Show that it is a partial order and not an equivalence reaction [closed]
$begingroup$
I've been given the question below and I'm not sure how to show that it is a partial order.
Show that the relation $R ={(a,b) mid a text{ divides } b}$ over the set $mathbb{Z}^+$ is a partial order and is not an equivalence reaction.
I understand that it would be called a partial order if is transitive, reflexive and anti-symmetric.
Would you say that is transitive as (1,2) is an element of Z+ and (2,3) is an element of Z+, then (1,3) is an element of Z+?
Would you say it is reflexive as every element is related to itself for being a member of the set Z+ ?
How would you show it is anti-symmetric?
I am new to this topic and so I may sound rather stupid but I genuinely have no idea how to show that is a partial order.
functions discrete-mathematics equivalence-relations
asked Dec 2 '18 at 23:57
anon2000anon2000
32
$endgroup$
closed as off-topic by amWhy, Jyrki Lahtonen, José Carlos Santos, ancientmathematician, Lord_Farin Dec 3 '18 at 14:30
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Jyrki Lahtonen, José Carlos Santos, ancientmathematician, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I've been given the question below and I'm not sure how to show that it is a partial order.
Show that the relation $R ={(a,b) mid a text{ divides } b}$ over the set $mathbb{Z}^+$ is a partial order and is not an equivalence reaction.
I understand that it would be called a partial order if is transitive, reflexive and anti-symmetric.
Would you say that is transitive as (1,2) is an element of Z+ and (2,3) is an element of Z+, then (1,3) is an element of Z+?
Would you say it is reflexive as every element is related to itself for being a member of the set Z+ ?
How would you show it is anti-symmetric?
I am new to this topic and so I may sound rather stupid but I genuinely have no idea how to show that is a partial order.
functions discrete-mathematics equivalence-relations
asked Dec 2 '18 at 23:57
anon2000anon2000
32
$endgroup$
closed as off-topic by amWhy, Jyrki Lahtonen, José Carlos Santos, ancientmathematician, Lord_Farin Dec 3 '18 at 14:30
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Jyrki Lahtonen, José Carlos Santos, ancientmathematician, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
So can you show that it is transitive, reflexive, and anti-symmetric? Where are you stuck?
$endgroup$
– angryavian
Dec 3 '18 at 0:01
$begingroup$
@angryavian i'm stuck in showing that is is reflexive, transitive and anti-symmetric. would you say that is transitive as (1,2) is an element of R and (2,3) is an element of R, then (1,3) is an element of R?
$endgroup$
– anon2000
Dec 3 '18 at 0:04
add a comment |
$begingroup$
I've been given the question below and I'm not sure how to show that it is a partial order.
Show that the relation $R ={(a,b) mid a text{ divides } b}$ over the set $mathbb{Z}^+$ is a partial order and is not an equivalence reaction.
I understand that it would be called a partial order if is transitive, reflexive and anti-symmetric.
Would you say that is transitive as (1,2) is an element of Z+ and (2,3) is an element of Z+, then (1,3) is an element of Z+?
Would you say it is reflexive as every element is related to itself for being a member of the set Z+ ?
How would you show it is anti-symmetric?
I am new to this topic and so I may sound rather stupid but I genuinely have no idea how to show that is a partial order.
functions discrete-mathematics equivalence-relations
asked Dec 2 '18 at 23:57
anon2000anon2000
32
$endgroup$
I've been given the question below and I'm not sure how to show that it is a partial order.
Show that the relation $R ={(a,b) mid a text{ divides } b}$ over the set $mathbb{Z}^+$ is a partial order and is not an equivalence reaction.
I understand that it would be called a partial order if is transitive, reflexive and anti-symmetric.
Would you say that is transitive as (1,2) is an element of Z+ and (2,3) is an element of Z+, then (1,3) is an element of Z+?
Would you say it is reflexive as every element is related to itself for being a member of the set Z+ ?
How would you show it is anti-symmetric?
I am new to this topic and so I may sound rather stupid but I genuinely have no idea how to show that is a partial order.
functions discrete-mathematics equivalence-relations
functions discrete-mathematics equivalence-relations
asked Dec 2 '18 at 23:57
anon2000anon2000
32
asked Dec 2 '18 at 23:57
anon2000anon2000
32
edited Dec 3 '18 at 0:11
anon2000
asked Dec 2 '18 at 23:57
anon2000anon2000
32
asked Dec 2 '18 at 23:57
anon2000anon2000
32
asked Dec 2 '18 at 23:57
anon2000anon2000
32
32
closed as off-topic by amWhy, Jyrki Lahtonen, José Carlos Santos, ancientmathematician, Lord_Farin Dec 3 '18 at 14:30
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Jyrki Lahtonen, José Carlos Santos, ancientmathematician, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Jyrki Lahtonen, José Carlos Santos, ancientmathematician, Lord_Farin Dec 3 '18 at 14:30
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Jyrki Lahtonen, José Carlos Santos, ancientmathematician, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
So can you show that it is transitive, reflexive, and anti-symmetric? Where are you stuck?
$endgroup$
– angryavian
Dec 3 '18 at 0:01
$begingroup$
@angryavian i'm stuck in showing that is is reflexive, transitive and anti-symmetric. would you say that is transitive as (1,2) is an element of R and (2,3) is an element of R, then (1,3) is an element of R?
$endgroup$
– anon2000
Dec 3 '18 at 0:04
add a comment |
$begingroup$
So can you show that it is transitive, reflexive, and anti-symmetric? Where are you stuck?
$endgroup$
– angryavian
Dec 3 '18 at 0:01
$begingroup$
@angryavian i'm stuck in showing that is is reflexive, transitive and anti-symmetric. would you say that is transitive as (1,2) is an element of R and (2,3) is an element of R, then (1,3) is an element of R?
$endgroup$
– anon2000
Dec 3 '18 at 0:04
$begingroup$
So can you show that it is transitive, reflexive, and anti-symmetric? Where are you stuck?
$endgroup$
– angryavian
Dec 3 '18 at 0:01
$begingroup$
So can you show that it is transitive, reflexive, and anti-symmetric? Where are you stuck?
$endgroup$
– angryavian
Dec 3 '18 at 0:01
$begingroup$
@angryavian i'm stuck in showing that is is reflexive, transitive and anti-symmetric. would you say that is transitive as (1,2) is an element of R and (2,3) is an element of R, then (1,3) is an element of R?
$endgroup$
– anon2000
Dec 3 '18 at 0:04
$begingroup$
@angryavian i'm stuck in showing that is is reflexive, transitive and anti-symmetric. would you say that is transitive as (1,2) is an element of R and (2,3) is an element of R, then (1,3) is an element of R?
$endgroup$
– anon2000
Dec 3 '18 at 0:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Reflexive is clear since $a$ divides $a$ for all $a in mathbb{Z}^+$. Now suppose that $a$ divides $b$ and $a neq b$ this means that $b$ does not divide $a$ (try this for yourself). Finally suppose $a$ divides $b$ and $b$ divides $c$. We want to show that $a$ divides $c$. We want to show there exists an integer $z$ such that $az = c$. From our hypothesis for some integers $x,y$ we have that
begin{gather}
ax = b hspace{20 pt} by = c.
end{gather}
Hence we can substitute to see that
begin{gather}
axy = c.
end{gather}
Now $xy$ is a natural number, say $z.$ Hence $az = c$ or $a$ divides $c$. Hence this is a partial ordering. Try as an additional exercise to show where the partial ordering breaks down if we take our set to be all integers instead of just positive ones.
answered Dec 3 '18 at 0:13
JosabanksJosabanks
936
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Reflexive is clear since $a$ divides $a$ for all $a in mathbb{Z}^+$. Now suppose that $a$ divides $b$ and $a neq b$ this means that $b$ does not divide $a$ (try this for yourself). Finally suppose $a$ divides $b$ and $b$ divides $c$. We want to show that $a$ divides $c$. We want to show there exists an integer $z$ such that $az = c$. From our hypothesis for some integers $x,y$ we have that
begin{gather}
ax = b hspace{20 pt} by = c.
end{gather}
Hence we can substitute to see that
begin{gather}
axy = c.
end{gather}
Now $xy$ is a natural number, say $z.$ Hence $az = c$ or $a$ divides $c$. Hence this is a partial ordering. Try as an additional exercise to show where the partial ordering breaks down if we take our set to be all integers instead of just positive ones.
answered Dec 3 '18 at 0:13
JosabanksJosabanks
936
$endgroup$
add a comment |
$begingroup$
Reflexive is clear since $a$ divides $a$ for all $a in mathbb{Z}^+$. Now suppose that $a$ divides $b$ and $a neq b$ this means that $b$ does not divide $a$ (try this for yourself). Finally suppose $a$ divides $b$ and $b$ divides $c$. We want to show that $a$ divides $c$. We want to show there exists an integer $z$ such that $az = c$. From our hypothesis for some integers $x,y$ we have that
begin{gather}
ax = b hspace{20 pt} by = c.
end{gather}
Hence we can substitute to see that
begin{gather}
axy = c.
end{gather}
Now $xy$ is a natural number, say $z.$ Hence $az = c$ or $a$ divides $c$. Hence this is a partial ordering. Try as an additional exercise to show where the partial ordering breaks down if we take our set to be all integers instead of just positive ones.
answered Dec 3 '18 at 0:13
JosabanksJosabanks
936
$endgroup$
add a comment |
$begingroup$
Reflexive is clear since $a$ divides $a$ for all $a in mathbb{Z}^+$. Now suppose that $a$ divides $b$ and $a neq b$ this means that $b$ does not divide $a$ (try this for yourself). Finally suppose $a$ divides $b$ and $b$ divides $c$. We want to show that $a$ divides $c$. We want to show there exists an integer $z$ such that $az = c$. From our hypothesis for some integers $x,y$ we have that
begin{gather}
ax = b hspace{20 pt} by = c.
end{gather}
Hence we can substitute to see that
begin{gather}
axy = c.
end{gather}
Now $xy$ is a natural number, say $z.$ Hence $az = c$ or $a$ divides $c$. Hence this is a partial ordering. Try as an additional exercise to show where the partial ordering breaks down if we take our set to be all integers instead of just positive ones.
answered Dec 3 '18 at 0:13
JosabanksJosabanks
936
$endgroup$
Reflexive is clear since $a$ divides $a$ for all $a in mathbb{Z}^+$. Now suppose that $a$ divides $b$ and $a neq b$ this means that $b$ does not divide $a$ (try this for yourself). Finally suppose $a$ divides $b$ and $b$ divides $c$. We want to show that $a$ divides $c$. We want to show there exists an integer $z$ such that $az = c$. From our hypothesis for some integers $x,y$ we have that
begin{gather}
ax = b hspace{20 pt} by = c.
end{gather}
Hence we can substitute to see that
begin{gather}
axy = c.
end{gather}
Now $xy$ is a natural number, say $z.$ Hence $az = c$ or $a$ divides $c$. Hence this is a partial ordering. Try as an additional exercise to show where the partial ordering breaks down if we take our set to be all integers instead of just positive ones.
answered Dec 3 '18 at 0:13
JosabanksJosabanks
936
edited Dec 3 '18 at 0:20
answered Dec 3 '18 at 0:13
JosabanksJosabanks
936
answered Dec 3 '18 at 0:13
JosabanksJosabanks
936
answered Dec 3 '18 at 0:13
JosabanksJosabanks
936
936
add a comment |
add a comment |
$begingroup$
So can you show that it is transitive, reflexive, and anti-symmetric? Where are you stuck?
$endgroup$
– angryavian
Dec 3 '18 at 0:01
$begingroup$
@angryavian i'm stuck in showing that is is reflexive, transitive and anti-symmetric. would you say that is transitive as (1,2) is an element of R and (2,3) is an element of R, then (1,3) is an element of R?
$endgroup$
– anon2000
Dec 3 '18 at 0:04