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Chapter 3: Problem 3

Assume that \(r\) and \(s\) are particular integers. a. Is \(4 r s\) even? b. Is \(6 r+4 s^{2}+3\) odd? c. If \(r\) and \(s\) are both positive, is \(r^{2}+2 r s+s^{2}\) composite?

Short Answer

Expert verified
a. Yes, 4rs is always even because 4 is even and any product involving an even number is even. b. Yes, 6r + 4s^2 + 3 is always odd because the sum of even and odd terms results in an odd expression. c. Yes, r^2 + 2rs + s^2 is composite when r and s are both positive integers, as it can be rewritten as (r + s)^2, which is always composite for r > 1 and s > 1 or when r = 1 and s = 1.

Step by step solution

01

a. Determine if 4rs is even

Recall that a number is considered even if it is divisible by 2. As 4 is a multiple of 2, then any product involving 4 will also be a multiple of 2. Therefore, 4rs will always be even, regardless of the values of r and s.
02

b. Determine if 6r + 4s^2 + 3 is odd

An expression is odd if the result is odd when substituted r and s with integers. Let's analyze the given expression: 1. 6r will always be even because 6 is a multiple of 2 2. 4s^2 will always be even because 4 is a multiple of 2 3. 3 is odd Now, let's combine these three parts: - Even (6r) + Even (4s^2) = Even - Even + Odd (3) = Odd Therefore, 6r + 4s^2 + 3 will always be odd, regardless of the values of r and s.
03

c. Determine if r^2 + 2rs + s^2 is composite when r and s are positive

We can rewrite the expression r^2 + 2rs + s^2 as a perfect square: \[ (r + s)^2 = r^2 + 2rs +s^2 \] Now, consider the following cases: 1. If r = 1 and s = 1, then (r + s)^2 = (1 + 1)^2 = 2^2 = 4, which is composite. 2. If r > 1 and s > 1, the result will be the square of a greater integer (since r + s > 2). As squares of integers greater than 2 are always composite, (r + s)^2 will also be composite in this case. In conclusion, r^2 + 2rs + s^2 is a composite number when r and s are both positive integers.

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Key Concepts

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

Even and Odd Integers
Integer numbers can be classified into two main categories based on their divisibility by 2: even and odd integers. An even integer is any number that can be divided by 2 without leaving a remainder. For instance, numbers like 2, 4, and 6 are even because when you divide them by 2, there's no leftover. On the other hand, odd integers are those which, when divided by 2, have a remainder of 1. Numbers such as 1, 3, and 5 are classic examples of odd integers.

In our exercise, understanding the definition of even and odd integers allows us to immediately determine that any product involving an even integer (like 4 in the term 4rs) will also be even. Similarly, when you add an odd number (like 3) to an even number, the result is always odd, as seen in the expression 6r + 4s^2 + 3.
Composite Numbers
Moving onto composite numbers, these are integers that have more divisibility options than just 1 and themselves. In other words, composite numbers have at least one divisor that is neither 1 nor the number itself. For example, 4, 6, 8, and 9 are all composite because they can be divided evenly by numbers other than 1 and the number itself. The concept of composite numbers is essential when analyzing expressions for composability, such as in the latter part of our exercise, where we evaluate if the expression \(r^2 + 2rs + s^2\) is composite. This expression simplifies to \((r + s)^2\), which is a perfect square and thus will always yield a composite number when \(r\) and \(s\) are positive integers greater than 1.
Perfect Square Expression
Finally, a perfect square expression represents the product of some integer with itself. For instance, 4 is a perfect square because it's the result of 2 multiplied by 2, written as 2^2. In our scenario, the expression \(r^2 + 2rs + s^2\) is recognized as the square of the binomial \((r + s)^2\), making it a perfect square.

Knowing this, we can deduce certain properties about the expression's value. Perfect squares are always non-negative, and if they originate from any integer other than 0 or 1, they're guaranteed to be composite numbers. The reason is that perfect squares have more than two divisors鈥攏amely, 1, the number itself, and at least the square root of the number.

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Most popular questions from this chapter

"Proof: Suppose \(r\) and \(s\) are rational numbers. By definition of rational, \(r=a / b\) for some integers \(a\) and \(b\) with \(b \neq 0\), and \(s=a / b\) for some integers \(a\) and \(b\) with \(b \neq 0\). Then \(r+s=a / b+a / b=2 a / b\). Let \(p=2 a\). Then \(p\) is an integer since it is a product of integers. Hence \(r+s=p / b\), where \(p\) and \(b\) are integers and \(b \neq 0\). Thus \(r+s\) is a rational number by definition of rational. This is what was to be shown."

Prove that if one solution for a quadratic equation of the form \(x^{2}+b x+c=0\) is rational (where \(b\) and \(c\) are ra-tional), then the other solution is also rational. (Use the fact that if the solutions of the equation are \(r\) and \(s\), then \(\left.x^{2}+b x+c=(x-r)(x-s) .\right)\)

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 \(7,609+n\).

For all real numbers \(x\), if \(x>1\) then \(x^{2}>x\).

A calculator display shows that \(\sqrt{2}=1.414213562\), and \(1.414213562=\frac{1414213562}{1000000000}\). This suggests that \(\sqrt{2}\) is a rational number, which contradicts Theorem 3.7.1. Explain the discrepancy.
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