/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Consider the following propositi... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 16

Consider the following proposition: There are no integers \(a\) and \(b\) such that \(b^{2}=4 a+2\) (a) Rewrite this statement in an equivalent form using a universal quantifier by completing the following: For all integers \(a\) and \(b, \ldots\) (b) Prove the statement in Part (a).

Short Answer

Expert verified
(a) For all integers \(a\) and \(b\), \(b^2 \neq 4a + 2\). (b) By contradiction, assume there exist integers \(a\) and \(b\) such that \(b^2 = 4a + 2\). Considering the parity of the equation, we conclude that \(b\) must be even. Let \(b = 2k\), for some integer \(k\). Rewriting the equation as \((2k)^2 = 4a + 2\) and simplifying, we get \(2k^2 = 2a + 1\). This implies that \(2a + 1\) is even, which contradicts the fact that \(2a + 1\) is always odd. Hence, there are no integers \(a\) and \(b\) such that \(b^2 = 4a + 2\).

Step by step solution

01

Rewrite the proposition in terms of universal quantifiers

In order to rewrite the proposition using a universal quantifier, we want to express that the given integer equation does not hold for any pair of integers \(a\) and \(b\). Thus, we can rewrite the statement as: For all integers \(a\) and \(b\), \(b^2 \neq 4a + 2\). #Step 2: Prove the statement in Part (a)#
02

Using Contrapositive and Parity#

To prove this statement, we will use the method of contradiction. We will assume that the given proposition is false and try to find a contradiction. So, let's assume that there exist integers \(a\) and \(b\) such that \(b^2 = 4a + 2\). Now, consider the parity of both sides of the equation. If \(b\) is even then \(b = 2k\) for some integer \(k\). Hence, \(b^2 = 4k^2\). Since every even integer can be expressed as 2 times an integer, then \(b^2\) is even. On the other hand, if \(b\) is odd then \(b = 2k + 1\) for some integer \(k\). So, \(b^2 = (2k + 1)^2 = 4k^2 + 4k + 1\). As you can see, \(b^2\) is odd in this case. However, the right-hand side of the equation \(b^2 = 4a + 2\) must be even, because \(4a + 2 = 2(2a + 1)\). Therefore, \(b^2\) should also be even. Thus, \(b\) must be even, and we can express it as \(b = 2k\) for some integer \(k\). Now, rewrite the equation as follows: \((2k)^2 = 4a + 2\) Simplifying this equation, we get: \(4k^2 = 4a + 2\) Divide both sides of the equation by 2, we get: \(2k^2 = 2a + 1\) This tells us that \(2a + 1\) is an even number, which is a contradiction because \(2a + 1\) is always odd for any integer \(a\). Therefore, our assumption is wrong, and there are no integers \(a\) and \(b\) such that \(b^2 = 4a + 2\).

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Key Concepts

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

Proof by Contradiction
Proof by contradiction is a technique used to establish the truth of a statement by assuming the opposite of what you are trying to prove. The idea is to show that this assumption leads to a contradiction. This contradiction implies that the original assumption is false, thus proving the statement.In the exercise, the proposition "There are no integers \(a\) and \(b\) such that \(b^{2}=4a+2\)" was assumed to be false. We supposed that there exist integers \(a\) and \(b\) satisfying the equation \(b^2 = 4a + 2\). Through logical steps, we showed this assumption led to a contradiction involving integer parity. Therefore, the initial proposition that no such integers exist must be true.
Integer Parity
Parity in mathematics refers to whether an integer is even or odd. An even integer can be expressed as \(2k\) for some integer \(k\), while an odd integer can be written as \(2k + 1\). Understanding parity can help simplify and solve equations, especially those involving even powers or multiples.In our case, we analyzed the parity of \(b\) to determine the consistency of the equation \(b^2 = 4a + 2\). The square of an even integer is even, and the square of an odd integer is odd. By assuming \(b^2\) could be odd or even and analyzing these conditions, we deduced contradictions. If \(b\) is odd, \(b^2\) cannot match the even form of \(4a+2\). This parity analysis helped in reaching the contradiction needed to prove our proposition by contradiction.
Logical Equivalence
Logical equivalence involves two statements having the same truth value in every model. This is key in rewriting statements and simplifying expressions using logical transformations and rules.In the exercise, our task was to rewrite the statement using a universal quantifier, maintaining logical equivalence. The original proposition "There are no integers \(a\) and \(b\) such that \(b^{2}=4a+2\)" was logically equivalent to "For all integers \(a\) and \(b\), \(b^2 eq 4a + 2\)." This universal statement expresses the same idea, asserting that for every combination of integers \(a\) and \(b\), the equation cannot hold true. Logical equivalence allows us to express complex ideas more clearly, helping in proofs and calculations.
Quantifier Negation
Quantifiers like "for all" (universal quantifier) and "there exists" (existential quantifier) are used to make general statements. Negating these quantifiers can help in rewriting statements and eventually simplifying proofs.In this exercise, the original statement "There are no integers \(a\) and \(b\) such that \(b^2 = 4a + 2\)" involves the negation of an existential quantifier. Negating "there exists" results in a universal statement, leading us to "For all integers \(a\) and \(b\), \(b^2 eq 4a + 2\)." Through quantifier negation, we shifted from considering possibilities to making a broad, sweeping assertion that simplifies proving the impossibility of the equation holding for any integers.

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Most popular questions from this chapter

Consider the following proposition: Proposition. For all integers \(m\) and \(n,\) if \(n\) is odd, then the equation $$ x^{2}+2 m x+2 n=0 $$ has no integer solution for \(x\). (a) What are the solutions of the equation when \(m=1\) and \(n=-1 ?\) That is, what are the solutions of the equation \(x^{2}+2 x-2=0 ?\) (b) What are the solutions of the equation when \(m=2\) and \(n=3\) ? That is, what are the solutions of the equation \(x^{2}+4 x+6=0 ?\) (c) Solve the resulting quadratic equation for at least two more examples using values of \(m\) and \(n\) that satisfy the hypothesis of the proposition. (d) For this proposition, why does it seem reasonable to try a proof by contradiction? (e) For this proposition, state clearly the assumptions that need to be made at the beginning of a proof by contradiction. (f) Use a proof by contradiction to prove this proposition.

Prove that for all integers \(a\) and \(m,\) if \(a\) and \(m\) are the lengths of the sides of a right triangle and \(m+1\) is the length of the hypotenuse, then \(a\) is an odd integer.

Consider the following proposition: For each integer \(a, a \equiv 3(\bmod 7)\) if and only if \(\left(a^{2}+5 a\right) \equiv 3(\bmod 7)\). (a) Write the proposition as the conjunction of two conditional statements. (b) Determine if the two conditional statements in Part (a) are true or false. If a conditional statement is true, write a proof, and if it is false, provide a counterexample. (c) Is the given proposition true or false? Explain.

This exercise is intended to provide another rationale as to why a proof by contradiction works. Suppose that we are trying to prove that a statement \(P\) is true. Instead of proving this statement, assume that we prove that the conditional statement "If \(\neg P\), then \(C^{\prime \prime}\) is true, where \(C\) is some contradiction. Recall that a contradiction is a statement that is always false. (a) In symbols, write a statement that is a disjunction and that is logically equivalent to \(\neg P \rightarrow C\). (b) Since we have proven that \(\neg P \rightarrow C\) is true, then the disjunction in Exercise (1a) must also be true. Use this to explain why the statement \(P\) must be true. (c) Now explain why \(P\) must be true if we prove that the negation of \(P\) implies a contradiction.

Let \(y_{1}, y_{2}, y_{3}, y_{4}\) be real numbers. The mean, \(\bar{y},\) of these four numbers is defined to be the sum of the four numbers divided by \(4 .\) That is, $$ \bar{y}=\frac{y_{1}+y_{2}+y_{3}+y_{4}}{4} $$ Prove that there exists a \(y_{i}\) with \(1 \leq i \leq 4\) such that \(y_{i} \geq \bar{y}\). Hint: One way is to let \(y_{\max }\) be the largest of \(y_{1}, y_{2}, y_{3}, y_{4}\).
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