Equivalence Relations on Z - Are There Infinite Equivalence Classes?

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 2K views
gtfitzpatrick
Messages
372
Reaction score
0

Homework Statement



Deciede if the following are equivalence relations on Z. If so desribe the eqivalence classes
i) a[itex]\equiv[/itex] b if [itex]\left|a\right|[/itex] = [itex]\left|b\right|[/itex]
ii) a[itex]\equiv[/itex] b if b=a-2

Homework Equations





The Attempt at a Solution



i) [itex]\left|a\right|[/itex] = [itex]\left|a\right|[/itex] so its reflexive

[itex]\left|a\right|[/itex] = [itex]\left|b\right|[/itex] is equivalent to [itex]\left|b\right|[/itex] = [itex]\left|a\right|[/itex] so its symmetric

[itex]\left|a\right|[/itex] = [itex]\left|b\right|[/itex] and [itex]\left|b\right|[/itex] = [itex]\left|c\right|[/itex] then [itex]\left|a\right|[/itex] = [itex]\left|c\right|[/itex] for all values a,b and c elemets of Z so its transitive.

Are there infinite equivalence classes??


ii) a=a so its reflexive
b=a-2 [itex]\neq[/itex] a=b-2 so its not symetric, am i right in thinking this?
Thanks for reading
 
Physics news on Phys.org
gtfitzpatrick said:

Homework Statement



Deciede if the following are equivalence relations on Z. If so desribe the eqivalence classes
i) a[itex]\equiv[/itex] b if [itex]\left|a\right|[/itex] = [itex]\left|b\right|[/itex]
ii) a[itex]\equiv[/itex] b if b=a-2

Homework Equations


The Attempt at a Solution



i) [itex]\left|a\right|[/itex] = [itex]\left|a\right|[/itex] so its reflexive

[itex]\left|a\right|[/itex] = [itex]\left|b\right|[/itex] is equivalent to [itex]\left|b\right|[/itex] = [itex]\left|a\right|[/itex] so its symmetric

[itex]\left|a\right|[/itex] = [itex]\left|b\right|[/itex] and [itex]\left|b\right|[/itex] = [itex]\left|c\right|[/itex] then [itex]\left|a\right|[/itex] = [itex]\left|c\right|[/itex] for all values a,b and c elemets of Z so its transitive.

Are there infinite equivalence classes??
Yes. Can you describe them? Simply listing a few to show the pattern would be sufficient.
ii) a=a so its reflexive
a=a-2?
b=a-2 [itex]\neq[/itex] a=b-2 so its not symetric, am i right in thinking this?
Yes.
 
For (i):
What elements(s) of Z is/are equivalent to 3?
What elements(s) of Z is/are equivalent to 7?
What elements(s) of Z is/are equivalent to 0?
What elements(s) of Z is/are equivalent to -5?
...​

For (ii):
This relation is not transitive either.​
 
Thanks for the replies.
So i need to say there are infinity equivalent classes such as -3 equivalent to 3; -5 equivalent to 5 or 10 is equivalent to -10 under the relation.

for ii) i only need to show 1 of the 3 properties doesn't hold, right? or should i show whether all 3 hold or not just for clarity?
 
gtfitzpatrick said:
Thanks for the replies.
So i need to say there are infinity equivalent classes such as -3 equivalent to 3; -5 equivalent to 5 or 10 is equivalent to -10 under the relation.
Basically, yes, though your instructor may cringe at your grammar. ;)

The equivalence classes are subsets consisting of all elements that are equivalent to each other. So in this case, they'd be {0}, {1,-1}, {2,-2}, and so on.
for ii) i only need to show 1 of the 3 properties doesn't hold, right? or should i show whether all 3 hold or not just for clarity?
Right. You need to show only one requirement doesn't hold to rule out the relation being an equivalence relation.
 
grammar isn't a strong point of mine :)
Thanks a mill