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Is the following proposition true or false? Justify your conclusion with a counterexample or a proof. For each integer \(n,\) if \(n\) is odd, then \(8 \mid\left(n^{2}-1\right)\).

Short Answer

Expert verified
The proposition is false. A counterexample is when n=5, as \(n^2 - 1 = 24\) which is not divisible by 8.

Step by step solution

01

Express n as an odd integer

Since n is an odd integer, we can express it as n = 2k + 1, where k is an integer. This is because any odd integer can be written in this form.
02

Compute n^2 - 1

Find the expression n^2 - 1 using the given representation of odd integers. n^2 - 1 = (2k + 1)^2 - 1
03

Simplify the expression

Expand (2k + 1)^2 and simplify the expression: (2k + 1)^2 - 1 = (4k^2 + 4k + 1) - 1 = 4k^2 + 4k
04

Factor the expression

Factor the expression to check if it is divisible by 8: 4k^2 + 4k = 4(k^2 + k)
05

Check divisibility by 8

We can observe that the expression is a multiple of 4, but not a multiple of 8. Therefore, we need to find a counterexample to show that the proposition is false.
06

Find a counterexample

Let's take an odd integer n=3, which can be expressed as 2(1) + 1. Now, compute the expression n^2 - 1: n^2 - 1 = (3)^2 - 1 = 9 - 1 = 8 In this case, 8 does divide (n^2 - 1), but this is not generally true for all odd integers. Let's take another odd integer, n=5: n^2 - 1 = (5)^2 - 1 = 25 - 1 = 24 Now, 8 does not divide (n^2 - 1) in this case. This shows that the given proposition is false. In conclusion, the proposition is false as there are odd integers for which 8 does not divide the expression (n^2 - 1). We have shown the counterexample with n=5.

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Key Concepts

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

Counterexample
A counterexample is a powerful tool used in mathematics to disprove a statement or proposition. It involves providing a specific example where the proposition does not hold true. This single instance shows that the proposed general statement is false. Consider the proposition, "For all odd integers \( n \), 8 divides \( n^2 - 1 \)." To disprove this, we need to find an odd integer \( n \) such that \( n^2 - 1 \) is not divisible by 8. As seen in the exercise above, the odd integer \( n = 5 \) serves as this counterexample, since for \( n = 5 \), \( n^2 - 1 = 24 \), and 8 does not divide 24 without a remainder. Thus, the proposition is false.
Odd Integers
Odd integers are whole numbers that cannot be evenly divided by 2. They can generally be represented by the formula \( n = 2k + 1 \), where \( k \) is any integer. This representation helps in mathematical proofs and problem-solving when dealing with properties specific to odd numbers. For example, when \( n \) is an odd integer, substituting into the formula \( n^2 - 1 \) helps us test various propositions, like the one in our exercise. By setting \( n = 2k + 1 \), you can perform algebraic manipulations to uncover properties related to odd integers and examine how they behave in different mathematical contexts.
Divisibility
Divisibility is a fundamental concept in mathematics that explains whether one integer can be divided by another without leaving a remainder. When a number \( a \) is divisible by another number \( b \), it means there exists an integer \( k \) such that \( a = b \cdot k \). In our context, we investigated whether the expression \( n^2 - 1 \) is divisible by 8 for odd integers. For \( n^2 - 1 \) to be divisible by 8, it should be able to be expressed as \( 8 \cdot m \), where \( m \) is an integer. However, our counterexample with \( n = 5 \) showed that \( n^2 - 1 = 24 \) is not divisible by 8, demonstrating the falsity of the proposition.
Proposition Testing
Proposition testing involves verifying if a given statement or proposition is universally true or false. In exercises involving mathematical logic, proposition testing requires checking the truth of the statement over all possible scenarios or cases. In our exercise, we started with the proposition: "For all odd integers \( n \), 8 divides \( n^2 - 1 \)." To test this, we explored it algebraically and numerically. Algebraically, we simplified \( n^2 - 1 \) for an odd integer \( n = 2k + 1 \), finding that it reduces to \( 4(k^2 + k) \), which wasn't always divisible by 8. Numerically, we used specific values for \( n \) to perform proposition testing via counterexamples, ultimately identifying an odd integer that disproves the proposition.

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Most popular questions from this chapter

Consider the following proposition: Proposition. For all integers \(m\) and \(n,\) if \(n\) is odd, then the equation $$ x^{2}+2 m x+2 n=0 $$ has no integer solution for \(x\). (a) What are the solutions of the equation when \(m=1\) and \(n=-1 ?\) That is, what are the solutions of the equation \(x^{2}+2 x-2=0 ?\) (b) What are the solutions of the equation when \(m=2\) and \(n=3\) ? That is, what are the solutions of the equation \(x^{2}+4 x+6=0 ?\) (c) Solve the resulting quadratic equation for at least two more examples using values of \(m\) and \(n\) that satisfy the hypothesis of the proposition. (d) For this proposition, why does it seem reasonable to try a proof by contradiction? (e) For this proposition, state clearly the assumptions that need to be made at the beginning of a proof by contradiction. (f) Use a proof by contradiction to prove this proposition.

Prove that for each integer \(a\), if \(a^{2}-1\) is even, then 4 divides \(a^{2}-1\).

(Exercise (18), Section 3.2) Prove the following proposition: Let \(a\) and \(b\) be integers with \(a \neq 0\). If \(a\) does not divide \(b\), then the equation \(a x^{3}+b x+(b+a)=0\) does not have a solution that is a natural number.

(a) Write the contrapositive of the following statement: For all positive real numbers \(a\) and \(b\), if \(\sqrt{a b} \neq \frac{a+b}{2},\) then \(a \neq b\). (b) Is this statement true or false? Prove the statement if it is true or provide a counterexample if it is false.

See the instructions for Exercise (19) on page 100 from Section 3.1 . (a) Proposition. For all integers \(a\) and \(b\), if \((a+2 b) \equiv 0(\) mod 3 ), then \((2 a+b) \equiv 0(\bmod 3)\) Proof. We assume \(a, b \in \mathbb{Z}\) and \((a+2 b) \equiv 0(\) mod 3\() .\) This means that 3 divides \(a+2 b\) and, hence, there exists an integer \(m\) such that \(a+2 b=3 m .\) Hence, \(a=3 m-2 b .\) For \((2 a+b)=0(\bmod 3),\) there exists an integer \(x\) such that \(2 a+b=3 x .\) Hence, $$ \begin{aligned} 2(3 m-2 b)+b &=3 x \\ 6 m-3 b &=3 x \\ 3(2 m-b) &=3 x \\ 2 m-b &=x . \end{aligned} $$ Since \((2 m-b)\) is an integer, this proves that 3 divides \((2 a+b)\) and hence, \((2 a+b) \equiv 0(\bmod 3)\). (b) Proposition. For each integer \(m, 5\) divides \(\left(m^{5}-m\right)\). Proof. Let \(m \in \mathbb{Z}\). We will prove that 5 divides \(\left(m^{5}-m\right)\) by proving that \(\left(m^{5}-m\right) \equiv 0(\bmod 5)\). We will use cases. For the first case, if \(m \equiv 0(\bmod 5),\) then \(m^{5} \equiv 0(\bmod 5)\) and, hence, \(\left(m^{5}-m\right) \equiv 0(\bmod 5)\) For the second case, if \(m \equiv 1(\bmod 5),\) then \(m^{5} \equiv 1(\bmod 5)\) and, hence, \(\left(m^{5}-m\right) \equiv(1-1)(\) mod 5\()\), which means that \(\left(m^{5}-m\right) \equiv\) \(0(\bmod 5)\) For the third case, if \(m \equiv 2(\bmod 5),\) then \(m^{5} \equiv 32(\bmod 5)\) and, hence, \(\left(m^{5}-m\right)=(32-2)(\bmod 5),\) which means that \(\left(m^{5}-m\right)=\) \(0(\bmod 5)\).

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