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Chapter 1: Problem 29

Solve the given equation or indicate that there is no solution. $$2 x+3=2 \text { in } \mathbb{Z}_{5}$$

Short Answer

Expert verified
The solution is \(x = 2\) in \(\mathbb{Z}_5\).

Step by step solution

01

Understand the Problem

We need to solve the equation \(2x + 3 = 2\) in \(\mathbb{Z}_5\). Here, \(\mathbb{Z}_5\) denotes the set of integers \(\{0, 1, 2, 3, 4\}\) with arithmetic done modulo 5.
02

Simplify the Equation

First, we simplify the equation by subtracting 3 from both sides:\[2x + 3 - 3 = 2 - 3 \]which gives us:\[2x = -1 \] In modulo 5, \(-1\) is equivalent to \(4\), because:\[-1 \equiv 4 \mod 5\] Thus the equation becomes \(2x \equiv 4 \mod 5\).
03

Find the Multiplicative Inverse of 2 in \(\mathbb{Z}_5\)

To solve \(2x \equiv 4 \mod 5\), we need the multiplicative inverse of 2 modulo 5. The inverse \(y\) satisfies \(2y \equiv 1 \mod 5\). Testing values of \(y\) from 1 to 4, we find:- \(y = 3\) works since \(2 \times 3 = 6 \equiv 1 \mod 5\). Thus, the inverse is \(3\).
04

Solve for x

Multiply both sides of the equation \(2x \equiv 4 \mod 5\) by the inverse of 2, which is 3:\[(3)2x \equiv (3)4 \mod 5\]This simplifies to:\[6x \equiv 12 \mod 5\]Since \(6 \equiv 1 \mod 5\) and \(12 \equiv 2 \mod 5\), the equation becomes:\[x \equiv 2 \mod 5\]
05

Conclusion

The solution to the equation is \(x = 2\) in \(\mathbb{Z}_5\).

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Key Concepts

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

Integers modulo 5
When we talk about integers modulo 5, we refer to a mathematical system where we only care about the remainder when numbers are divided by 5. In this system, written as \( \mathbb{Z}_5 \), we use only the integers 0, 1, 2, 3, and 4.
This means if a number exceeds 4, it wraps around starting again from 0. Arithmetic operations like addition, subtraction, and multiplication are performed as usual, but after getting the result, we divide by 5 and take the remainder.
For instance:
  • 5 modulo 5 is 0 because 5 divided by 5 is 1 with a remainder of 0.
  • 7 modulo 5 is 2 because 7 divided by 5 is 1 with a remainder of 2.
  • -1 modulo 5 is 4 because -1 + 5 = 4.
Understanding this modular system is crucial when dealing with equations, like in our exercise, where solutions must fit within this set.
Multiplicative Inverse
The concept of a multiplicative inverse is similar to finding the reciprocal of a number but within a modular system. In mathematical terms, the multiplicative inverse of a number \( a \) modulo \( n \) is a number \( b \) such that:\[ ab \equiv 1 \mod n \]
In the context of \( \mathbb{Z}_5 \), we needed the inverse of 2.
To find this, we looked for a number such that when multiplied by 2, the result is congruent to 1 under modulo 5.
By checking possible values from our set {0, 1, 2, 3, 4}, we discovered that:
  • When \( 2 \times 3 = 6 \equiv 1 \mod 5 \), 3 is the multiplicative inverse of 2 in \( \mathbb{Z}_5 \).
Finding the inverse allows us to "divide" in modular arithmetic. This is key for solving equations where multiplication by a number is involved.
Solving linear equations modulo n
Solving linear equations modulo \( n \) involves finding a value for the unknown variable that satisfies the equation, similar to solving regular linear equations, but within the confines of modular arithmetic.
The primary difference is that, instead of a solution being any number, it must be a member of the set \( \{0, 1, 2, \ldots, n-1\} \).
Let's see how this works through the example:
  • The original equation was \( 2x + 3 \equiv 2 \mod 5 \).
  • First, we simplified it to \( 2x \equiv -1 \mod 5 \).
  • Converting \(-1\) to a positive equivalent in \( \mathbb{Z}_5 \) gave us \( 2x \equiv 4 \mod 5 \).
  • Next, we multiplied throughout by the multiplicative inverse of 2, which is 3, to isolate \( x \).
  • This simplified the equation to \( x \equiv 2 \mod 5 \).
This solution, \( x = 2 \), means that substituting \( x \) with 2 satisfies the equation when computed in the integers modulo 5.

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