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

Consider the partition \(P=\\{\\{0\\},\\{-1,1\\},\\{-2,2\\},\\{-3,3\\},\\{-4,4\\}, \ldots\\}\) of \(\mathbb{Z} .\) Describe the equivalence relation whose equivalence classes are the elements of \(P\).

Short Answer

Expert verified
The equivalence relation \(R\) corresponding to the partition \(P\) can be defined as follows: for any two integers \(a\) and \(b\), \(aRb\) if and only if \(a = b\) or \(a = -b\), or both are zero. This relation is reflexive, symmetric and transitive, hence it is an equivalence relation. The equivalence classes of \(R\) are the subsets of \(P\).

Step by step solution

01

Define the equivalence relation based on the partition P

An equivalence relation can be defined by identifying how two elements are related in each subset of the partition \(P\). In this case, each subset pairs an integer with its negation, or contains zero. The equivalence relation \(R\) can therefore be defined as: \((a, b) \in R\) if and only if \(a = b\), \(a = -b\), or both \(a\) and \(b\) are zero.
02

Verify the reflexivity of the equivalence relation

A relation is reflexive if every element is related to itself. In our case, each integer \(a\) satisfies \(a = a\), and therefore satisfies the condition for \(R\). Zero is also related to itself, so the equivalence relation is reflexive.
03

Verify the symmetry of the equivalence relation

A relation is symmetric if \(aRb\) implies that \(bRa\). In our case, if \(a = b\) or \(a = -b\), then obviously \(b = a\) and \(b = -a\). Therefore, the equivalence relation is symmetric.
04

Verify the transitivity of the equivalence relation

A relation is transitive if \(aRb\) and \(bRc\) implies that \(aRc\). In this case, given that only absolute values or negatives matter, it's clear that \(a = c\) or \(a = -c\) as soon as \(a = b\) or \(a = -b\) and \(b = c\) or \(b = -c\). So, the equivalence relation is also transitive.
05

Conclude the equivalence relation

As \(R\) is reflexive, symmetric, and transitive, it is indeed an equivalence relation. The equivalence relation corresponding to the partition \(P\) can hence be defined as \(aRb\) if and only if \(a = b\) or \(a = -b\) or both are zero. If \(a\) and \(b\) are in the same subset of \(P\), then \(aRb\). Conversely, if \(aRb\), then \(a\) and \(b\) are in the same subset. Therefore, we can say that the equivalence classes of the relation \(R\) are the subsets of the partition \(P\).

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