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Chapter 1: Problem 10

Prove each directly. The product of any two odd integers is odd.

Short Answer

Expert verified
The product of two odd integers, \(a = 2k_1 + 1\) and \(b = 2k_2 + 1\), can be computed as \(c = a \cdot b = (2k_1 + 1)(2k_2 + 1)\). Expanding and factoring out 2, we get \(c = 2(2k_1k_2 + k_1 + k_2) + 1\), which is also in the form of an odd integer \(2m + 1\) with \(m = 2k_1k_2 + k_1 + k_2\). Therefore, the product of any two odd integers is odd.

Step by step solution

01

Define odd integers

Let's represent two odd integers as \(a = 2k_1 + 1\) and \(b = 2k_2 + 1\), where \(k_1\) and \(k_2\) are integers. It is important to note that both the odd integers follow the definition \(2k + 1\).
02

Compute the product of the odd integers

Next, we need to calculate the product of these two odd integers, which can be represented as \(c = a \cdot b\). Substituting the expressions for \(a\) and \(b\) from Step 1, we get: \(c = (2k_1 + 1)(2k_2 + 1)\)
03

Expand the expression for the product

In order to verify whether the product of these two odd integers is odd, we need to expand the expression for \(c\): \[c = (2k_1 + 1)(2k_2 + 1) = 4k_1k_2 + 2k_1 + 2k_2 + 1\]
04

Factor out 2 from the expression

Now, we will factor out 2 from the terms with \(k_1\) and \(k_2\): \[c = 4k_1k_2 + 2(k_1) + 2(k_2) + 1 = 2(2k_1k_2 + k_1 + k_2) + 1\]
05

Confirm the product is odd

We can now see that the expression for \(c\) fits the definition of an odd integer, which is in the form \(2m + 1\), where \(m = 2k_1k_2 + k_1 + k_2\) is an integer. Thus, we have proved that the product of any two odd integers is odd.

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Key Concepts

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

Odd Integers
Odd integers are numbers that cannot be evenly divided by 2. They are numbers like 1, 3, 5, 7, and so forth. These numbers leave a remainder of 1 when divided by 2. You can represent any odd integer with the formula:
  • \(a = 2k + 1\)
Here, \(k\) is any integer. When you multiply \(k\) by 2, you get an even number, and by adding 1 to it, the number becomes odd. This formula ensures that every odd number has the same characteristic鈥攖he remainder is 1 when divided by 2. By representing odd integers this way, mathematics can easily describe and manipulate them without losing their odd nature. This is useful, especially in proofs like the one we're discussing.
Direct Proof
A direct proof is a method of proving mathematical statements. It involves straightforward steps that solve or derive a mathematical truth from given premises. In the proof presented here, the goal is to show that the product of two odd integers is also odd. To execute a direct proof:
  • Start with known information or premises.
  • Employ logical steps to arrive at the conclusion.
In our case, we began by defining the integers, followed by calculating their product. Finally, we showed the resulting expression fit the definition of an odd integer. Direct proofs are powerful as they offer a clear and systematic way to arrive at conclusions.
Integer Properties
Integers are a key concept in mathematics. They include whole numbers and their negatives, like -3, -2, -1, 0, 1, 2, 3. Two important properties of integers are evenness and oddness.
  • Even integers can be divided by 2 with no remainder.
  • Odd integers have a remainder of 1 when divided by 2.
In our proof, we used the property of odd integers through mathematical expressions. Integer properties, like closure when adding, subtracting, or multiplying integers, guarantee their outcomes will still be integers. This is crucial when dealing with expressions that require maintaining integer properties throughout calculations.
Algebraic Expressions
Algebraic expressions allow us to generalize mathematical concepts. They use variables like \(k_1\) and \(k_2\) and constants to express arithmetic relationships. In our proof, algebraic expressions helped demonstrate that a product of odd integers results in another odd integer. Here's how it went:
  • The odd integers were represented as \(a = 2k_1 + 1\) and \(b = 2k_2 + 1\).
  • The product \(c = (2k_1 + 1)(2k_2 + 1)\) was expanded and simplified.
  • The expression \(c = 2(2k_1k_2 + k_1 + k_2) + 1\) emerged.
The ability to manipulate these expressions is crucial in algebra. They serve as tools to verify properties, like ensuring the result conforms to expectations. Algebraic expressions thus offer a reliable framework for proving mathematical statements.

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Most popular questions from this chapter

The Sheffer stroke / is a binary operator" defined by the following truth table.(Note: On page 25 we used the vertical bar \(|\) to mean is a factor of. The actual meaning should be clear from the context. So be careful.) Verify each. (Note: Exercise 58 shows that the logical operators \(|\) and \(\mathrm{NAND}\) are the same. (TABLE CAN'T COPY) $$p \rightarrow q \equiv((p | p)|(p | p))|(q | q)$$

Exercises \(65-78\) deal with propositions in fuzzy logic. Let \(p, q,\) and \(r\) be simple propositions with \(t(p)=1, t(q)=0.3,\) and \(t(r)=\) 0.5 . Compute the truth value of each, where \(s^{\prime}\) denotes the negation of the statement \(s\) . $$ p \vee r $$

Use De Morgan's laws to verify each. (Hint: \(p \rightarrow q \equiv \sim p \vee q\) ).

Determine whether or not the assignment statement \(x \leftarrow x+1\) will be executed in each sequence of statements, where \(i \leftarrow 2, j \leftarrow 3, k \leftarrow 6,\) and \(x \leftarrow 0\). $$ \begin{array}{l} \text { If }(i4) \text { then } \\ x \leftarrow x-1 \end{array} $$ else $$ x

Construct a truth table for each proposition. $$(p \wedge q) \rightarrow \sim p$$
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