Mathematics 67 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  = {x in Z: 2x, x = 3k}  = {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):

$=\{y\in \mathbb{Z}:0\sim y\}$ $=\{y\in \mathbb{Z}: 3|0+2y\}=\{y\in \mathbb{Z}: 3|2y\}=\{y\in \mathbb{Z}: y\text{ is a multiple of 3}\}$ OpenStudy (anonymous):

so is that how you write the proof of the eq class? OpenStudy (zarkon):

$=\{y\in \mathbb{Z}: 3|1+2y\}=\{y\in \mathbb{Z}: 1+2y=3k\}=\{\cdots, -2,1,4,7,\cdots\}$ 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  and , 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  is an equivalence class different from ; 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): OpenStudy (zarkon):

no OpenStudy (anonymous):

don't tell me OpenStudy (zarkon):

= OpenStudy (anonymous):

it must be 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

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