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Chapter 3: Problem 12

Give a reason for your answer in each of 1-13. Assume that all variables represent integers. If \(n=4 k+1\), does 8 divide \(n^{2}-1\) ?

Short Answer

Expert verified
Yes, \(8\) divides \(n^2-1\) when \(n\) is of the form \(4k+1\), as we can write the expression as \(8(2k^2 + k)\), showing that it is divisible by \(8\).

Step by step solution

01

Squaring n

Begin by squaring the given expression for \(n\), which is \(4k+1\): \[ (4k+1)^2 \]
02

Expanding the square

Expand the square: \[ (4k+1)^2 = (4k)^2 + 2(4k)(1) + 1^2 \]
03

Simplify the expression

Simplify the squared expression: \[ (4k+1)^2 = 16k^2 + 8k + 1 \] Now, let's analyze \(n^2 - 1\), where \(n^2\) is given by the expression from step 3.
04

Subtract 1 from the expression

Subtract 1 from the simplified squared expression: \[ 16k^2 + 8k + 1 - 1 = 16k^2 + 8k \]
05

Check divisibility by 8

Now, we need to check if 8 divides the expression \(16k^2 + 8k\). We can first factor 8 from the expression: \[ 8(2k^2 + k) \] Since we can write the expression as a product with 8, it's clear that 8 does divide \(n^2 - 1\) when \(n\) is of the form \(4k + 1\).

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Key Concepts

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

Integer Division
Integer division deals with dividing one integer by another, resulting in a quotient and possibly a remainder. In mathematics, understanding whether one integer divides another without leaving a remainder is a key concept. In our exercise, we handle this by determining if 8 evenly divides the expression \(n^2 - 1\).- Given that \(n = 4k + 1\), we start by squaring \(n\) to examine \(n^2 - 1\).- After expanding and simplifying \((4k + 1)^2\), we reach \(16k^2 + 8k + 1\).- Subtracting 1 results in the expression \(16k^2 + 8k\). We can factor out 8, simplifying it to \(8(2k^2 + k)\).This factorization shows that 8 indeed divides \(n^2 - 1\) since the entire expression can be expressed as a multiple of 8. In integer division, showing that a factor can be extracted in this way confirms divisibility, highlighting the nature of dividing perfectly into an integer.
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying expressions to solve equations or prove statements. It's a fundamental tool in algebra and critical to solving the given exercise.In our problem, we apply algebraic manipulation to the expression \(n = 4k + 1\):- We begin by squaring the expression to get \((4k + 1)^2\).- This expansion uses the distributive property: \((4k)^2 + 2(4k)(1) + 1^2\), resulting in \(16k^2 + 8k + 1\).- Subtraction of 1 simplifies the expression further to \(16k^2 + 8k\).By rearranging and factoring expressions, algebraic manipulation streamlines problem-solving, guiding us towards solutions like determining divisibility by 8 in this context. Each step helps break down complex expressions into simpler forms that are easier to handle and solve.
Mathematical Proof
Mathematical proof is a logical argument that validates the truth of a mathematical statement. In exercises like the one we explored, proofs offer certainty and demonstrate why a particular solution holds.In the problem presented:- We proved that \(8\) divides \(n^2 - 1\) for \(n = 4k + 1\) by first expanding and simplifying \((4k + 1)^2\).- By subtracting 1, we got the expression \(16k^2 + 8k\), factored as \(8(2k^2 + k)\).- The factorization conclusively shows that the expression is divisible by 8, proving our original statement.Proofs like these confirm our findings through step-by-step logical reasoning. By progressively breaking down mathematical elements, such proofs validate solutions and ensure accuracy in algebraic conclusions.

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Most popular questions from this chapter

The difference of any two odd integers is even.

If \(m\) and \(n\) are perfect squares, then \(m+n+2 \sqrt{m n}\) is also a perfect square. Why?

A matrix \(\mathbf{M}\) has 3 rows and 4 columns.$$ \left[\begin{array}{llll} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \end{array}\right] $$The 12 entries in the matrix are to be stored in row major form in locations 7,609 to 7,620 in a computer's memory. This means that the entries in the first row (reading left to right) are stored first, then the entries in the second row, and finally the entries in the third row. a. Which location will \(a_{22}\) be stored in? b. Write a formula (in \(i\) and \(j\) ) that gives the integer \(n\) so that \(a_{i j}\) is stored in location \(7,609+n\). c. Find formulas (in \(n\) ) for \(r\) and \(s\) so that \(a_{r s}\) is stored in location \(7,609+n\).

"Proof: Suppose \(r\) and \(s\) are rational numbers. If \(r+s\) is rational, then by definition of rational \(r+s=a / b\) for some integers \(a\) and \(b\) with \(b \neq 0\). Also since \(r\) and \(s\) are rational, \(r=i / j\) and \(s=m / n\) for some integers \(i, j, m\), and \(n\) with \(j \neq 0\) and \(n \neq 0\). It follows that \(r+s=i / j+m / n=\) \(a / b\), which is a quotient of two integers with a nonzero denominator. Hence it is a rational number. This is what was to be shown. \(^{.1}\)

Prove that if a real number \(c\) satisfies a polynomial equation of the form $$ r_{3} x^{3}+r_{2} x^{2}+r_{1} x+r_{0}=0, $$ where \(r_{0}, r_{1}, r_{2}\), and \(r_{3}\) are rational numbers, then \(c\) satisfies an equation of the form $$ n_{3} x^{3}+n_{2} x^{2}+n_{1} x+n_{0}=0 \text {, } $$ where \(n_{0}, n_{1}, n_{2}\), and \(n_{3}\) are integers.
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