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I dont know how to start this proof? Also, our professor did not explain equivalence classes fully so I am not understanding them very well.

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  • $\begingroup$ What do you know about equivalence classes? $\endgroup$ – Asaf Karagila May 27 '14 at 19:27
  • $\begingroup$ An equivalence relation needs to sattisfy reflexitivity, symmetry and transitivity, so you need to check whether those conditions hold for this relation. en.wikipedia.org/wiki/Equivalence_class. $\endgroup$ – Marc May 27 '14 at 19:28
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You need to prove that $\sim$ is reflexive, symmetric, and transitive on $\mathbb Z$.

Then, an equivalence class of $a \in \mathbb Z$ is defined as $$[a] = \{b\mid b \in \mathbb Z \text { and } a\sim b\}$$: That is, $[a]$ is the class containing $a$ and all elements in $\mathbb Z$ that are related to $a$.

Hint: There are two equivalence classes that together, partition $\mathbb Z$.

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  • $\begingroup$ But the relation is not reflexive because when you subtract a number from itself you get 0? $\endgroup$ – user02398409 May 27 '14 at 19:35
  • $\begingroup$ It is reflexive. Note that zero is an even number. $\endgroup$ – amWhy May 27 '14 at 19:54
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    $\begingroup$ @gatorgurl4 Did you get this question figured out? You might want to think about accepting a helpful answer to each of your questions. You can accept only one answer per question, but you get two reputation points for each accepted answer. Just click on the grey $\checkmark$ to the left of the answer you'd like to accept. $\endgroup$ – amWhy May 28 '14 at 15:36
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In order to prove some relation $\sim$ is an equivalence relation on some set $A$ we need to show that for $a,b,c \in A$ we have the following properties:

  • $a \sim a$
  • $a \sim b \implies b \sim a$
  • $a \sim b, b \sim c \implies a \sim c$

Now I will give hints to to each step for your equivalence relation:

  • Does $2$ divide $0$?
  • If $x$ is even, is $-x$ even?
  • Since we know that $a-b, b-c$ are even, we know that $a-b = 2n, b-c=2k$ for some $k,n$

Now the equivalence class of some element $a$ are defined as:

$[a] = \{ b \mid a \sim b\}$

So think about the different elements of $\mathbb{Z}$, and what their equivlence classes will be. Hint: there are only 2 distinct classes.

Hope this helps

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  • $\begingroup$ What I don't understand is how a can NOT be itself? I get completely how a can NOT be b (symmetrical) and if a is NOT b and b is NOT c THEN a is NOT c (transitive), but the reflexive relation baffles me. a is not a statement, it's an element of a set, so a ~ a doesn't make any sense...? $\endgroup$ – Benjamin R Aug 11 '15 at 9:07
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    $\begingroup$ @BenjaminR I don't quite follow your question. $\sim$ can be defined in many different ways, even non-symmetric ways. For example, consider $\sim$ to be $<$ $\endgroup$ – DanZimm Aug 12 '15 at 1:39
  • $\begingroup$ Hi Dan, I am new to logical symbols and notation. Whatever Discrete math book or reference I look at uses ~ to mean negation (or literally "not" some statement or value). That's why I find the equivalence relation questions on Math.SE and the reference on Wikipedia to equivalence classes confusing. It seems like ~ is literally used to represent the equivalence relation, am I understanding that right? $\endgroup$ – Benjamin R Aug 12 '15 at 2:24
  • $\begingroup$ Okay, I couldn't see the MathJax in your comment before (I was using the iOS SE app and it doesn't render it by default). I think you are essentially saying "Consider the relation $a\:\rm R\: \it b$ if $a<b$. Is $\rm R$ an equivalence relation?" $\endgroup$ – Benjamin R Aug 12 '15 at 2:50
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    $\begingroup$ @BenjaminR correct that's what I was saying in my previous comment - we are simply starting with a relation, and we want to prove that its an equivalence relation $\endgroup$ – DanZimm Aug 12 '15 at 3:01

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