Mathematics 18 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (zarkon):

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

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

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

OpenStudy (anonymous):

just need to understand what i'm doin

OpenStudy (zarkon):

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

OpenStudy (anonymous):

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

OpenStudy (zarkon):

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

OpenStudy (zarkon):

what are you trying to prove?

OpenStudy (jamesj):

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?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (zarkon):

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

OpenStudy (jamesj):

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

OpenStudy (anonymous):

oh i see

OpenStudy (anonymous):

so could i write:

OpenStudy (anonymous):

[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?

OpenStudy (zarkon):

no

OpenStudy (zarkon):

that is wrong on a couple levels

OpenStudy (zarkon):

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

OpenStudy (zarkon):

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.

OpenStudy (anonymous):

ok so I think the other class is [3}

OpenStudy (anonymous):

[3]

OpenStudy (zarkon):

no

OpenStudy (anonymous):

don't tell me

OpenStudy (zarkon):

[0]=[3]

OpenStudy (anonymous):

it must be [2]

OpenStudy (zarkon):

correct

OpenStudy (anonymous):

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

OpenStudy (anonymous):

k is a multiple of 2

OpenStudy (zarkon):

k is some integer

OpenStudy (zarkon):

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

OpenStudy (anonymous):

because the remainder is 2?

OpenStudy (anonymous):

right, so it's even multiples of 3

OpenStudy (zarkon):

you could also write out the set in roster notation

OpenStudy (anonymous):

6, 12, 18

OpenStudy (anonymous):

2k+2

OpenStudy (anonymous):

?

OpenStudy (zarkon):

$\{\ldots,-1,2,5,8,\ldots\}$

OpenStudy (anonymous):

sigh

OpenStudy (anonymous):

K+3