What is the Union and Intersection of Modulo Relations?

nistaria
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Homework Statement



Let R1 and R2 be the "congruent modulo 3" and "congruent modulo 4" relations on the set of integers.


Homework Equations


Find:
a) R1 \cupR2
b)R1 \cap R2
There is also problem c, d but I won't write these here. If I am able to solve this, then the rest should be cake.


The Attempt at a Solution


my problem with this question is this: I'm not sure I understand what the question wants
Does it want all possible elements in the relation?
such as for a)
R1 \cupR2 ={(a,b)| (3|a-b) or (4|a-b)}
Is that a valid answer?
b) R1 \cap R2= {(a,b)| 12|a-b}

PS: I have already proved in a previous problem that
a\equivb(mod m) is an equivalence relation as it is transitive, symmetric and reflexive.

Thanks for reading
 
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welcome to pf!

hi nistaria! welcome to pf! :wink:

yes, that all looks fine :smile:

(a relation on X is just a subset of X x X, so yes you use the ordinary set union and intersection)
nistaria said:
… the rest should be cake.

hmm … candy, coffee, and now cake :rolleyes:

are you food-motivated? :biggrin:
 


tiny-tim said:
hi nistaria! welcome to pf! :wink:

yes, that all looks fine :smile:

(a relation on X is just a subset of X x X, so yes you use the ordinary set union and intersection)


hmm … candy, coffee, and now cake :rolleyes:

are you food-motivated? :biggrin:

AHAHAHAHA
I guess you read my other posts! Give me food and you got yourself a happy camper!

According to this http://books.google.ca/books?id=guh...t modulo 3" and "congruent modulo 4"&f=false"on googlebooks, it has the same question the answer to R1 U R2 is {(a,b):a-b is congruent to 0,3,4,6,8, or 9 (mod 12)}
My school book states the same as well.
I've been trying to figure out how they came down to this conclusion.
 
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hi nistaria! :smile:
nistaria said:
According to this http://books.google.ca/books?id=guh...t modulo 3" and "congruent modulo 4"&f=false"on googlebooks, it has the same question the answer to R1 U R2 is {(a,b):a-b is congruent to 0,3,4,6,8, or 9 (mod 12)}

yes, i was wondering whether to point that out, and i looked at the question, and i couldn't see any reason to do so …

especially since there's no point in trying to identify equivalence classes since R1 U R2 simply isn't an https://www.physicsforums.com/library.php?do=view_item&itemid=151"

personally, i think the way you wrote it is both shorter and clearer than the mod 12 way

(but maybe the rest of the question make a mod 12 answer more appropriate?)

it is 0,3,4,6,8, or 9 (mod 12) because anything divisible by 3 is 0,3,6, or 9 (mod 12), and anything divisible by 4 is 0,4 or 8 (mod 12) :wink:

… something for you to chew over! :biggrin:
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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