the magic happens in mathematical induction. It shows the logical progression from one instance to the next, ensuring every integer after the base case falls into place:
  • Here's how it works: assuming the statement is true for \(n = k\) (inductive hypothesis), you must show it is also true for \(n = k + 1\).
  • In our situation, we manipulate the expression \(7^{k+1} - 1\) using the hypothesis by extending \(7^{k+1} - 1\) to \(7 \times 7^k - 7 + 6\), which simplifies to \(6 \times (7^k + j + 1)\), where \(j\) is an integer from the hypothesis \(7^k = 6j + 1\).
  • This transformation and simplification prove \(7^{k+1} - 1\) is divisible by 6, just like \(7^k - 1\), thus proving the pattern continues for every subsequent integer.
Performing this step correctly ensures the power and universality of induction across all natural numbers starting from our base case.

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Most popular questions from this chapter

Prove or disprove: There exist rational numbers \(a\) and \(b\) such that \(a^{b}\) is irrational. What kind of proof did you give?

Suppose that a real number \(a\) has the property that \(a^{n}\) is irrational for every positive integer \(n\). Prove that \(a^{r}\) is irrational for every positive rational number \(r\). (There are real numbers such as \(a\). The real number \(e=2.718 \ldots\) the base of the nutural logarithm, has the property that \(e^{r}\) is irrational for every rational number \(r\), An elementary proof, using only calculus, may be found in [Aigner].).

The ordered pair \((a, b)\) can be defined in tems of sets as $$ (a, b)=\\{\mid a\\},\\{a, b|| .$$ Taking the preceding equation as the definition of ordered pair, prove that \((a, b)=(c, d)\) if and only if \(a=c\) and \(b=d\).

Prove that for all \(x, y \in \mathbf{R},\) if \(x\) is rational and \(y\) is irrational, then \(x+y\) is irrational.

Concern a variant of the Josephus Problem in which every second person is eliminated. We assume that \(n\) people are arranged in a circle and numbered \(1,2, \ldots, n\) clockwise. Then, proceeding clockwise, 2 is eliminated, 4 is eliminated, and so on, until there is one survivor, denoted \(J(n) .\) Use induction to show that \(J\left(2^{i}\right)=1\) for all \(i \geq 1\).
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