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Prove that for all integers \(m\) and \(n\), if \(m\) is odd and \(n\) is even, then \(m n\) is even.

Short Answer

Expert verified
Given that \(m\) and \(n\) are odd and even respectively, their product \(m \cdot n\) will always be represented as \(2z\) where \(z\) is an integer, hence \(m \cdot n\) will always be an even number.

Step by step solution

01

Expressing m and n

Express \(m\) and \(n\) in terms of their definitions. Assume \(m = 2k+1\) for some integer \(k\), since \(m\) is odd. Also, let \(n = 2j\) for some integer \(j\), since \(n\) is even.
02

Multiplying m and n

Performing the multiplication \(m \cdot n = (2k+1) \cdot 2j = 4kj + 2j = 2(2kj + j)\).
03

Representing Result as Even Number

Notice that \(2kj + j\) is an integer say \(z\), since integers are closed under addition and multiplication. So, it can be written as \(m \cdot n = 2z\), where \(z\) = \(2kj + j\).

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Key Concepts

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

Even and Odd Integers
Understanding the difference between even and odd integers is essential in integer arithmetic. An even integer is any integer that can be divided by 2 without leaving a remainder. It can be expressed in the form of (2k) where k is any integer. This means every even integer can be evenly divided into pairs.

On the other hand, an odd integer cannot be evenly divided by 2, leaving a remainder of 1. We express odd integers as (2k + 1), again with k being any integer. By definition, when you divide an odd integer by 2, a part always remains 'unpaired', hence the remainder.

Take note of this key idea: when you multiply an even number by any other integer, the product is always even. This is because division by 2 is 'built-in' to the even number, and multiplying it by another integer doesn't change that characteristic. But an odd number, with its intrinsic remainder, behaves differently under multiplication, especially when combined with an even integer.
Integer Arithmetic
When performing arithmetic operations with integers, certain rules apply that enable us to predict outcomes for operations like addition, subtraction, multiplication, and division. In the case of multiplication, which is our focus here, we saw this rule in action: multiplying two integers always results in another integer. This notion is referred to as the closure property of integers under multiplication.

Another important operation is multiplication involving even and odd numbers. As seen in the exercise, the product of an even and an odd number results in an even number. This consistent result is rooted in the very definitions of even and odd numbers and how we handle multiplication of numbers. The property can be extended to addition and subtraction as well, where the sum or difference of two even numbers, or two odd numbers, is always even, while the sum or difference of an even and an odd number is always odd.
Mathematical Proofs
In mathematics, proofs are the gold standard for verifying that a statement or proposition is true. When we say we've 'proven' something mathematically, it means we've shown it to be true beyond any doubt, using logical reasoning and accepted mathematical principles.

In our exercise, we used a direct proof method where we started by accepting the definitions of odd and even integers, expressed them as (2k + 1) and (2j) respectively, and then followed a logical progression of steps to show that their product is indeed even. A proof generally proceeds by taking assumed truths (like definitions or previously proven theorems) and, through a series of logically sound steps, arrives at the conclusion we set out to prove.

Effective proofs avoid unnecessary complexities and strive to be as clear and concise as possible. They are structured so that each step logically leads to the next, without taking any leaps that might confuse the reader or leave room for doubt. Just as the proof in our exercise clearly shows how manipulating the initial expressions of an odd and even integer gives us a final product that can be expressed as 2 times an integer, thereby confirming its evenness.

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Most popular questions from this chapter

A 3D-septomino is a three-dimensional \(2 \times 2 \times 2\) cube with one \(1 \times 1 \times 1\) corner cube removed. \(A\) deficient cube is \(a k \times k \times k\) cube with one \(1 \times 1 \times 1\) cube removed. Suppose that \(S_{n}=(n+2)(n-1)\) is (incorrectly) proposed as a formula for $$ 2+4+\cdots+2 n $$ (a) Show that the Inductive Step is satisfied but that the Basis Step fails. (b) If \(S_{n}^{\prime}\) is an arbitrary expression that satisfies the Inductive Step, what form must \(S_{n}^{\prime}\) assume?

Prove that for all sets \(A\) and \(B, A \subseteq B\) if and only if \(\bar{B} \subseteq \bar{A}\).

Suppose that \(n>1\) people are positioned in a field (Euclidean plane) so that each has a unique nearest neighbor: Suppose further that each person has a pie that is hurled at the nearest neighbor: A survivor is a person that is not hit by a pie. Prove or disprove: If \(n\) is odd, one of two persons farthest apart is a survivor.

Define the sequence \(c_{1}, c_{2}, \ldots\) by the equations $$c_{1}=c_{2}=0, \quad c_{n}=c_{\lfloor n / 3\rfloor}+n \text { for all } n>2$$ Suppose that we want to prove a statement for all \(n \geq 2\) involving \(c_{n} .\) The Inductive Step will assume the truth of the statement involving \(c_{\lfloor n / 3 j} .\) What are the Basis Steps?

Refer to the sequence \(c_{1}, c_{2}, \ldots\) defined by the equations $$c_{1}=0, \quad c_{n}=c_{\lfloor n / 2\rfloor}+n^{2} \text { for all } n>1$$ Suppose that we want to prove a statement for all \(n \geq 3\) involving \(c_{n}\). The Inductive Step will assume the truth of the statement involving \(c_{\lfloor n / 2\rfloor}\). What are the Basis Steps?

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