Question

Ok, so I want to show that there are 3 eq classes for eq relation xRy 3|x+2y

Answer 1

so I have [0] = {x in Z: 2x, x = 3k}
[3] = {x in Z: 2x+3, x = 3k}
[x} = {x in Z: x=x}

Answer 2

but I'm not sure if I'm writing this correctly

Answer 3

so your relation is x is equivalent to y iff 3 divides x+2y

correct?

Answer 4

yes

Answer 5

So, i don't want the answer given to me by the way

Answer 6

just need to understand what i'm doin

Answer 7

\[[0]=\{y\in \mathbb{Z}:0\sim y\}\]

\[[0]=\{y\in \mathbb{Z}: 3|0+2y\}=\{y\in \mathbb{Z}: 3|2y\}=\{y\in \mathbb{Z}: y\text{ is a multiple of 3}\}\]

Answer 8

so is that how you write the proof of the eq class?

Answer 9

\[[1]=\{y\in \mathbb{Z}: 3|1+2y\}=\{y\in \mathbb{Z}: 1+2y=3k\}=\{\cdots, -2,1,4,7,\cdots\}\]

Answer 10

what are you trying to prove?

Answer 11

He's writing out carefully for you the equivalence classes, and has given you two of them so far.

If the question is what are all of the equivalence classes, the next question is: what other classes are there beyond these two, if any?

Answer 12

ok, i already figured out the [0] and [1], but I wrote them differently

Answer 13

but i want to show that if x and y are the same, that it works too

Answer 14

remember that the equivalence classes form a partition of the underlying set.

Answer 15

Notice that you wrote above in your original statement that [3] is an equivalence class different from [0]; that is not the case.

Answer 16

oh i see

Answer 17

so could i write:

Answer 18

[n] = {y in Z: n + 3n = 3k} to show that when x and y are the same it's another class or does that not count?

Answer 19

no

Answer 20

that is wrong on a couple levels

Answer 21

there are only 3 EQ classes...I gave you two of them

Answer 22

note that 1~1 and 2~2 but that doesn't mean that they are in the same EQ class. this is just a consequence of the relation being reflexive.

Answer 23

ok
so I think the other class is [3}

Answer 24

[3]

Answer 25

no

Answer 26

don't tell me

Answer 27

[0]=[3]

Answer 28

it must be [2]

Answer 29

correct

Answer 30

so i would say y is in z: 2 + 2y = 3k

Answer 31

k is a multiple of 2

Answer 32

k is some integer

Answer 33

it will have to be true that it will be a multiple of 2....but i would still say k is some integer

Answer 34

because the remainder is 2?

Answer 35

right, so it's even multiples of 3

Answer 36

you could also write out the set in roster notation

Answer 37

6, 12, 18

Answer 38

2k+2

Answer 39

?

Answer 40

\[\{\ldots,-1,2,5,8,\ldots\}\]

Answer 41

sigh

Answer 42

K+3