Relation and order relation question

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The discussion centers around determining whether the set G = {(4,2), (4,1), (4,3), (2,1), (2,3), (1,3)} is a relation and an order relation based on the definitions provided. It is established that G is indeed a relation because it is a subset of AxA, where A = {1, 2, 3, 4}. The participants analyze the conditions for G to be an order relation, concluding that it satisfies both necessary conditions: each pair only satisfies one of the three possible relations, and the transitive property holds true for the pairs in G. Despite initial confusion about identifying the specific order, it is agreed that G qualifies as an order relation based on the definitions and conditions discussed. The complexity of understanding these concepts is acknowledged, highlighting the challenge of defining relations intuitively.
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Homework Statement


Let A = {1, 2, 3, 4}
Let G = {(4,2), (4,1), (4,3), (2,1), (2,3), (1,3)}

Is G a relation? is G an order relation?


Homework Equations



i think i should put the definitions of a relation and an order relation here. also this is from the book Elementary geometry from an advanced standpoint 3rd edition. section 3.2.

The definition of a relation is
A relation defined on a set A is a subset of AxA (A^2)

or in other words, a relation * is defined by the ordered pairs within a given set, G, which is a subset of AxA

example:
A = {1,2,3}
* = {(1,2), (1,3), (2,3)}
because * is a subset of AxA (which is basically all 9 combinations of 1, 2, 3 in ordered pairs)
* is a relation.

in this case * is actually the < relation

a relation * is a an ordered relation if it satisfies these two conditions
1
for every (a,b) only one of these conditions can be satisfied
a*b, a = b, b * a
2
if a * b and b * c then a * c

The Attempt at a Solution



So i know that it is a relation because G is a subset of AxA, which is all 16 combinations of 1, 2, 3, 4

however i am unsure of how i would determine whether or not its an order relation. i am leaning towards "no" since if G was > then it would only contain {(a,b)|a > b} however, (2,3) contradicts that. it also can't be < since that would only contain {(a,b)|a<b} however because it contains (4,1) that can't be true either. therefore i want to say it isn't an order relation, but i am not sure if this is sufficient proof or not, or if i even approached it correctly.
 
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hi ironman1478! :smile:
ironman1478 said:
… if G was > then …

no, that's not a sensible way to start, the relation could be anything

the best way is just to attempt to write them in an order that fits …

if you can, then that's the proof,

and if you can't, then you can probably see how to prove that you can't :wink:
 
The "defining" property of an order relation is the "transitive" property: If (a, b) and (b, c) are in the relation then so is (a, c). (4, 2) and (2, 1) are in the relation. Is (4, 1)?
 
ah i think i get it.

so then it is an order relation

it satisfies conditions 1 since each (a,b) only satisfies only one of the three
there is only aGb
for example there is only 4G1, but there is not 4 = 4 (4,4) nor is there a 1G4 (1,4)

then it also satisfies condition 2 since if we try to apply the transitive property to every ordered pair in which it is applicable, we get an (a,c) which is in the set G.

therefore, since it satisfies those two conditions G is an order relation.

i hope i did this right, it is a bit hard to think of these problems since i find it more intuitive to use a relation to define a set rather than a set define a relation.
 
ironman1478 said:
ah i think i get it.

so then it is an order relation

it satisfies conditions 1 since each (a,b) only satisfies only one of the three
"One of the three"? Are you referring to the conditions for an "equivalence relation"?

there is only aGb
for example there is only 4G1, but there is not 4 = 4 (4,4) nor is there a 1G4 (1,4)

then it also satisfies condition 2 since if we try to apply the transitive property to every ordered pair in which it is applicable, we get an (a,c) which is in the set G.

therefore, since it satisfies those two conditions G is an order relation.

i hope i did this right, it is a bit hard to think of these problems since i find it more intuitive to use a relation to define a set rather than a set define a relation.
I will repeat what I said before- G includes (4,2) and (2,3). What does the "transitive property" say G must include if it is an order relation?
 
HallsofIvy said:
"One of the three"? Are you referring to the conditions for an "equivalence relation"?


I will repeat what I said before- G includes (4,2) and (2,3). What does the "transitive property" say G must include if it is an order relation?

under "relevant equations" in my first post, at the end the two conditions given by my textbook for a relation being an order relation are given.

a relation * (* can be anything) is an order relation if it satisfies these two conditions


condition 1

for every (a,b) only one of these conditions can be satisfied
a*b, a = b, b * a
condition 2
if a * b and b * c then a * c

also, i understand your last point. (4,3) should be in the set, which it is. if we apply the transitive property to any 2 ordered pairs then we get ordered pairs which are also already in G.
there are no two ordered pairs that when the transitive property used, we get something out of set so therefore, the transitive property holds (i think)

god, this stuff is so difficult to understand sometimes
 
ironman1478, if it's an order, what is the order? :smile:
 
tiny-tim said:
ironman1478, if it's an order, what is the order? :smile:

well honestly i don't know lol.

according to the two conditions given by my book for a relation to be an order relation, this set of ordered pairs should be an order relation since it satisfies both. however i just can't figure it out what it is exactly, if it even is an order relation. it seems like the this question is easier than i think it is, but i just don't see the relation, if there is one
 
just write the four numbers on four separate pieces of paper,

and shuffle them around until you find it! :rolleyes:
 

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