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Chapter 3: Problem 13

For all integers \(m\) and \(n\), if \(2 m+n\) is odd then \(m\) and \(n\) are both odd.

Short Answer

Expert verified
From the analysis, we found that the only combinations that satisfy the given condition (\(2m + n\) is odd) are when either both \(m\) and \(n\) are odd integers or \(m\) is even and \(n\) is odd. However, since \(2m\) will always be even, we can conclude that if \(2m + n\) is odd, then both \(m\) and \(n\) must be odd integers.

Step by step solution

01

Understanding Even and Odd Integers

An even integer is any integer that can be represented as \(2k\), where \(k\) is also an integer. An odd integer is any integer that can be represented as \(2k + 1\), where \(k\) is an integer.
02

Analyzing the given condition

We are given that \(2m + n\) is odd, which means that it can be represented as \(2k + 1\), where \(k\) is an integer. So, we have the equation: \(2m + n = 2k + 1\)
03

Finding possible combinations for \(m\) and \(n\)

Now we need to find possible combinations of \(m\) and \(n\) which makes the given equation true. There are 4 possible combinations for \(m\) and \(n\): 1. \(m\) is even, and \(n\) is even 2. \(m\) is even, and \(n\) is odd 3. \(m\) is odd, and \(n\) is even 4. \(m\) is odd, and \(n\) is odd
04

Evaluating each combination

1. If \(m\) is even and \(n\) is even, then \(2m\) is even, and \(2m + n\) will be even. Thus, this condition contradicts the given statement, and it's not possible. 2. If \(m\) is even and \(n\) is odd, then \(2m\) is even. However, an even number plus an odd number is an odd number, so this satisfies the condition \(2m + n\) is odd. 3. If \(m\) is odd and \(n\) is even, then \(2m\) is even, and an even number plus an even number is an even number. This condition contradicts the given statement, and it's not possible. 4. If \(m\) is odd and \(n\) is odd, then \(2m\) is even. However, an even number plus an odd number is an odd number, so this satisfies the condition \(2m + n\) is odd.
05

Conclusion

From the previous step, we can see that the only combinations that satisfy the condition \(2m + n\) is odd are when both \(m\) and \(n\) are odd integers. Thus, we can conclude: If "For all integers \(m\) and \(n\), \(2m + n\) is odd," then \(m\) and \(n\) are both odd integers.

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Key Concepts

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

Integer Properties
Understanding the properties of integers is fundamental in mathematics. Integers are whole numbers that can be positive, negative, or zero. They do not include fractions or decimals. One crucial property of integers is their divisibility by 2, which divides them into two categories: even and odd. An even integer can be divided by 2 with no remainder (e.g., 2, 4, 6), while an odd integer will have a remainder of 1 when divided by 2 (e.g., 1, 3, 5).

In relation to the exercise, the idea that an integer can be expressed as either of the form of 2k or 2k + 1, with k being another integer, demonstrates this fundamental property of even and odd integers, respectively. This concept is not just a rule but an intrinsic property of integers that we use to simplify and solve complex mathematical problems.
Proofs in Mathematics
Mathematical proofs are logical arguments that establish the truth or falsehood of a proposition. They use a combination of axioms, definitions, theorems, and previously proven statements to reach a conclusion. Unlike empirical evidence, proofs provide certainty and finality because they're based on logical reasoning.

For instance, when the exercise asserts 'For all integers m and n, if 2m + n is odd then m and n are both odd,' it's making a general claim that needs to be proven for all possible cases. As shown in the step-by-step solution, the exercise methodically examines all possible combinations of even and odd integers for m and n, ruling out the combinations that don't satisfy the provided condition. This systematic approach is crucial to proving mathematical statements definitively.
Parity of Integers
Parity is a concept that defines whether an integer is even or odd. It is a vital analytic tool in various branches of mathematics and computer science. Understanding the parity of integers allows us to make predictions about the outcomes of operations involving whole numbers and informs us about their properties at a fundamental level.

In our context, the exercise demonstrates that the sum of an even and an odd integer is odd, aligning with the rules of integer parity. This fact is used as a tool to dissect the problem and identify the conditions under which m and n must both be odd to satisfy the given statement. The exploration of parity in such exercises doesn't just solve a singular problem; it also builds a conceptual toolkit that students can use in wider mathematical reasoning.

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Most popular questions from this chapter

Consider the statement "For all integers \(n\), if \(n^{2}\) is odd then \(n\) is odd." a. Write what you would suppose and what you would need to show to prove this statement by contradiction. b. Write what you would suppose and what you would nced to show to prove this statement by contraposition.

a. Prove that for all integers \(a\), if \(a^{3}\) is even then \(a\) is even. b. Prove that \(\sqrt[3]{2}\) is irrational.

Prove that there exists a unique prime number of the form \(n^{2}-1\), where \(n\) is an integer that is greater than or equal to 2 .

There is a real number \(x\) such that \(x>1\) and \(2^{x}>x^{10}\).

An alternative proof of the infinitude of the prime numbers begins as follows: Proof: Suppose there are only finitely many prime numbers. Then one is the largest. Call it \(p\). Let \(M=p !+1\). We will show that there is a prime number \(q\) such that \(q>p\). Complete this proof.
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