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We define the relation R on the set $A=\{ -8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8 \}.$

so that $xRy$ if and only if $x+4y$ is dividable with $5$.

Ok so how should i define this $R$ with $x$ and $y$?

Should I try with every element = $x$ and with every element = $y$? Is not that very many combinations to consider?

cmk
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Daniel Andersson
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1 Answers1

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HINT: $5|(x+4y)$ if and only if $x \equiv y \mod 5$. Then can you find the (equivalence) classes (if you did not learn about it, you can try to find the elements which are equivalent modulo $5$)?

As an example, $\{-5,0,5\}$ is one of the classes because $-5 \equiv 0 \equiv 5 \mod 5$. There are four more such classes and when you find them, you are basically done because you know that $x \equiv y \mod 5$ so $x$ and $y$ should be chosen from the same class.

ArsenBerk
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  • Can you elaborate on that? :P – Daniel Andersson Jun 04 '19 at 12:46
  • Maybe I can add one more hint which makes it easier to continue. – ArsenBerk Jun 04 '19 at 12:46
  • Yeah please do :) – Daniel Andersson Jun 04 '19 at 12:49
  • As I understand it I pick x = 0 then using the formula for y = 0-8. Then i pick x =1 and y=0-8 and so on. So i need to check 9*9=81 different combinations of x and y? – Daniel Andersson Jun 04 '19 at 12:54
  • Not really. If you pick an arbitrary $x$, then you can add/subtract $5$ to find the equivalent elements ($x-5, x+5$, etc.) without checking everything. Can you see why? – ArsenBerk Jun 04 '19 at 12:56
  • Hmm well if I add/subtract 5 to an arbitrary that then gets me another element in the set, I guess you mean that that element is dividable by 5 or something? – Daniel Andersson Jun 04 '19 at 13:00
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    No, if $x = 1$ for instance, adding/subtracting $5$ will still give you an element with remainder $1$ when divided by $5$. So they will be in the same class. If $x$ is divisible by $5$, then adding/subtracting $5$ will also yield an element that is divisible by $5$ as well. – ArsenBerk Jun 04 '19 at 13:03
  • −6≡-1≡4 and −5≡0≡5 and −4≡1≡6 and -8≡−3≡2≡7 and -7≡−2≡3≡8 ? How do i translate these to R? – Daniel Andersson Jun 05 '19 at 08:27
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    You can either write $R$ instead of $\equiv$'s by using transitivity of the relation you have or, since a relation is nothing but a set, you can use set notation such as $R = {(x,y)\in A\times A\ |\ x \equiv y \mod 5}$. – ArsenBerk Jun 05 '19 at 08:54