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Chapter 1: Problem 12

Show that 8 divides \(k^{2}-1\) for \(k \in\\{1,3,5,7]\).

Short Answer

Expert verified
8 divides \(k^2 - 1\) for each \(k\) in \{1, 3, 5, 7\}.

Step by step solution

01

Understanding the Problem

To show that 8 divides \(k^2 - 1\) for each \(k\) in the set \{1, 3, 5, 7\}, we need to compute \(k^2 - 1\) for each value and verify if the result is divisible by 8. If a number is divisible by 8, then it can be written as \(8n\) for some integer \(n\).
02

Calculate \(k^2 - 1\) for each \(k\)

We'll calculate \(k^2 - 1\) for each value in the set.\- For \(k = 1\): \(1^2 - 1 = 0\).\- For \(k = 3\): \(3^2 - 1 = 9 - 1 = 8\).\- For \(k = 5\): \(5^2 - 1 = 25 - 1 = 24\).\- For \(k = 7\): \(7^2 - 1 = 49 - 1 = 48\).
03

Verify Divisibility by 8 for Each Result

Now check each result to see if it is divisible by 8.\- \(0\) is divisible by 8 since \(0 = 8 imes 0\).\- \(8\) is divisible by 8 since \(8 = 8 imes 1\).\- \(24\) is divisible by 8 since \(24 = 8 imes 3\).\- \(48\) is divisible by 8 since \(48 = 8 imes 6\).
04

Conclusion

Since each value of \(k^2 - 1\) is divisible by 8 for every \(k\) in the set \{1, 3, 5, 7\}, we have shown that 8 divides \(k^2 - 1\) for each \(k\) in the given set.

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Key Concepts

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

Number Theory
Number theory is a branch of mathematics dedicated to understanding the properties and relationships of numbers, particularly integers. One of the central activities in number theory is investigating divisibility, which helps determine how one number can be evenly divided by another. This topic is not only theoretical but has practical applications ranging from cryptography to coding theory. In our exercise, when we show that 8 divides \( k^2 - 1 \), we explore a classic number theory problem involving divisibility rules.
  • We use specific sets of numbers, known as congruence classes, to test for divisibility by calculating expressions like \( k^2 - 1 \).
  • In our example, these numbers were chosen based on their form, helping us demonstrate divisibility without testing infinite numbers.
Modular Arithmetic
Modular arithmetic might sound intimidating, but it's essentially arithmetic involving remainders. In this framework, numbers "wrap around" after reaching a certain point, known as the modulus. Think of it like the clock on the wall making a full circle every 12 hours.
When showing that \(8\) divides \(k^2 - 1\), using modular arithmetic can simplify this process by focusing on remainders after division by 8.
  • For example, \(3^2 - 1\) simplifies as 9 mod 8 leaves a remainder of 1, showing the expression equals 8, effortlessly proving divisibility by 8.
  • Modular arithmetic is useful here because it provides a direct way to understand the behavior of numbers under specific conditions.
It simplifies complex calculations and is a powerful proof tool in mathematics.
Proof Techniques
Mathematics thrives on proofs, which are logical arguments that establish the truth of a statement. In the context of divisibility problems like ours, several techniques are commonly used, including direct computation, induction, and contradiction.
  • Direct computation, as we used in this exercise, involves straightforward calculation and verification for specific examples within a given set.
  • This approach can also highlight patterns or properties that could be generalized to larger sets, paving the way for deeper insights.
  • However, more complex problems might require different proof techniques. For instance, mathematical induction might be used to prove statements true for infinitely many numbers.
Proofs build confidence in results and ensure that mathematical statements hold true under all stated conditions.

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Most popular questions from this chapter

Draw a combinatorial circuit for each of the following boolean expressions: (a) \((x \wedge y) \vee \neg z\) (b) \((x \wedge y) \vee(\neg x \wedge y)\) (c) \(\neg(\neg x \vee y) \vee(x \wedge z)\) (d) \(((x \wedge y) \vee(y \wedge z)) \vee \neg z\) (e) \((x \vee \neg(x \vee y)) \vee(\neg x \wedge \neg y)\)

Write out the information that describes what the inductive step assumes and what the step must prove for proving $$1^{5}+2^{5}+3^{5}+\cdots+n^{5}=\frac{1}{6} n^{6}+\frac{1}{2} n^{5}+\frac{5}{12} n^{4}-\frac{1}{12} n^{2}$$ with \(n_{0}\) given.

For (a) and (b), prove the stated result. For (c) and (d), find a counterexample to show that these conjcctures are false. (a) \(A \oplus B=(A \cup B)-(A \cap B)\) (b) \(A \cap(B \oplus C)=(A \cap B) \oplus(A \cap C)\) (c) \((A \cap B) \oplus(C \cap D) \subseteq(A \oplus C) \cap(B \oplus D)\) (d) \((A \cup B) \oplus(C \cup D) \subseteq(A \cup C) \oplus(B \cup D)\)

Using the Principle of Mathematical Induction, prove each of the following different forms of the principle; (a) Induction with a possibly negative starting point: Suppose that \(S \subseteq \mathbb{Z}\), that some integer \(n_{0} \in S,\) and that for every \(n \in \mathbb{Z},\) if \(n \in S\) and \(n \geq n_{0},\) then \(n+1 \in S .\) Then, for every integer \(n \geq n_{0}\), we have \(n \in S\). (b) Induction downward: Suppose that \(S \subseteq \mathbb{Z}\), that some integer \(n_{0} \in S\), and that for every \(n \in \mathbb{Z},\) if \(n \in S\) and \(n \leq n_{0},\) then \(n-1 \in S .\) Then, for every integer \(n \leq n_{0}\) we have \(n \in S\) (c) Finite induction upward: Let \(n_{0}, n_{1} \in \mathbb{Z}, n_{0} \leq n_{1} .\) Suppose that \(S \subseteq \mathbb{Z}, n_{0} \in S\). and for every \(n \in \mathbb{Z},\) if \(n \in S, n \geq n_{0},\) and \(n

Find the expression tree for the following formulas: (a) \(\neg p \wedge(\neg q \vee r)\) (b) \(p \vee(\neg q \wedge \neg r)\) (c) \(((p \vee q) \leftrightarrow r) \leftrightarrow p\) (d) \((\neg q \wedge \neg r) \leftrightarrow(p \rightarrow(q \vee r))\)
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