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Sep 20 2015 03:37pm
Hey so I am having trouble with some stuff here:



For (a) I am having trouble proving R is an equivalence relation (in particular, proving R is transitive).

In addition, I'm having trouble wrapping my head around what a partition is. For example, in question (b) would the partition be P = {{x,x+2pi*k}| x in R and k in Z}?
For the partition in (d) I'm not really sure how to start that since its a coordinate pair. I have that it is an equivalence relation and sqrt(x1^2+y1^2) = sqrt(x2^2+y2^2) but don't know how to proceed.

Could anyone help? Thanks
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Sep 20 2015 04:13pm
for a y dont u check if its reflexive then u'll see u dont have to check transitivity
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Sep 20 2015 04:14pm
(a) Does that mean a ~ b if and only if gcd(a;b) is even ? Because there is no relation written in (a), but just a subset of N².

In the case of a ~ b if and only if gcd(a;b) is even, then no, it is not a relation of equivalence, because not every a satisties : a ~ a.
In other words, that relation is not reflexive.

(b) Same guess : a ~ b if and only if cos (a) = cos (b).
This is a relation of equivalence, because :
a ~ a for every a,
if a ~ b, then b ~ a,
if a ~ b and b ~ c, then a ~ c.

The partition of the set of real numbers that arise from it is :
R = ∪ { θ + 2k.pi ; - θ + 2k.pi, k ∈ Z } , union with θ lying in [0;pi]
in other words, for every possible value of cos between -1 and 1, choose 1 real θ whose cos equals that value.

(c) ~ is not a relation of equivalence.
(x₁ ; y₁) ~ (x₂ ; y₂) simply means that the vectors are colinear.
This is not transitive :
any vector V is colinear to the nul vector, the nul vector is colinear to any vector W, but that doesn't mean that V and W are colinear.

(d) is a relation of equivalence.
The partition of R² is the union of concentric circles of radius R > 0, and the set containing only (0;0).
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Sep 21 2015 01:26pm
Oh okay I totally missed reflexivity on (a). The partitions make more sense now thanks!
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