# true\false claims in relations and equivalence relations

i don't understand this basic concept and i wrote those questions to better understand two concepts i'm having trouble with: a)relations and classes b)equivalence relation. i'll try to write what i think, please help me if you can.

the questions are true/false type

1)for every relation over A ={1,2,3,4,5,6} there's a class with odd number of variables.

2)for every class over a ={1,2,3,4,5,6,7} there's a class with odd number of variables.

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

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

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

6)the function f:P(N) -> P(N) that is defined by f(x)=x-{1} for every $x \in P(N)$ is injective

7)the function f:P(N) -> P(N) that is defined by f(x)=x-{1} for every $x \in P(N)$ is surjective

what i don't understand is: a)regarding the first two questions - does the addition of one number, {7}, changes the result? b)same regarding question 3-5, does the addition of one variable changes the results in regards to the equivalence relation? {1,2,3} over {1,2} c)what is the major difference between R and $R^2$ in regards to the equivalence relation?

what i know and tried:

1-2) from what i understand, i need to check how to arrange the set A into several groups and to see if in all cases there would be a class with odd number of variables. so in the case |A|=6, it can't always be odd, because for instance: {1,2} {3,4} {5,6}, but in the case of |A|= 7 there always would be an odd group.

3-5)in order to become an equivalence relation, i know that a relation should be: transitive, reflexive and symmetrical. given that it is supplied that R is reflexive, and symmetrical, i believe that the question is about the transitivity of the set. now what i don't understand is how the addition of an variable to the set {1,2} versus {1,2,3} and the difference between R and $R^2$ influences the transitivity of the sets. i think that in the case {1,2} it is transitive, but i am not sure if {1,2,3} is transitive, but i'm quite sure that for $R^2$ A={1,2,3} is transitive.

6-7) the function f:P(N) -> P(N) that is defined f(x)=x-{1} is surjective, because for every element y in the codomain Y of f there is at least one element x in the domain X of f such that f(x) = y, but is not injective, because of the "-{1}" part.

so please tell me if i'm correct or not:

1)for every relation over A ={1,2,3,4,5,6} there's a class with odd number of variables. -- FALSE

2)for every class over a ={1,2,3,4,5,6,7} there's a class with odd number of variables. --TRUE

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

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

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

6)the function f:P(N) -> P(N) that is defined by f(x)=x-{1} for every $x \in P(N)$ is injective -- FALSE

7)the function f:P(N) -> P(N) that is defined by f(x)=x-{1} for every $x \in P(N)$ is surjective --TRUE

thank you very much for helping me, i hope i provided enough information and i tried to elaborate the best i can what i tried to do and why.

if i'm wrong, please, if you can, mention it and tell me why.

• meta.stackexchange.com/a/39224/345058 one question per post if possible. – Siong Thye Goh Aug 13 '17 at 19:17
• next time i would do that, thank you for explaining me the rules. – BeginningMath Aug 13 '17 at 19:25
• I'm not familiar with 'class' and I can't seem to find it in a search either. Presumably it isn't the same as 'equivalence class'? On your point about 3-5, transitivity may still be true and needs to be proven if so. You need to show a counterexample to show it isn't (and 5 is slightly different anyway). – Shuri2060 Aug 13 '17 at 20:02
• can you pleae tell me what the difference i in 5? i don't need to prove, i only need to deicde whether the claim is true or not. – BeginningMath Aug 13 '17 at 20:13