/*! 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 22 See the instructions for Exercis... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 22

See the instructions for Exercise (19) on page 100 from Section 3.1 . (a) Proposition. For all integers \(a\) and \(b\), if \((a+2 b) \equiv 0(\) mod 3 ), then \((2 a+b) \equiv 0(\bmod 3)\) Proof. We assume \(a, b \in \mathbb{Z}\) and \((a+2 b) \equiv 0(\) mod 3\() .\) This means that 3 divides \(a+2 b\) and, hence, there exists an integer \(m\) such that \(a+2 b=3 m .\) Hence, \(a=3 m-2 b .\) For \((2 a+b)=0(\bmod 3),\) there exists an integer \(x\) such that \(2 a+b=3 x .\) Hence, $$ \begin{aligned} 2(3 m-2 b)+b &=3 x \\ 6 m-3 b &=3 x \\ 3(2 m-b) &=3 x \\ 2 m-b &=x . \end{aligned} $$ Since \((2 m-b)\) is an integer, this proves that 3 divides \((2 a+b)\) and hence, \((2 a+b) \equiv 0(\bmod 3)\). (b) Proposition. For each integer \(m, 5\) divides \(\left(m^{5}-m\right)\). Proof. Let \(m \in \mathbb{Z}\). We will prove that 5 divides \(\left(m^{5}-m\right)\) by proving that \(\left(m^{5}-m\right) \equiv 0(\bmod 5)\). We will use cases. For the first case, if \(m \equiv 0(\bmod 5),\) then \(m^{5} \equiv 0(\bmod 5)\) and, hence, \(\left(m^{5}-m\right) \equiv 0(\bmod 5)\) For the second case, if \(m \equiv 1(\bmod 5),\) then \(m^{5} \equiv 1(\bmod 5)\) and, hence, \(\left(m^{5}-m\right) \equiv(1-1)(\) mod 5\()\), which means that \(\left(m^{5}-m\right) \equiv\) \(0(\bmod 5)\) For the third case, if \(m \equiv 2(\bmod 5),\) then \(m^{5} \equiv 32(\bmod 5)\) and, hence, \(\left(m^{5}-m\right)=(32-2)(\bmod 5),\) which means that \(\left(m^{5}-m\right)=\) \(0(\bmod 5)\).

Short Answer

Expert verified
Short Answer: Proof for Proposition (a): Assume \(a, b \in \mathbb{Z}\) and \((a+2b) \equiv 0(\bmod 3)\). From this, find the integer \(m\) such that \(a+2b = 3m\). Express \(a\) in terms of \(m\) and \(b\) as \(a = 3m - 2b\). Then, show that there exists an integer \(x\) such that \(2a+b = 3x\). By substituting \(a\) and simplifying, we find that \(2m - b = x\). Since \((2m - b)\) is an integer, this proves \((2a+b) \equiv 0(\bmod 3)\). Proof for Proposition (b): To prove 5 divides \(\left(m^5 - m\right)\) for any integer \(m\), show that \(\left(m^5 - m\right) \equiv 0(\bmod 5)\) for various cases. For cases where \(m \equiv 0, 1,\) or \(2 (\bmod 5)\), we demonstrate that \(\left(m^5 - m\right) \equiv 0(\bmod 5)\). Thus, 5 divides \(\left(m^5 - m\right) \) for any integer \(m\).

Step by step solution

01

Assume \((a+2b) \equiv 0(\) mod 3\()\) and find the integer m such that \(a+2b = 3m\)

Since we're given that the condition \((a+2b) \equiv 0 (\bmod 3)\) holds for integers \(a\) and \(b\), this means that 3 divides \((a+2b)\). So there exists an integer \(m\) such that \(a+2b = 3m\).
02

Express \(a\) in terms of \(m\) and \(b\)

From the equation \(a+2b = 3m\), we can express \(a\) as \(a = 3m - 2b\).
03

Prove \((2a+b) \equiv 0(\bmod 3)\)

To prove that \((2a+b) \equiv 0(\bmod 3)\), we need to show that there exists an integer \(x\) such that \(2a+b = 3x\). Substitute \(a\) from Step 2 into this equation as follows: \[ \begin{aligned} 2(3m - 2b) + b &= 3x \\ 6m - 3b &= 3x \\ 3(2m - b) &= 3x \\ 2m - b &= x \end{aligned} \] Since \((2m - b)\) is an integer, we have successfully demonstrated that there exists an integer \(x\) such that \(2a + b = 3x\). Therefore, \((2a+b) \equiv 0(\bmod 3)\). For the second proposition:
04

Prove that \(\left(m^5 - m\right) \equiv 0(\bmod 5)\) using Cases

Our goal is to show that for any integer \(m\), 5 divides \(\left(m^5 - m\right)\). This can be proven by demonstrating that \(\left(m^5 - m\right) \equiv 0(\bmod 5)\) for various cases with different values of \(m\) modulo 5.
05

Case 1: If \(m \equiv 0(\bmod 5)\)

In this case, \(m^5 \equiv 0(\bmod 5)\), and thus \(\left(m^5 - m\right) \equiv 0(\bmod 5)\).
06

Case 2: If \(m \equiv 1(\bmod 5)\)

In this case, \(m^5 \equiv 1(\bmod 5)\), and thus \(\left(m^5 - m\right) \equiv (1-1)(\bmod 5)\), which means that \(\left(m^5 - m\right) \equiv 0(\bmod 5)\).
07

Case 3: If \(m \equiv 2(\bmod 5)\)

In this case, \(m^5 \equiv 32(\bmod 5)\), and thus \(\left(m^5 - m\right) \equiv (32-2)(\bmod 5)\), which means that \(\left(m^5 - m\right) \equiv 0(\bmod 5)\). Since each of the cases demonstrates that \(\left(m^5 - m\right) \equiv 0(\bmod 5)\) holds, this establishes that 5 divides \(\left(m^5 - m\right)\) for any integer \(m\).

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Key Concepts

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

Integer Division
Integer division is the process of dividing one integer by another, yielding a quotient and possibly a remainder. In mathematical notation, if you have two integers, say \( a \) and \( b \), and you're dividing \( a \) by \( b \), the quotient can be denoted as \( q \) and the remainder as \( r \). This is written as:\[a = bq + r\]where the remainder \( r \) satisfies \( 0 \leq r < |b| \).
This concept is useful in modular arithmetic, where we often focus solely on the remainder (modulus) part. For example, when we say \( a \equiv b \pmod{m} \), we mean the remainder of \( a \) divided by \( m \) is the same as the remainder of \( b \) divided by \( m \). Let鈥檚 break that down more:
  • Quotient (q): The number of times one number divides another without leftovers.
  • Remainder (r): What's left after all the whole number division is done.
Understanding integer division is crucial when working with divisibility rules and constructing proofs using modular arithmetic.
Proposition Proof
A proposition proof is a mathematical statement that establishes a truth by using logical steps and known facts. It involves demonstrating that if certain conditions are met, a particular conclusion necessarily follows. In other words, it's like building a logical staircase where each step must be correct and valid.
A good way to start understanding proposition proofs is to realize it generally involves:
  • Assumption: Start with an assumed statement or condition to be true.
  • Logical Steps: Use known theorems, definitions or previously proved results to establish each step.
  • Conclusion: Arrive at a new statement or conclusion, verifying the original proposition.
In our exercise, the proposition starts by assuming that \((a+2b) \equiv 0(\bmod 3)\) and proves logically that it leads to \((2a+b) \equiv 0(\bmod 3)\). This proof relies on manipulating the given equations and using integer properties to arrive at the correct logical steps. It鈥檚 important to clearly understand each progression to ensure precision in the conclusion.
Divisibility Rules
Divisibility rules are shortcuts or patterns that help determine if one number can be divided by another without a remainder. These rules aid in simplifying calculations, especially when working with large numbers.
Several common divisibility rules are:
  • By 2: A number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8).
  • By 3: A number is divisible by 3 if the sum of its digits is divisible by 3.
  • By 5: For a number to be divisible by 5, its last digit must be 0 or 5.
In our context, the exercise utilizes the divisibility by 3 and 5, where we see whether expressions like \((a+2b)\) and \((m^5-m)\) could be divided by these numbers without leaving a remainder, thus effectively explored through modular arithmetic.
Mathematical Reasoning
Mathematical reasoning is the process of using logical thinking to make sense of mathematical concepts and solve problems. It's like using a detective mindset to uncover truths and establish connections between seemingly unrelated ideas.
To demonstrate, let's consider how reasoning is applied in our solutions:
  • Understanding Assumptions: Clearly defining assumptions allows you to create a logical base.
  • Logical Progression: Every step builds upon the previous, reinforcing the validity of each deduction.
  • Conclusion: The final deduction ties back to the assumption, showing that the conclusion is in fact true.
Mathematical reasoning is not only vital in proving propositions; it also deepens comprehension and promotes critical thinking, helping students grasp complex mathematical concepts thoroughly.

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Most popular questions from this chapter

Using a Logical Equivalency. Consider the following proposition: Proposition. For all integers \(a\) and \(b,\) if 3 does not divide \(a\) and 3 does not divide \(b,\) then 3 does not divide the product \(a \cdot b\). (a) Notice that the hypothesis of the proposition is stated as a conjunction of two negations (" 3 does not divide \(a\) and 3 does not divide \(b\) "). Also, the conclusion is stated as the negation of a sentence ("3 does not divide the product \(a \cdot b\) "). This often indicates that we should consider using a proof of the contrapositive. If we use the symbolic form \((\neg Q \wedge \neg R) \rightarrow \neg P\) as a model for this proposition, what is \(P,\) what is \(Q,\) and what is \(R ?\) (b) Write a symbolic form for the contrapositive of \((\neg Q \wedge \neg R) \rightarrow \neg P\). (c) Write the contrapositive of the proposition as a conditional statement in English. We do not yet have all the tools needed to prove the proposition or its contrapositive. However, later in the text, we will learn that the following proposition is true. Proposition \(\mathbf{X}\). Let \(a\) be an integer. If 3 does not divide \(a\), then there exist integers \(x\) and \(y\) such that \(3 x+a y=1\) (d) i. Find integers \(x\) and \(y\) guaranteed by Proposition \(\mathrm{X}\) when \(a=5\). ii. Find integers \(x\) and \(y\) guaranteed by Proposition \(\mathrm{X}\) when \(a=2\). iii. Find integers \(x\) and \(y\) guaranteed by Proposition \(\mathrm{X}\) when \(a=-2\). (e) Assume that Proposition \(\mathrm{X}\) is true and use it to help construct a proof of the contrapositive of the given proposition. In doing so, you will most likely have to use the logical equivalency \(P \rightarrow(Q \vee R) \equiv\) \((P \wedge \neg Q) \rightarrow R\)

Are the following statements true or false? Justify your conclusions. (a) For each \(a \in \mathbb{Z},\) if \(a \equiv 2(\bmod 5),\) then \(a^{2} \equiv 4(\bmod 5)\). (b) For each \(a \in \mathbb{Z},\) if \(a^{2} \equiv 4(\bmod 5),\) then \(a \equiv 2(\bmod 5)\) (c) For each \(a \in \mathbb{Z}, a \equiv 2(\bmod 5)\) if and only if \(a^{2} \equiv 4(\bmod 5)\).

Evaluation of proofs This type of exercise will appear frequently in the book. In each case, there is a proposed proof of a proposition. However, the proposition may be true or may be false. \- If a proposition is false, the proposed proof is, of course, incorrect. In this situation, you are to find the error in the proof and then provide a counterexample showing that the proposition is false. \- If a proposition is true, the proposed proof may still be incorrect. In this case, you are to determine why the proof is incorrect and then write a correct proof using the writing guidelines that have been presented in this book. \- If a proposition is true and the proof is correct, you are to decide if the proof is well written or not. If it is well written, then you simply must indicate that this is an excellent proof and needs no revision. On the other hand, if the proof is not well written, then you must then revise the proof by writing it according to the guidelines presented in this text.(a) Proposition. If \(m\) is an even integer, then \((5 m+4)\) is an even integer. \mathrm{\\{} \text { Proof. We see that } 5 m + 4 = 1 0 n + 4 \(=2(5 n+2)\). Therefore, \((5 m+4)\) is an even integer. (b) Proposition. For all real numbers \(x\) and \(y,\) if \(x \neq y, x>0,\) and \(y>0,\) then \(\frac{x}{y}+\frac{y}{x}>2\) \mathrm{\\{} P r o o f . ~ S i n c e ~ \(x\) and \(y\) are positive real numbers, \(x y\) is positive and we can multiply both sides of the inequality by \(x y\) to obtain $$ \begin{aligned} \left(\frac{x}{y}+\frac{y}{x}\right) \cdot x y &>2 \cdot x y \\ x^{2}+y^{2} &>2 x y \end{aligned} $$ By combining all terms on the left side of the inequality, we see that \(x^{2}-2 x y+y^{2}>0\) and then by factoring the left side, we obtain \((x-y)^{2}>0 .\) Since \(x \neq y,(x-y) \neq 0\) and so \((x-y)^{2}>0 .\) This proves that if \(x \neq y, x>0,\) and \(y>0,\) then \(\frac{x}{y}+\frac{y}{x}>2\) (c) Proposition. For all integers \(a, b,\) and \(c,\) if \(a \mid(b c),\) then \(a \mid b\) or \(a \mid c\). \mathrm{\\{} P r o o f . ~ W e ~ a s s u m e ~ t h a t ~ \(a, b,\) and \(c\) are integers and that \(a\) divides \(b c\). So, there exists an integer \(k\) such that \(b c=k a\). We now factor \(k\) as \(k=m n,\) where \(m\) and \(n\) are integers. We then see that $$ b c=m n a . $$ This means that \(b=m a\) or \(c=n a\) and hence, \(a \mid b\) or \(a \mid c\). (d) Proposition. For all positive integers \(a, b,\) and \(c,\left(a^{b}\right)^{c}=a^{\left(b^{c}\right)}\). This proposition is false as is shown by the following counterexample: If we let \(a=2, b=3,\) and \(c=2,\) then $$ \begin{aligned} \left(a^{b}\right)^{c} &=a^{\left(b^{c}\right)} \\ \left(2^{3}\right)^{2} &=2^{\left(3^{2}\right)} \\ 8^{2} &=2^{9} \\ 64 & \neq 512 \end{aligned} $$

(a) Prove that for each integer \(a\), if \(a \neq 0(\bmod 7),\) then \(a^{2} \neq 0(\bmod 7)\). (b) Prove that for each integer \(a\), if 7 divides \(a^{2}\), then 7 divides \(a\). (c) Prove that the real number \(\sqrt{7}\) is an irrational number.

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).
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