Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Define a relation \(R\) on \(\mathbb{Z}\) as \(x R y\) if and only if \(3 x-5 y\) is even. Prove \(R\) is an equivalence relation. Describe its equivalence classes.

Short Answer

Expert verified
The relation \(R\) is an equivalence relation because it's reflexive, symmetric, and transitive. The equivalence classes are sets of integers that leave the same remainder when divided by \(5\).

Step by step solution

01

Prove Reflexivity

An equivalence relation needs to be reflexive. This means that every element must be related to itself. In other words, for all \(x\) in \(\mathbb{Z}\), you should check that \(x R x\) holds true. Which means \(3x - 5x\) should be an even number, and since \(3x - 5x = -2x\), it is even as any integer multiplied by 2 is even.
02

Prove Symmetry

An equivalence relation also needs to be symmetric. This means if \(x R y\), then \(y R x\). You need to prove that if \(3x - 5y\) is even, then \(3y - 5x\) is also even. Since \(3x - 5y\) is even, it can be written as \(2m\) for some integer \(m\). When you rearrange this to get \(3y - 5x\), you find that it is also even.
03

Prove Transitivity

Finally, an equivalence relation needs to be transitive. This means if \(x R y\) and \(y R z\), then \(x R z\). Start by assuming that \(3x - 5y\) and \(3y - 5z\) are both even, which means they can be written as \(2m\) and \(2n\) respectively for some integers \(m\) and \(n\). This implies that \(3x - 5z = 2(m + n)\), therefore, \(x R z\)
04

Describe The Equivalence Classes

The last part of the exercise is to describe the equivalence classes. The equivalence class of an element - say \(a\) - under a relation \(R\) is the set of all elements that are related to \(a\). In this case, since \(x R y\) if and only if \(3x -5y\) is even, then \(x\) and \(y\) belong to the same equivalence class if and only if \(3x\) and \(3y\) leave the same remainder when divided by \(5\). This is because the difference of these two will be divisible by \(5\), hence, \(5\) times an integer, so it will be even. Such equivalence classes would be all the integers which leave the same remainder when divided by \(5\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free