Question

a. Show that if \(3 a=0\) in \(\mathbb{Z}_{10},\) then necessarily \(a=0\) in \(\mathbb{Z}_{10}\) b. Show that \(2 a=0\) in \(\mathbb{Z}_{10}\) holds in \(\mathbb{Z}_{10}\) if and only if \(a=0\) or \(a=5\)

Short answer

a. Only \(a = 0\) satisfies \(3a = 0 \) in \(\mathbb{Z}_{10}\). b. \(a = 0\) or \(a = 5\) satisfies \(2a = 0\) in \(\mathbb{Z}_{10}\).

Step-by-step solution

Understand the problem

We need to understand what operations in \( \mathbb{Z}_{10} \) mean. \( \mathbb{Z}_{10} \) is the set of integers \( \{0, 1, 2, ..., 9\} \) with addition and multiplication performed modulo 10. This means operations wrap around after reaching 10.

Solve part (a)

For part (a), we start with the equation \(3a = 0\) in \(\mathbb{Z}_{10}\). We need to find for which values of \(a\) the result of multiplying by 3 is congruent to 0 modulo 10. We test values from 0 to 9.1. \(3 \times 0 = 0\)2. \(3 \times 1 = 3\)3. \(3 \times 2 = 6\)4. \(3 \times 3 = 9\)5. \(3 \times 4 = 12 \equiv 2 \pmod{10}\)6. \(3 \times 5 = 15 \equiv 5 \pmod{10}\)7. \(3 \times 6 = 18 \equiv 8 \pmod{10}\)8. \(3 \times 7 = 21 \equiv 1 \pmod{10}\)9. \(3 \times 8 = 24 \equiv 4 \pmod{10}\)10. \(3 \times 9 = 27 \equiv 7 \pmod{10}\)We find that only when \(a = 0\), we have \(3a = 0\mod{10}\). Therefore, \(a = 0\) is the only solution to \(3a = 0\) in \(\mathbb{Z}_{10}\).

Solve part (b)

For part (b), we start with the equation \(2a = 0\) in \(\mathbb{Z}_{10}\). Similar to part (a), we multiply \(a\) by 2 and find the results modulo 10, from 0 to 9.1. \(2 \times 0 = 0\)2. \(2 \times 1 = 2\)3. \(2 \times 2 = 4\)4. \(2 \times 3 = 6\)5. \(2 \times 4 = 8\)6. \(2 \times 5 = 10 \equiv 0 \pmod{10}\)7. \(2 \times 6 = 12 \equiv 2 \pmod{10}\)8. \(2 \times 7 = 14 \equiv 4 \pmod{10}\)9. \(2 \times 8 = 16 \equiv 6 \pmod{10}\)10. \(2 \times 9 = 18 \equiv 8 \pmod{10}\)The result \(2a = 0 \mod{10}\) holds when \(a = 0\) and \(a = 5\). Therefore, the solutions are \(a = 0\) or \(a = 5\).

Key concepts

Congruence Relations

Congruence relations are at the heart of modular arithmetic, a key concept in number theory. It involves expressions like \( a \equiv b \pmod{n} \), meaning that when \( a \) is divided by \( n \), it leaves the same remainder as \( b \). In simpler terms, two numbers are congruent modulo \( n \) if they differ by a multiple of \( n \). This allows us to handle integers within a fixed range, wrapping around once the end is reached, much like the hands of a clock.

For example, in the case of the modular equation \(3a = 0 \mod 10\), we seek values of \( a \) such that a multiple of 3 results in a number that is divisible by 10. In learning and applying congruence, remember that it's more about personalizing operations to a set limit—like a circular track that resets on reaching a boundary. Exploring these concepts can deepen understanding of how numbers behave under rotation using mod arithmetic.

Modulo Operation

The modulo operation is crucial for performing calculations under modular arithmetic. It can be thought of as the remainder of a division operation. When you compute \( a \mod n \), you are essentially finding what remains after taking away as many full \( n \)'s from \( a \) as possible. This operation is widely used in programming and cryptography, providing a way to keep numbers within a specific range.

For the problems given, consider how \( 3 \times a \mod 10 = 0 \) and \( 2 \times a \mod 10 = 0 \) translate to real calculations. For \( 3a = 0 \mod 10\), only \( a = 0 \) works because the result of \( 3 \times a \) does not loop all numbers back to zero except when \( a = 0 \). Similarly, \( 2a = 0 \mod 10\) is true when \( a = 0 \) or \( a = 5 \). Understanding what each iteration via multiplication results in—or cycles through—helps master the concept of modulo.

Set of Integers \(\mathbb{Z}_{10}\)

The set \(\mathbb{Z}_{10}\) consists of all integers from 0 to 9. It is a cyclic group under addition, meaning operations within this set "wrap around" once they hit 10. In practical terms, when you add 1 to 9 in this set, instead of getting 10, you end up back at 0, similar to completing a lap in a race.

This characteristic is critical for solving modular equations, like those presented in the original exercise. To verify a solution, calculations are performed within this set by testing each value from 0 through 9 in the equation. Only those that satisfy the given congruences, such as \(3a = 0\) only having \(a = 0\) as a valid solution, fit the mold formed by \(\mathbb{Z}_{10}\). Similarly, the equation \(2a = 0\) is solved by verifying which multiples in this limited set align perfectly at zero modulo 10, discovering that \(a = 0\) or \(a = 5\) are the only solutions. Familiarizing yourself with this finite, looping set makes operations in modular arithmetic intuitive and clear.