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Can someone give me an example for this question on reflexivity please. How is this reflexive? Can you show an example with numbers from the set?

Question 6
Let $X = \{0,1,2,3,\dotsc, 9\}$. Define a relation $\mathcal R$ from $X$ by "$x$ is related to $y$ if $x$ and $y$ gives the same remainder on division by $3$".
(a) Show that $\mathcal R$ is an equivalence relation.
Anwser:
(a) $\mathcal R$ is clearly both reflexive and symmetric, for every number has the same (unique) remainder on division by $3$ as itself and if $a$ has the same remainder on division by $3$ as does $b$, then $b$ has the same remainder on division by $3$ as $a$ also. Similarly $\mathcal R$ is transitive, for if $a\mathcal R b$ and $b\mathcal R c$, then $a$, $b$, and $c$ all have the same remainder on division by $3$, so $a\mathcal R c$ also.

Em.
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Sanone
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    I vote to close this question which is about a definition that can be found everywhere, for example in (https://en.wikipedia.org/wiki/Reflexive_relation). – Jean Marie May 08 '17 at 08:22
  • Please type out all images. Formatting tips here. – Em. May 08 '17 at 08:28
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    JeanMarie is the police force I guess. Do you know I have all of my study guides and sometimes some questions hard to grasp? If you can know the answers just by reading the definition then we don't need math forums. Also I am asking for example numbers from the set 0,1,2,3...9. – Sanone May 08 '17 at 08:38
  • I guess you need work in the combination of reading and interpreting a definition. It takes a while, but really becomes easy with practice. Let's see. If you divide, say 4, by 3 you get some remainder. Now what happens if you start over and divide 4 by 3 again. Do you get the same remainder? Does this happen to numbers other than 4 as well? If it does, then this relation is reflexive. And, yes, this is what the definition says. – Jyrki Lahtonen May 08 '17 at 09:00
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    You have completely modified your question, and in a good way ! I congratulate you for this effort. – Jean Marie May 08 '17 at 09:10
  • No, someone else did. I guess Max. Thanks Max. – Sanone May 08 '17 at 09:11

2 Answers2

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The reflexive property, in many cases, is very subtle. This is where your difficulty lies. To show that an element x in the set X is reflexive, we need to show x R x.

So, does 0 R 0 ? (It may help to think of the zero on the left and zero on the right as being two different entities, that is, we are picking two elements from your set, where we can pick the same element twice.)

The answer is yes, since 0/3 = 0 and 0/3 = 0. That is, the 'left' zero and the 'right' zero have the same remainder after dividing by 3. The same argument applies for the rest of {1,...,9}.

Now, this has nothing to do with reflexivity, but may help in understanding the relation R.

Does 3R9? Well, 3/3 = 1 (remainder 0) and 9/3 = 3 (remainder 0), so yes, 3R9.

Does 4R9? Well, 4/3 = 1 + 1/3 (remainder 1) and 9/3 = 3 (remainder 0), so 4 "notR" 9.

D. Nichols
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  • thanks a lot. I appreciate your time and explanation. These things are very subtle and just by learning definitions you can't grasp the idea. I have a hard time with questions like these. Some are easy some are difficult to grasp. – Sanone May 08 '17 at 08:58
  • Equivalence relations/ classes were difficult for me to learn because I could not see the bigger picture. When you start looking at partitions (and later on, quotient spaces), this stuff will make more sense. Thinking in terms of a bag of skittles (candy) helped me. Let S = {bag of skittles}. Suppose there are five colors: Red, Orange, Yellow, Green, Purple. To make it easy, let there be 10 of each color in the bag. You can label them R_1,R_2,...,R_10 and so on. Define a relation R on the set S where x R y if two skittles have the same color. (Equiv. classes are used to partition sets.) – D. Nichols May 08 '17 at 09:07
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Let us show that $8\mathcal R8.$ By definition, $8\mathcal R8$ holds if $8$ and $8$ give same remainder on division by $3.$

Divide $8$ by $3$: $$8=2\cdot3+2$$ $3$ goes into $8$ twice, with a remainder of $\boxed2.$

Now divide $8$ by $8$: $$8=2\cdot3+2$$ $3$ goes into $8$ twice, with a remainder of $\boxed2.$

We got the same remainder, namely $2,$ both times. Therefore $8\mathcal R8.$

bof
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