Partial Orders

Hi,
A = {1,3,5,7,9,10,12,15,8,21}

R = {(x,y) | y divided by x}


Im trying to prove that R is a partial order (reflexsive, antisymmetric and transitive).

Reflexsive: I said that it is true, because for all x, xRx. That is, every number in A is divided into itself once.

Antisymmetric: is false. Because if we have (5, 10) and (10, 5) then x is not equal to y. Is that correct? Because we get 2 and 0.5 and thats not equal?

Transitive:: if we have (5, 10, 15) then thats true but (15, 10, 5) false? Hence not transitive?

Can somebody help me out on this question?

I drew a hasse diagram for this but i dont know how to draw one on this forum. If someone can draw me one on the forum so i can check mine it would be of great help?

My maximal was 12 and 15 and minimal element was 1. Correct?

Quote:

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Originally Posted by kurac

Hi,
A = {1,3,5,7,9,10,12,15,8,21}

R = {(x,y) | y divided by x}

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clarify here. do you means y "divides" x?, as in y|x??

if so, you have certain flaws in your logic. for instance, (5,10) would not be in the relation. etc etc

Thats what confusing me alittle.

All is said that "y is divided by x". I need prove that its a partial order.

Quote:

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Originally Posted by kurac

Thats what confusing me alittle.

All is said that "y is divided by x". I need prove that its a partial order.

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that is what it says in words? that makes no sense. it should be y "divides" x, or maybe something like "y divided by x is an integer". just saying y divided by x can mean many things. divides exactly? has a remainder of a certain amount?...etc

if there was a symbol used, say so. do not type the english if you are not sure how the notation is read

Quote:

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Originally Posted by kurac

Hi,
A = {1,3,5,7,9,10,12,15,8,21} R = {(x,y) | y divided by x}
Im trying to prove that R is a partial order (reflexive, anti-symmetric and transitive).
Reflexsive: I said that it is true, because for all x, xRx. That is, every number in A is divided into itself once.
Antisymmetric: is false. Because if we have (5, 10) and (10, 5) then x is not equal to y. Is that correct? Because we get 2 and 0.5 and thats not equal?
Transitive:: if we have (5, 10, 15) then thats true but (15, 10, 5) false? Hence not transitive?

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Reflexive: I said that it is true, because for all x, xRx. That is, every number in A is divided into itself once.

Anti-symmetric: is false. Wrong
Transitive:: if we have (5, 10, 15) then thats true but (15, 10, 5) false? Wrong again.
You seem to have a most odd understanding of this relation.

2 divides 4 but 4 does not divide 2.
So if x divides y and y divides x then x=y: anti-symmetric.

So if x divides y and y divides z then clearly x divides z: transitive.

Plato, so your saying my relation definition will always only apply too numbers where x and y are the same; for e.g (2,2) (3,3). Hence x and y will divide and y and x will divide?

I tried drawing a hasse diagram of the partial order R to try get better understanding of it...

Can someone who knows how to draw these diagrams check if 12 and 15 is the maximal and 1 the minimal?

Thanks for the help, im not very good at maths but im trying my best!
Kurac.