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Chapter 3: Problem 5

(a) Let \(n \in \mathbb{N}\) and let \(a \in \mathbb{Z}\). Explain why \(n\) divides \(a\) if and only if \(a \equiv 0(\bmod n)\) (b) Let \(a \in \mathbb{Z}\). Explain why if \(a \neq 0(\bmod 3),\) then \(a \equiv 1(\bmod 3)\) or \(a \equiv 2(\bmod 3)\) (c) Is the following proposition true or false? Justify your conclusion. For each \(a \in \mathbb{Z},\) if \(a \neq 0(\bmod 3),\) then \(a^{2} \equiv 1(\bmod 3)\).

Short Answer

Expert verified
In conclusion: a) The statement "n divides a if and only if a 鈮� 0 (mod n)" is true. This is because it has been shown that the divisibility of a and congruence to 0 modulo n are equivalent, as their definitions imply each other. b) If a 鈮� 0 (mod 3), then a must be congruent to either 1 or 2 modulo 3, as these are the only remaining congruence classes modulo 3. c) The proposition "For each a in integers, if a 鈮� 0 (mod 3), then \(a^2 \equiv 1 \pmod{3}\)" is true, as demonstrated by testing the cases where a 鈮� 0 (mod 3).

Step by step solution

01

Definition of divisibility and congruence

Divisibility: By definition, an integer n divides an integer a if there exists an integer k such that a = nk. In other words, a is divisible by n if a can be written as a multiple of n. Congruence: Two integers a and b are congruent modulo n if their difference (a - b) is divisible by n. We denote this as a 鈮� b (mod n).
02

Show n divides a implies a 鈮� 0 (mod n)

If n divides a, there is an integer k such that a = nk. Therefore, a - 0 = nk, which implies (a - 0) is divisible by n. According to the definition of congruence, we have a 鈮� 0 (mod n).
03

Show a 鈮� 0 (mod n) implies n divides a

If a 鈮� 0 (mod n), then by definition, n divides (a - 0) or n divides a. Hence, a is divisible by n. Since both implications have been proven, we conclude that n divides a if and only if a 鈮� 0 (mod n). b) Congruence classes modulo 3
04

Definition of congruence classes

Congruence classes modulo n are the sets of integers that give the same remainder when divided by n. There are n possible congruence classes modulo n: 0, 1, 2, ..., n-1. In this case, we have the following congruence classes modulo 3: 0, 1, and 2.
05

Examine the cases

If a is any integer, the possible remainders when it is divided by 3 are 0, 1, and 2. Thus, a can belong to one of the following congruence classes: 1. a 鈮� 0 (mod 3) 2. a 鈮� 1 (mod 3) 3. a 鈮� 2 (mod 3) Now, the question states that a 鈮� 0 (mod 3), which means a cannot belong to the first congruence class. We are left with the possibilities that a 鈮� 1 (mod 3) or a 鈮� 2 (mod 3). c) Proposition evaluation
06

Test the proposition

The proposition states: "For each a in integers, if a 鈮� 0 (mod 3), then a^2 鈮� 1 (mod 3)." To evaluate the proposition, we can test the cases where a 鈮� 0 (mod 3) as derived in part (b). Case 1: a 鈮� 1 (mod 3) \(a^2 \equiv 1^2 \equiv 1 \pmod{3}\) Case 2: a 鈮� 2 (mod 3) \(a^2 \equiv 2^2 \equiv 4 \equiv 1 \pmod{3}\)
07

Conclusion

For both cases of a not being congruent to 0 modulo 3, we can see that \(a^2 \equiv 1 \pmod{3}\). Therefore, the proposition is true and can be justified by testing the cases where a 鈮� 0 (mod 3).

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Key Concepts

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

Modular Arithmetic
Modular arithmetic is a system of arithmetic for integers, where numbers wrap around upon reaching a certain value鈥攖he modulus. Think of it as the arithmetic of clock where, after 12 hours, the cycle starts again. In formal terms, for any integer a, modulus n, and integer b, we say that a is congruent to b modulo n (denoted as a 鈮� b (mod n)) if n divides the difference a - b. This means that a and b leave the same remainder when divided by n.

For example, let's take 17 and 5 with a modulus of 6. The remainders when dividing these numbers by 6 are 5 and 5, respectively. Hence, 17 and 5 are congruent modulo 6. This can be a powerful tool for solving a variety of problems in number theory and cryptography.
Integer Division
Integer division is the process of dividing one integer by another and rounding down to the nearest integer if necessary. The result of this division is known as the quotient. An important aspect of integer division in the context of divisibility is the remainder. When an integer a is divided by another integer n, the division yields a quotient q and a remainder r, such that a = nq + r and 0 鈮� r < n. In terms of congruence, we can relate this to the statement that a is congruent to r modulo n.

In the context of the exercise, this is why if n divides a, then a is congruent to 0 modulo n, because the remainder is zero.
Congruence Classes
A congruence class is a complete set of numbers that are all congruent to one another modulo n. In other words, all members of a congruence class share the same remainder when divided by n. The set of all congruence classes modulo n forms a complete system that covers all the integers with no overlaps.

Consider the modulus 3 from the exercise. There are three congruence classes: those numbers that, when divided by 3, leave a remainder of 0, 1, or 2. These classes are represented by 3k, 3k+1, and 3k+2 respectively, for any integer k. The exercise demonstrates that if an integer does not leave a remainder of 0 when divided by 3 (a 鈮� 0 (mod 3)), it must belong to one of the other two congruence classes: either 3k+1 or 3k+2, which correspond to the remainders 1 and 2 respectively.
Proofs in Number Theory
Proofs are the backbone of number theory, serving to establish the truthfulness of mathematical statements. These proofs often rely on understanding properties of integers, such as divisibility and congruence relationships. In the exercise, a simple proof is constructed to show a given proposition about squares of integers and their congruence modulo 3.

The approach taken is by case analysis. We examine all cases that stem from the given conditions鈥攊n this case, the integers that are not congruent to 0 modulo 3, which narrows it down to two cases: those congruent to 1 and those congruent to 2. By squaring these representatives and applying the modulo operation, we can establish that, indeed, all such squares are congruent to 1 modulo 3. This logical method of breaking down the problem and dealing with each piece individually often simplifies complex problems in number theory and leads to elegant proofs.

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Most popular questions from this chapter

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.

One of the most famous unsolved problems in mathematics is a conjecture made by Christian Goldbach in a letter to Leonhard Euler in 1742. The conjecture made in this letter is now known as Goldbach's Conjecture. The conjecture is as follows: Every even integer greater than 2 can be expressed as the sum of two (not necessarily distinct) prime numbers. Currently, it is not known if this conjecture is true or false. (a) Write \(50,142,\) and 150 as a sum of two prime numbers. (b) Prove the following: If Goldbach's Conjecture is true, then every integer greater than 5 can be written as a sum of three prime numbers. (c) Prove the following: If Goldbach's Conjecture is true, then every odd integer greater than 7 can be written as a sum of three odd prime numbers.

Prove the following proposition: Let \(a\) and \(b\) be integers with \(a \neq 0\). If \(a\) does not divide \(b\), then the equation \(a x^{3}+b x+(b+a)=0\) does not have a solution that is a natural number. Hint: It may be necessary to factor a sum of cubes. Recall that $$ u^{3}+v^{3}=(u+v)\left(u^{2}-u v+v^{2}\right) $$

The Last Two Digits of a Large Integer. Notice that \(7,381,272 \equiv 72(\) mod 100\()\) since \(7,381,272-72=7,381,200\). which is divisible by 100 . In general, if we start with an integer whose decimal representation has more than two digits and subtract the integer formed by the last two digits, the result will be an integer whose last two digits are \(00 .\) This result will be divisible by 100 . Hence, any integer with more than 2 digits is congruent modulo 100 to the integer formed by its last two digits. (a) Start by squaring both sides of the congruence \(3^{4} \equiv 81\) (mod 100 ) to prove that \(3^{8}=61\) (mod 100 ) and then prove that \(3^{16}=21\) (mod 100 ). What does this tell you about the last two digits in the decimal representation of \(3^{16} ?\) (b) Use the two congruences in Part (24a) and laws of exponents to determine \(r\) where \(3^{20} \equiv r(\bmod 100)\) and \(r \in \mathbb{Z}\) with \(0 \leq r<100 .\) What does this tell you about the last two digits in the decimal representation of \(3^{20} ?\) (c) Determine the last two digits in the decimal representation of \(3^{400}\). (d) Determine the last two digits in the decimal representation of \(4^{804}\). Hint: One way is to determine the "mod 100 values" for \(4^{2}, 4^{4}, 4^{8}\), \(4^{16} \cdot 4^{32}, 4^{64},\) and so on. Then use these values and laws of exponents to determine \(r,\) where \(4^{804} \equiv r(\) mod 100\()\) and \(r \in \mathbb{Z}\) with \(0 \leq r<\) \(100 .\) (e) Determine the last two digits in the decimal representation of \(3^{3356}\). (f) Determine the last two digits in the decimal representation of \(7^{403}\).

Prove each of the following: (a) For each nonzero real number \(x,\left|x^{-1}\right|=\frac{1}{|x|}\). (b) For all real numbers \(x\) and \(y,|x-y| \geq|x|-|y|\) Hint: An idea that is often used by mathematicians is to add 0 to an expression "intelligently". In this case, we know that \((-y)+y=0\). Start by adding this "version" of 0 inside the absolute value sign of \(|x|\). (c) For all real numbers \(x\) and \(y, \| x|-| y|| \leq|x-y|\).
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