My Math Forum several questions about relations and functions

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 August 19th, 2017, 04:38 PM #1 Newbie   Joined: Aug 2017 From: mathmatistan Posts: 9 Thanks: 0 several questions about relations and functions Hello, I'm having trouble with the concept of relations and functions. I don't understand some concepts and I'd like your help to do that if you can. Let's say we have a simple set M that is the relation set of A={1,2,3}. a) the number of elements of M? Is it 1M3,1M2,2M1,2M3,3M1,3M2,3M1 or do you not count the "opposites", like 1M3,3M1? Let's say that s:m->m is the function that assigns for every $\displaystyle R \in M$ its symmetrical complement (I don't know the term in English, so I'll provide the mathematical definition: $R \cup R^{-1} = R \cup \{\langle a,b\rangle\,|\,\langle b,a\rangle \in R\}$). b) is M injective? It seems that not since every element in the domain has more than one link in the range. (x doesn't have a unique y), how do I write it mathematically? c) is M surjective? Every value of f(x) seems to have at least one value of x. d) for every [MAT]R1,R2 \in M[/MATH] is $\displaystyle s(R_1 R_2)=s(R_1)s(R_2)$ (relation multipication) seems to be true. since if we have {<1M3>,<1M2>,<2M1>,<2M3>,<3M1>,<3M2>,<3M1>}, after assigning the multiplication through function s (as explained, s is defined s:m->m and assigns its symmetrical complement), as since we don't write each element twice, it holds. is it okay to write that as a proof? e) for every $\displaystyle R \in M$, $\displaystyle s(s(R))=s(R)$ seems to be very trivial, since you assign the complement and then assign the complement of it, it's like $\displaystyle (x^{-1})^{-1}$ or don't I understand it correctly? If you can, please help me with this as it's very important for a test I'll have on Monday. I've elaborated and explained my thoughts, so please correct me if I'm mistaken. Thank you in advance! Last edited by skipjack; August 19th, 2017 at 06:51 PM.
August 23rd, 2017, 05:00 AM   #2
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Quote:
 Originally Posted by mathsnoob Hello, I'm having trouble with the concept of relations and functions. I don't understand some concepts and I'd like your help to do that if you can. Let's say we have a simple set M that is the relation set of A={1,2,3}. a) the number of elements of M? Is it 1M3,1M2,2M1,2M3,3M1,3M2,3M1 or do you not count the "opposites", like 1M3,3M1?
A "relation on set A" is defined as "a subset of A X A". If A has n members then A x A has n^2 members. A subset may be empty or contain some or all n^2 pairs. An individual relation on a set with 3 members may have anywhere from 0 to 9 members.

Quote:
 Let's say that s:m->m is the function that assigns for every $\displaystyle R \in M$ its symmetrical complement (I don't know the term in English, so I'll provide the mathematical definition: $R \cup R^{-1} = R \cup \{\langle a,b\rangle\,|\,\langle b,a\rangle \in R\}$). b) is M injective?
You have already defined "M" as the set of all relations on A. You should not use "M" to mean this particular relation on M.

Quote:
 It seems that not since every element in the domain has more than one link in the range. (x doesn't have a unique y), how do I write it mathematically? c) is M surjective? Every value of f(x) seems to have at least one value of x. d) for every [MAT]R1,R2 \in M[/MATH] is $\displaystyle s(R_1 R_2)=s(R_1)s(R_2)$ (relation multipication) seems to be true. since if we have {<1M3>,<1M2>,<2M1>,<2M3>,<3M1>,<3M2>,<3M1>}, after assigning the multiplication through function s (as explained, s is defined s:m->m and assigns its symmetrical complement), as since we don't write each element twice, it holds. is it okay to write that as a proof? e) for every $\displaystyle R \in M$, $\displaystyle s(s(R))=s(R)$ seems to be very trivial, since you assign the complement and then assign the complement of it, it's like $\displaystyle (x^{-1})^{-1}$ or don't I understand it correctly? If you can, please help me with this as it's very important for a test I'll have on Monday. I've elaborated and explained my thoughts, so please correct me if I'm mistaken. Thank you in advance!

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