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August 13th, 2017, 05:48 AM   #1
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several truth\false from set theory

Apparently, I've solved the wrong set of exercises and now I have to solve a different bunch (lol). the questions are true/false types.
I'll write what I did/tried or got wrong for every question:

1) for every equivalence relation over A={1,2,3,4,5,6} there is a class in which the number of variables is odd.

Even though the number of variables is not odd, there could be an odd class with odd number of variables. but I don't know whether it's suffice to assert it's a true claim.

2) for every equivalence relation over A={1,2,3,4,5,6,7} there is a class in which the number of variables is odd.

Seems true, since the number of variables inside A is odd.

3) if R is symmetrical and reflexive over A={1,2} then R is equivalence relation.

To be called an equivalence relation, it should hold: reflexive relation, transitive relation and and symmetric relation. in this case, I think it's a true claim.

4) if R is symmetrical and reflexive over A={1,2,3} then R is equivalence relation.

I think that in this case, it is not not a true claim because of the odd number of variables.

5) if $\displaystyle R^2$ is symmetrical and reflexive over A={1,2,3}, then R is equivalence relation.

R^2 is defined as R x R, but I still don't know whether it's a true claim or not. I think that not. if I'm wrong, please show me why.

6) if S is equivalence relation, then every one of its classes has the same number of variables.

Logically, it seems to be true. I don't see a reason why this sentence would be wrong.

7) there exist even number $\displaystyle m,n \in Z$ so that $\displaystyle (n,m) \in S$

Seems to be true. Again, I don't see any reason why it would be false or example that would contradict it.

8) for every $\displaystyle n \in Z$, $\displaystyle (-n-1,n)\in S^2$

Since we're talking about $\displaystyle S^2$, then negative numbers become positive, and thus it seems logical that this sentence will be true.


I've tried to give detailed reasoning for my questions so you could understand why I chose a certain answer. Please correct me if I'm wrong and help me improve and become better.

Last edited by skipjack; August 13th, 2017 at 07:11 AM.
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August 13th, 2017, 07:23 AM   #2
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Quote:
Originally Posted by mathsnoob View Post
Apparently, I've solved the wrong set of exercises and now I have to solve a different bunch (lol). the questions are true/false types.
I'll write what I did/tried or got wrong for every question:

1) for every equivalence relation over A={1,2,3,4,5,6} there is a class in which the number of variables is odd.

Even though the number of variables is not odd, there could be an odd class with odd number of variables. but I don't know whether it's suffice to assert it's a true claim.


2) for every equivalence relation over A={1,2,3,4,5,6,7} there is a class in which the number of variables is odd.

Seems true, since the number of variables inside A is odd.
The point is whether it is true for ANY equivalence relation. So if I subdivides the A in (2) in {1, 2, 3}, {4, 5}, {6, 7}, then it's true for this one. But is it true for any one?

In (1) can you not see any way to subdivide A into disjoint classes such that all are even?

Quote:
3) if R is symmetrical and reflexive over A={1,2} then R is equivalence relation.

To be called an equivalence relation, it should hold: reflexive relation, transitive relation and and symmetric relation. in this case, I think it's a true claim.
You need to check whether a~b and b~c implies a~c. You can easy do this by brute force. Find all symmetric and reflexive relations on A. Check whether they are transitive. There aren't a lot of them.

Quote:

4) if R is symmetrical and reflexive over A={1,2,3} then R is equivalence relation.

I think that in this case, it is not not a true claim because of the odd number of variables.
It has nothing to do with odd and even. Can you find a symmetric and reflexive relation that is not transitive? There aren't a lot.

Quote:
5) if $\displaystyle R^2$ is symmetrical and reflexive over A={1,2,3}, then R is equivalence relation.

R^2 is defined as R x R, but I still don't know whether it's a true claim or not. I think that not. if I'm wrong, please show me why.
What are the possibilities for R?

Quote:
6) if S is equivalence relation, then every one of its classes has the same number of variables.

Logically, it seems to be true. I don't see a reason why this sentence would be wrong.
What about {1,2}, {3} on A={1,2,3}?

Quote:

7) there exist even number $\displaystyle m,n \in Z$ so that $\displaystyle (n,m) \in S$

Seems to be true. Again, I don't see any reason why it would be false or example that would contradict it.
What is Z, what is S?

Quote:


8) for every $\displaystyle n \in Z$, $\displaystyle (-n-1,n)\in S^2$

Since we're talking about $\displaystyle S^2$, then negative numbers become positive, and thus it seems logical that this sentence will be true.
What is Z? What is S?
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August 13th, 2017, 10:14 AM   #3
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Sorry for not providing the full information regarding questions 6,7,8:

S is a relation on the integer set(Z) that applies $\displaystyle (n,m) \in Z \text{ iff } m^2+m=n^2+n$

Last edited by skipjack; August 13th, 2017 at 10:24 AM.
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August 14th, 2017, 11:54 AM   #4
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Hi,

Did you manage to solve from (1) to (6) with the advice Micrm@ss gave you?

Quote:
Originally Posted by mathsnoob
S is a relation on the integer set(Z) that applies $\displaystyle (n,m) \in Z \text{ iff } m^2+m=n^2+n$

7) there exist even number $\displaystyle m,n \in Z$ so that $\displaystyle (n,m) \in S$

Seems to be true. Again, I don't see any reason why it would be false or example that would contradict it.
Don't you think $\displaystyle S$ is reflexive? Then choose any even number, it will be in relation with itself.

Quote:
Originally Posted by mathsnoob
8) for every $\displaystyle n \in Z$, $\displaystyle (-n-1,n)\in S^2$

Since we're talking about $\displaystyle S^2$, then negative numbers become positive, and thus it seems logical that this sentence will be true.
You're just asked whether for all $\displaystyle n\in Z$ you have $\displaystyle (-n-1)^2+(-n-1)=n^2+n$
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August 27th, 2017, 04:25 AM   #5
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An obvious counter-example for (1) is the relation in which every member of A is related to all others. The only equivalence classes are A itself and the empty set. They contain an even number of elements.
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