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Chapter 3: Problem 5

Is the following proposition true or false? For all integers \(a\) and \(b,\) if \(a b\) is even, then \(a\) is even or \(b\) is even. Justify your conclusion by writing a proof if the proposition is true or by providing a counterexample if it is false.

Short Answer

Expert verified
The proposition is true. If the product of two integers a and b is even, then at least one of them must be even. We have proven this using contradiction. We showed that when both a and b are odd, their product is also odd, contradicting the original statement. Therefore, if a 脳 b is even, either a or b must be even.

Step by step solution

01

Understanding the Given Proposition

The proposition claims that the product of two integers a and b can only be even if at least one of them is even. We need to prove this or provide a counterexample if it's false.
02

Proving the Proposition

To prove this statement, we will use contradiction. Suppose both a and b are odd integers. In other words, let a = 2m + 1 and b = 2n + 1, where m and n are integers. Now, let's calculate their product: Product = a 脳 b = (2m + 1) 脳 (2n + 1) Let's expand the product: Product = 4mn + 2m + 2n + 1 Notice that the product can be rewritten as: Product = 2(2mn + m + n) + 1 Since 2mn + m + n is an integer, the product is equal to 2 times an integer plus 1. So, the product is odd. This contradicts the original statement that a 脳 b is even. Therefore, we have proven that when a 脳 b is even, at least one of the integers a and b must be even.

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Key Concepts

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

Integer Properties
Understanding integer properties is crucial in mathematics because integers are the foundation of number theory and many mathematical proofs. Integers include all whole numbers and their negatives, such as
  • -3, -2, -1, 0, 1, 2, 3, and so on.

Some important properties of integers that are often used in proofs are their identity elements and their closure property under addition and multiplication.
  • The identity elements mean adding zero or multiplying by one does not change the value of an integer.
  • The closure property implies that when integers are added or multiplied together, the result is always another integer.

In proofs, these properties help to logically structure and understand arguments about numbers.
Even and Odd Numbers
Even and odd numbers are fundamental classifications of integers.
  • An even number is any integer that is divisible by 2, such as 0, 2, 4, 6, and 8.
  • An odd number, such as 1, 3, 5, 7, and 9, is not divisible by 2.

In mathematical proofs, knowing whether a number is even or odd helps determine outcomes, such as in our exercise about products of numbers.
When you multiply two integers:
  • If both numbers are even, the product is even.
  • If one number is even and the other is odd, the product is still even.
  • But if both numbers are odd, the product is always odd.

This understanding is key when exploring propositions that involve predicting whether a product will be even or odd.
Proof by Contradiction
Proof by contradiction is a powerful mathematical technique used to establish the truth of a proposition by assuming the opposite is true and then showing that this assumption leads to a contradiction.
In the exercise, we are asked to prove that if the product of two integers is even, then at least one of them must be even.
To do this, assume both integers, \(a\) and \(b\), are odd. We express these odd integers as \(a = 2m + 1\) and \(b = 2n + 1\), where \(m\) and \(n\) are integers.
Their product would be \((2m + 1)(2n + 1) = 4mn + 2m + 2n + 1\), which simplifies to \(2(2mn + m + n) + 1\).
This formula shows the product is actually odd.
Since this contradicts the original proposition that the product is even, the assumption that both \(a\) and \(b\) are odd must be false. Thus, at least one must be even, proving the original statement.
Proof by contradiction is especially useful when faced with clear binary conditions, helping verify mathematical truths effectively.

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Most popular questions from this chapter

Determine if each of the following statements is true or false. If a statement is true, then write a formal proof of that statement, and if it is false, then provide a counterexample that shows it is false. (a) For all integers \(a, b,\) and \(c\) with \(a \neq 0,\) if \(a \mid b,\) then \(a \mid(b c)\). (b) For all integers \(a\) and \(b\) with \(a \neq 0,\) if \(6 \mid(a b),\) then \(6 \mid a\) or \(6 \mid b\). (c) For all integers \(a, b,\) and \(c\) with \(a \neq 0,\) if \(a\) divides \((b-1)\) and \(a\) divides \((c-1),\) then \(a\) divides \((b c-1)\) (d) For each integer \(n,\) if 7 divides \(\left(n^{2}-4\right),\) then 7 divides \((n-2)\). (e) For every integer \(n, 4 n^{2}+7 n+6\) is an odd integer. ? (f) For every odd integer \(n, 4 n^{2}+7 n+6\) is an odd integer. (g) For all integers \(a, b,\) and \(d\) with \(d \neq 0,\) if \(d\) divides both \(a-b\) and \(a+b,\) then \(d\) divides \(a\) (h) For all integers \(a, b,\) and \(c\) with \(a \neq 0,\) if \(a \mid(b c),\) then \(a \mid b\) or \(a \mid c .\)

Determine if each of the following statements is true or false. Provide a counterexample for statements that are false and provide a complete proof for those that are true. (a) For all real numbers \(x\) and \(y, \sqrt{x y} \leq \frac{x+y}{2}\). (b) For all real numbers \(x\) and \(y, x y \leq\left(\frac{x+y}{2}\right)^{2}\). (c) For all nonnegative real numbers \(x\) and \(y, \sqrt{x y} \leq \frac{x+y}{2}\).

Prove that there do not exist three consecutive natural numbers such that the cube of the largest is equal to the sum of the cubes of the other two.

In Preview Activity \(2,\) we proved that if \(n\) is an integer, then \(n^{2}+n\) is an even integer. We define two integers to be consecutive integers if one of the integers is one more than the other integer. This means that we can represent consecutive integers as \(m\) and \(m+1,\) where \(m\) is some integer.

Let \(a\) and \(b\) be integers. Prove that if \(a \equiv 2(\bmod 3)\) and \(b \equiv 2(\bmod 3)\), then (a) \(a+b \equiv 1(\bmod 3) ;\) (b) \(a \cdot b \equiv 1(\bmod 3)\).
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