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Chapter 1: Problem 7

Perform the indicated calculations. $$2+2+2 \text { in } \mathbb{Z}_{3}$$

Short Answer

Expert verified
0 in \(\mathbb{Z}_{3}\).

Step by step solution

01

Understand the Problem

We need to calculate the sum of \(2+2+2\) in the integers modulo 3, denoted by \(\mathbb{Z}_{3}\). This means we perform the addition in such a way that we take the remainder when divided by 3.
02

Add the Numbers Normally

Start by adding the numbers as if there were no modulus: \(2 + 2 + 2 = 6\).
03

Apply Modulus Operation

To find the sum in \(\mathbb{Z}_{3}\), we divide the result by 3 and take the remainder. \(6 \div 3 = 2\) with a remainder of 0, so the result is \(0\).
04

Verification

Double-check the calculations by considering another approach: add \(2\) three times in modulo 3. First, \(2 + 2 = 4\), then apply modulus operation to get \(4 \equiv 1 \pmod{3}\). Add \(2\) again: \(1 + 2 = 3\), and \(3 \equiv 0 \pmod{3}\), confirming the result.

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Key Concepts

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

Integers modulo n
When we say "integers modulo \(n\)," we refer to a system of arithmetic for the integers where numbers reset at \(n\). Imagine a clock that only has \(n\) hours; once you reach \(n\), it goes back to zero. For example:
  • If \(n = 3\), we have integers mod 3 which means we only use numbers 0, 1, and 2.
  • The arithmetic operations wrap around after these numbers.
    • This reminds us of the days of the week cycling every 7 days!
This kind of arithmetic is crucial in various fields, including computer science and cryptography, due to its predictable pattern and simplicity.
Modulus operation
The modulus operation is all about finding the remainder when one number is divided by another. In mathematical notation, it is often shown as \(a \mod n\). This operation gives us:
  • The remainder of \(a\) when divided by \(n\).
  • The range of results is always between 0 and \(n-1\).
For instance, when applying the modulus operation to 6 in the context of 3, we have \(6 \mod 3 = 0\) because dividing 6 by 3 leaves a remainder of 0. The modulus operation is simple yet powerful, providing a method to handle calculations that naturally repeat or cycle.
Remainder
The remainder is what is left after division. It's the part of the number that does not evenly fit into our divisor. Consider how division works:
  • When we divide 6 by 3, for instance, it fits perfectly, and thus the remainder is 0.
  • In cases where division isn't perfect, the remainder is what's left over.
In our exercise, the remainder after dividing 6 by 3 was 0, signaling a perfect division. Understanding remainders is key in modular arithmetic since it directly tells us the end result of our operation within the modulus system. This concept ensures everything stays within the range defined by the modulus.

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Most popular questions from this chapter

Perform the indicated calculations. $$[2,0,3,2] \cdot([3,1,1,2]+[3,3,2,1]) \text { in } \mathbb{Z}_{4}^{4} \text { and in } \mathbb{Z}_{5}^{4}$$

Perform the indicated calculations. $$8(6+4+3) \text { in } \mathbb{Z}_{9}$$

(a) Prove that \(\|\mathbf{u}+\mathbf{v}\|=\|\mathbf{u}-\mathbf{v}\|\) if and only if \(\mathbf{u}\) and \(\mathbf{v}\) are orthogonal. (b) Draw a diagram showing \(\mathbf{u}, \mathbf{v}, \mathbf{u}+\mathbf{v},\) and \(\mathbf{u}-\mathbf{v}\) in \(\mathbb{R}^{2}\) and use \((a)\) to deduce a result about parallelograms.

Perform the indicated calculations. $$2(2+1+2) \text { in } \mathbb{Z}_{3}$$

The Cauchy-Schwarz Inequality \(|\mathbf{u} \cdot \mathbf{v}| \leq\|\mathbf{u}\|\|\mathbf{v}\|\) is equivalent to the inequality we get by squaring both sides: \((\mathbf{u} \cdot \mathbf{v})^{2} \leq\|\mathbf{u}\|^{2}\|\mathbf{v}\|^{2}\) (a) \(\operatorname{In} \mathbb{R}^{2},\) with \(\mathbf{u}=\left[\begin{array}{l}u_{1} \\ u_{2}\end{array}\right]\) and \(\mathbf{v}=\left[\begin{array}{l}v_{1} \\ v_{2}\end{array}\right],\) this becomes \\[ \left(u_{1} v_{1}+u_{2} v_{2}\right)^{2} \leq\left(u_{1}^{2}+u_{2}^{2}\right)\left(v_{1}^{2}+v_{2}^{2}\right) \\] Prove this algebraically. [Hint: Subtract the left-hand side from the right- hand side and show that the difference must necessarily be nonnegative. (b) Prove the analogue of \((a)\) in \(\mathbb{R}^{3}\).
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