Mixing
Show that $R$ is an order relation in $mathbb{R^2}$
In $mathbb{R^2}$ consider the relation $R$ as: $(x,y)R(x',y') Longleftrightarrow xleq x' and y leq y'$
show that $R$ is an order relation.
1) $ forall (x,y) in mathbb{R^2} (x,y)R(x,y) implies x leq x and y leq y$
which is true because $x = x$ and $ y = y$ $ implies $ $R$ is reflexive.
2) $( (x,y)R(x',y') implies x leq x' and y' leq y )$ and $( (x',y')R(x,y) implies x geq x' and y' geq y )$
which implies $x = x'$ and $ y = y' $ $implies$ $R$ is anti-symmetric
3) $( (x,y)R(x',y') implies x leq x' and y' leq y )$ and $( (x',y')R(z,z') implies x' leq z and y' leq z' )$ $implies x leq z and y leq z'$
which implies that $R$ is transitive
So $R$ is an order relation.
is this answer correct ?
proof-verification relations alternative-proof
asked Nov 20 '18 at 23:28
Fibi
484
|
show 1 more comment
In $mathbb{R^2}$ consider the relation $R$ as: $(x,y)R(x',y') Longleftrightarrow xleq x' and y leq y'$
show that $R$ is an order relation.
1) $ forall (x,y) in mathbb{R^2} (x,y)R(x,y) implies x leq x and y leq y$
which is true because $x = x$ and $ y = y$ $ implies $ $R$ is reflexive.
2) $( (x,y)R(x',y') implies x leq x' and y' leq y )$ and $( (x',y')R(x,y) implies x geq x' and y' geq y )$
which implies $x = x'$ and $ y = y' $ $implies$ $R$ is anti-symmetric
3) $( (x,y)R(x',y') implies x leq x' and y' leq y )$ and $( (x',y')R(z,z') implies x' leq z and y' leq z' )$ $implies x leq z and y leq z'$
which implies that $R$ is transitive
So $R$ is an order relation.
is this answer correct ?
proof-verification relations alternative-proof
asked Nov 20 '18 at 23:28
Fibi
484
What is your question? Are you wondering whether you are correct?
– Batominovski
Nov 20 '18 at 23:30
@Batominovski the post has a "proof-verification" tag so the answer is yes.
– Fibi
Nov 20 '18 at 23:32
1
There should be a question in your text no matter what. The proof-verification tag is sometimes used to ask about unclear points in certain proofs. Don't let readers read your mind.
– Batominovski
Nov 20 '18 at 23:33
1
While I do feel like your proof would be better communicated with more words that would clarify some of the underlying details, the core logic is intact. I guess you could say that the proof is "right," it would just be easier to parse through with more actual explanation. Granted, this might be a personal thing. $R$ indeed does form an order relation, specifically a partial order.
– Eevee Trainer
Nov 20 '18 at 23:34
1
At a glance you seem to have the right ideas, however you have written things in incorrect orders. For reflexivity for example it should read "Suppose $(x,y)in Bbb R^2$. Then since $xleq x$ and $yleq y$ is true it follows that $(x,y)R(x,y)$ and therefore $R$ is reflexive." You instead began with what you wanted to prove (or at least didn't make clear that this was the end goal rather than an assumption or a leap in logic).
– JMoravitz
Nov 20 '18 at 23:37
|
show 1 more comment
In $mathbb{R^2}$ consider the relation $R$ as: $(x,y)R(x',y') Longleftrightarrow xleq x' and y leq y'$
show that $R$ is an order relation.
1) $ forall (x,y) in mathbb{R^2} (x,y)R(x,y) implies x leq x and y leq y$
which is true because $x = x$ and $ y = y$ $ implies $ $R$ is reflexive.
2) $( (x,y)R(x',y') implies x leq x' and y' leq y )$ and $( (x',y')R(x,y) implies x geq x' and y' geq y )$
which implies $x = x'$ and $ y = y' $ $implies$ $R$ is anti-symmetric
3) $( (x,y)R(x',y') implies x leq x' and y' leq y )$ and $( (x',y')R(z,z') implies x' leq z and y' leq z' )$ $implies x leq z and y leq z'$
which implies that $R$ is transitive
So $R$ is an order relation.
is this answer correct ?
proof-verification relations alternative-proof
asked Nov 20 '18 at 23:28
Fibi
484
In $mathbb{R^2}$ consider the relation $R$ as: $(x,y)R(x',y') Longleftrightarrow xleq x' and y leq y'$
show that $R$ is an order relation.
1) $ forall (x,y) in mathbb{R^2} (x,y)R(x,y) implies x leq x and y leq y$
which is true because $x = x$ and $ y = y$ $ implies $ $R$ is reflexive.
2) $( (x,y)R(x',y') implies x leq x' and y' leq y )$ and $( (x',y')R(x,y) implies x geq x' and y' geq y )$
which implies $x = x'$ and $ y = y' $ $implies$ $R$ is anti-symmetric
3) $( (x,y)R(x',y') implies x leq x' and y' leq y )$ and $( (x',y')R(z,z') implies x' leq z and y' leq z' )$ $implies x leq z and y leq z'$
which implies that $R$ is transitive
So $R$ is an order relation.
is this answer correct ?
proof-verification relations alternative-proof
proof-verification relations alternative-proof
asked Nov 20 '18 at 23:28
Fibi
484
asked Nov 20 '18 at 23:28
Fibi
484
edited Nov 20 '18 at 23:34
asked Nov 20 '18 at 23:28
Fibi
484
asked Nov 20 '18 at 23:28
Fibi
484
asked Nov 20 '18 at 23:28
Fibi
484
484
What is your question? Are you wondering whether you are correct?
– Batominovski
Nov 20 '18 at 23:30
@Batominovski the post has a "proof-verification" tag so the answer is yes.
– Fibi
Nov 20 '18 at 23:32
1
There should be a question in your text no matter what. The proof-verification tag is sometimes used to ask about unclear points in certain proofs. Don't let readers read your mind.
– Batominovski
Nov 20 '18 at 23:33
1
While I do feel like your proof would be better communicated with more words that would clarify some of the underlying details, the core logic is intact. I guess you could say that the proof is "right," it would just be easier to parse through with more actual explanation. Granted, this might be a personal thing. $R$ indeed does form an order relation, specifically a partial order.
– Eevee Trainer
Nov 20 '18 at 23:34
1
At a glance you seem to have the right ideas, however you have written things in incorrect orders. For reflexivity for example it should read "Suppose $(x,y)in Bbb R^2$. Then since $xleq x$ and $yleq y$ is true it follows that $(x,y)R(x,y)$ and therefore $R$ is reflexive." You instead began with what you wanted to prove (or at least didn't make clear that this was the end goal rather than an assumption or a leap in logic).
– JMoravitz
Nov 20 '18 at 23:37
|
show 1 more comment
What is your question? Are you wondering whether you are correct?
– Batominovski
Nov 20 '18 at 23:30
@Batominovski the post has a "proof-verification" tag so the answer is yes.
– Fibi
Nov 20 '18 at 23:32
1
There should be a question in your text no matter what. The proof-verification tag is sometimes used to ask about unclear points in certain proofs. Don't let readers read your mind.
– Batominovski
Nov 20 '18 at 23:33
1
While I do feel like your proof would be better communicated with more words that would clarify some of the underlying details, the core logic is intact. I guess you could say that the proof is "right," it would just be easier to parse through with more actual explanation. Granted, this might be a personal thing. $R$ indeed does form an order relation, specifically a partial order.
– Eevee Trainer
Nov 20 '18 at 23:34
1
At a glance you seem to have the right ideas, however you have written things in incorrect orders. For reflexivity for example it should read "Suppose $(x,y)in Bbb R^2$. Then since $xleq x$ and $yleq y$ is true it follows that $(x,y)R(x,y)$ and therefore $R$ is reflexive." You instead began with what you wanted to prove (or at least didn't make clear that this was the end goal rather than an assumption or a leap in logic).
– JMoravitz
Nov 20 '18 at 23:37
What is your question? Are you wondering whether you are correct?
– Batominovski
Nov 20 '18 at 23:30
What is your question? Are you wondering whether you are correct?
– Batominovski
Nov 20 '18 at 23:30
@Batominovski the post has a "proof-verification" tag so the answer is yes.
– Fibi
Nov 20 '18 at 23:32
@Batominovski the post has a "proof-verification" tag so the answer is yes.
– Fibi
Nov 20 '18 at 23:32
1
1
There should be a question in your text no matter what. The proof-verification tag is sometimes used to ask about unclear points in certain proofs. Don't let readers read your mind.
– Batominovski
Nov 20 '18 at 23:33
There should be a question in your text no matter what. The proof-verification tag is sometimes used to ask about unclear points in certain proofs. Don't let readers read your mind.
– Batominovski
Nov 20 '18 at 23:33
1
1
While I do feel like your proof would be better communicated with more words that would clarify some of the underlying details, the core logic is intact. I guess you could say that the proof is "right," it would just be easier to parse through with more actual explanation. Granted, this might be a personal thing. $R$ indeed does form an order relation, specifically a partial order.
– Eevee Trainer
Nov 20 '18 at 23:34
While I do feel like your proof would be better communicated with more words that would clarify some of the underlying details, the core logic is intact. I guess you could say that the proof is "right," it would just be easier to parse through with more actual explanation. Granted, this might be a personal thing. $R$ indeed does form an order relation, specifically a partial order.
– Eevee Trainer
Nov 20 '18 at 23:34
1
1
At a glance you seem to have the right ideas, however you have written things in incorrect orders. For reflexivity for example it should read "Suppose $(x,y)in Bbb R^2$. Then since $xleq x$ and $yleq y$ is true it follows that $(x,y)R(x,y)$ and therefore $R$ is reflexive." You instead began with what you wanted to prove (or at least didn't make clear that this was the end goal rather than an assumption or a leap in logic).
– JMoravitz
Nov 20 '18 at 23:37
At a glance you seem to have the right ideas, however you have written things in incorrect orders. For reflexivity for example it should read "Suppose $(x,y)in Bbb R^2$. Then since $xleq x$ and $yleq y$ is true it follows that $(x,y)R(x,y)$ and therefore $R$ is reflexive." You instead began with what you wanted to prove (or at least didn't make clear that this was the end goal rather than an assumption or a leap in logic).
– JMoravitz
Nov 20 '18 at 23:37
|
show 1 more comment
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What is your question? Are you wondering whether you are correct?
– Batominovski
Nov 20 '18 at 23:30
@Batominovski the post has a "proof-verification" tag so the answer is yes.
– Fibi
Nov 20 '18 at 23:32
1
There should be a question in your text no matter what. The proof-verification tag is sometimes used to ask about unclear points in certain proofs. Don't let readers read your mind.
– Batominovski
Nov 20 '18 at 23:33
1
While I do feel like your proof would be better communicated with more words that would clarify some of the underlying details, the core logic is intact. I guess you could say that the proof is "right," it would just be easier to parse through with more actual explanation. Granted, this might be a personal thing. $R$ indeed does form an order relation, specifically a partial order.
– Eevee Trainer
Nov 20 '18 at 23:34
1
At a glance you seem to have the right ideas, however you have written things in incorrect orders. For reflexivity for example it should read "Suppose $(x,y)in Bbb R^2$. Then since $xleq x$ and $yleq y$ is true it follows that $(x,y)R(x,y)$ and therefore $R$ is reflexive." You instead began with what you wanted to prove (or at least didn't make clear that this was the end goal rather than an assumption or a leap in logic).
– JMoravitz
Nov 20 '18 at 23:37