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 August 19th, 2017, 03:38 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 05:51 PM. August 23rd, 2017, 04:00 AM   #2
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 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! Tags functions, injective, questions, relation, relations, set-theory, surjective Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post arybhatta01 Abstract Algebra 14 July 13th, 2017 08:01 PM Gurasees Singh Math Books 5 December 7th, 2014 02:18 AM mohitpd Applied Math 1 October 29th, 2013 05:16 PM mohitpd Applied Math 3 October 26th, 2013 07:53 PM tsl182forever8 Applied Math 1 November 26th, 2012 10:38 PM

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