Partial ordering relation

Question

Consider the following system of sets and operations an them:

$(1)(Z,\geq ),$ where $'Z'$ is the set of an integer

$(2) (Z^{+},1),$where $'1'$ is the divisibility relation

$(3) (P(S),\subseteq ),$where $'P(S)'$ is the power set of $'S'$

Which of the above is/are partial ordering?

$A) (1)$ only

$B) (2)$ and $(3)$ only

$C) (1)$ and $(3)$ only

$D)$ All are partial ordering

Comments

all are partial orders! All sets are reflexive, antisymmetric and transitive!!

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Yeah, but I have doubt to the (2)

can you explain?

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okay! all positive integers are divisible by themselves so, reflexive.

8 is divisible by 4 but 4 is not divisible by 8. (i don't know how to prove this for all pairs of elements but after trying some pairs you can realize that relation is antisymmetric.)

again for transitivity, 16 is divisible by 8, 8 is by 4 so 16 is divisible by 4. hence transitive.

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Ohh I did the silly mistake here, I try 4 is divisible by 2, but 2 is not divisible by 4,(Actually, I apply Symmetric relation property but in partial order relation Antisymmetric present)

thank you for give a example

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wc

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i don't know how to prove this for all pairs of elements

let a and b are the elements, if (a,b) present in the relation means, a can divide b

it means b should be multiple of a ===> b = k.a ===>$\frac{b}{a}=k$

then (b,a) could be in the relation? (forget k=1, due to it leads reflexivity)

if (b,a) in the relation, then b should divide a ==> $\frac{a}{b}=p$

$\frac{a}{b}=p$

$\frac{a}{k.a}=p$

$\frac{1}{k}=p$ ===> $\frac{1}{k}$ can't be a integer except k=1, but note that our domain is integers only.

∴ b can't divides a ===> (b,a) can't be in the relation ===> given is anti-symmetric

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where ′P(S)′ is the power set of ′S′

here P(S) is only one set so how we can apply the subset relation on this

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@Shaik Masthan

In the question of (2)i.e. with divisibility operation ,Is Z-, i.e. set of -ve integers, is partial ordered?