Understand math | notation within a problem
The question:
Let R be a relation over the positive integers defined as follows:
$ { (a,b) mid $ gcd$(a,b) > 1 $ but $ a nmid b $ and $b nmid a } $
Determine whether or not R satisfies the following properties. reflexive, irreflexive, symmetric, anti-symmetric, and transitive. Give a brief justification for each of your answers.
My attempt at figuring out what the relation statement meant:
$(a,b)$ exists such that the greatest common divisor of $(a,b)$ is greater than 1. But, a doesn't divide into b and b doesn't divide into a.
notation relations
asked Nov 28 '18 at 1:25
Arthur Green
776
|
show 3 more comments
The question:
Let R be a relation over the positive integers defined as follows:
$ { (a,b) mid $ gcd$(a,b) > 1 $ but $ a nmid b $ and $b nmid a } $
Determine whether or not R satisfies the following properties. reflexive, irreflexive, symmetric, anti-symmetric, and transitive. Give a brief justification for each of your answers.
My attempt at figuring out what the relation statement meant:
$(a,b)$ exists such that the greatest common divisor of $(a,b)$ is greater than 1. But, a doesn't divide into b and b doesn't divide into a.
notation relations
asked Nov 28 '18 at 1:25
Arthur Green
776
What‘s the question?
– Lukas Kofler
Nov 28 '18 at 1:28
1
That's a correct interpretation yes.
– Ovi
Nov 28 '18 at 1:28
1
$R$ is a relation defined as the set of all ordered pairs of positive integers $(a,b)$ such that $a,b$ satisfy the property that the greatest common divisor of $a$ and $b$ is greater than $1$ while also satisfying the property that $a$ is not a multiple of $b$ as well as $b$ is not a multiple of $a$.
– JMoravitz
Nov 28 '18 at 1:28
@LukasKofler I posted the question, but I haven't given it a shot as I wasn't sure if my interpretation of the relation statement was correct.
– Arthur Green
Nov 28 '18 at 1:31
2
Examples of pairs in the relation would be things such as $(6,9)$, $(15,10)$ and $(20,15)$ etc... while examples of pairs not in the relation would be things such as $(1,5)$, $(2,7)$, $(5,5)$ and $(10,20)$
– JMoravitz
Nov 28 '18 at 1:31
|
show 3 more comments
The question:
Let R be a relation over the positive integers defined as follows:
$ { (a,b) mid $ gcd$(a,b) > 1 $ but $ a nmid b $ and $b nmid a } $
Determine whether or not R satisfies the following properties. reflexive, irreflexive, symmetric, anti-symmetric, and transitive. Give a brief justification for each of your answers.
My attempt at figuring out what the relation statement meant:
$(a,b)$ exists such that the greatest common divisor of $(a,b)$ is greater than 1. But, a doesn't divide into b and b doesn't divide into a.
notation relations
asked Nov 28 '18 at 1:25
Arthur Green
776
The question:
Let R be a relation over the positive integers defined as follows:
$ { (a,b) mid $ gcd$(a,b) > 1 $ but $ a nmid b $ and $b nmid a } $
Determine whether or not R satisfies the following properties. reflexive, irreflexive, symmetric, anti-symmetric, and transitive. Give a brief justification for each of your answers.
My attempt at figuring out what the relation statement meant:
$(a,b)$ exists such that the greatest common divisor of $(a,b)$ is greater than 1. But, a doesn't divide into b and b doesn't divide into a.
notation relations
notation relations
asked Nov 28 '18 at 1:25
Arthur Green
776
asked Nov 28 '18 at 1:25
Arthur Green
776
edited Nov 28 '18 at 1:29
asked Nov 28 '18 at 1:25
Arthur Green
776
asked Nov 28 '18 at 1:25
Arthur Green
776
asked Nov 28 '18 at 1:25
Arthur Green
776
776
What‘s the question?
– Lukas Kofler
Nov 28 '18 at 1:28
1
That's a correct interpretation yes.
– Ovi
Nov 28 '18 at 1:28
1
$R$ is a relation defined as the set of all ordered pairs of positive integers $(a,b)$ such that $a,b$ satisfy the property that the greatest common divisor of $a$ and $b$ is greater than $1$ while also satisfying the property that $a$ is not a multiple of $b$ as well as $b$ is not a multiple of $a$.
– JMoravitz
Nov 28 '18 at 1:28
@LukasKofler I posted the question, but I haven't given it a shot as I wasn't sure if my interpretation of the relation statement was correct.
– Arthur Green
Nov 28 '18 at 1:31
2
Examples of pairs in the relation would be things such as $(6,9)$, $(15,10)$ and $(20,15)$ etc... while examples of pairs not in the relation would be things such as $(1,5)$, $(2,7)$, $(5,5)$ and $(10,20)$
– JMoravitz
Nov 28 '18 at 1:31
|
show 3 more comments
What‘s the question?
– Lukas Kofler
Nov 28 '18 at 1:28
1
That's a correct interpretation yes.
– Ovi
Nov 28 '18 at 1:28
1
$R$ is a relation defined as the set of all ordered pairs of positive integers $(a,b)$ such that $a,b$ satisfy the property that the greatest common divisor of $a$ and $b$ is greater than $1$ while also satisfying the property that $a$ is not a multiple of $b$ as well as $b$ is not a multiple of $a$.
– JMoravitz
Nov 28 '18 at 1:28
@LukasKofler I posted the question, but I haven't given it a shot as I wasn't sure if my interpretation of the relation statement was correct.
– Arthur Green
Nov 28 '18 at 1:31
2
Examples of pairs in the relation would be things such as $(6,9)$, $(15,10)$ and $(20,15)$ etc... while examples of pairs not in the relation would be things such as $(1,5)$, $(2,7)$, $(5,5)$ and $(10,20)$
– JMoravitz
Nov 28 '18 at 1:31
What‘s the question?
– Lukas Kofler
Nov 28 '18 at 1:28
What‘s the question?
– Lukas Kofler
Nov 28 '18 at 1:28
1
1
That's a correct interpretation yes.
– Ovi
Nov 28 '18 at 1:28
That's a correct interpretation yes.
– Ovi
Nov 28 '18 at 1:28
1
1
$R$ is a relation defined as the set of all ordered pairs of positive integers $(a,b)$ such that $a,b$ satisfy the property that the greatest common divisor of $a$ and $b$ is greater than $1$ while also satisfying the property that $a$ is not a multiple of $b$ as well as $b$ is not a multiple of $a$.
– JMoravitz
Nov 28 '18 at 1:28
$R$ is a relation defined as the set of all ordered pairs of positive integers $(a,b)$ such that $a,b$ satisfy the property that the greatest common divisor of $a$ and $b$ is greater than $1$ while also satisfying the property that $a$ is not a multiple of $b$ as well as $b$ is not a multiple of $a$.
– JMoravitz
Nov 28 '18 at 1:28
@LukasKofler I posted the question, but I haven't given it a shot as I wasn't sure if my interpretation of the relation statement was correct.
– Arthur Green
Nov 28 '18 at 1:31
@LukasKofler I posted the question, but I haven't given it a shot as I wasn't sure if my interpretation of the relation statement was correct.
– Arthur Green
Nov 28 '18 at 1:31
2
2
Examples of pairs in the relation would be things such as $(6,9)$, $(15,10)$ and $(20,15)$ etc... while examples of pairs not in the relation would be things such as $(1,5)$, $(2,7)$, $(5,5)$ and $(10,20)$
– JMoravitz
Nov 28 '18 at 1:31
Examples of pairs in the relation would be things such as $(6,9)$, $(15,10)$ and $(20,15)$ etc... while examples of pairs not in the relation would be things such as $(1,5)$, $(2,7)$, $(5,5)$ and $(10,20)$
– JMoravitz
Nov 28 '18 at 1:31
|
show 3 more comments
1 Answer
1
active
oldest
votes
Almost. I wouldn't say "$(a,b)$ exists".
This relation is the set of pairs with gcd greater than $1$ where neither divides the other. So it contains $(6,15)$ but not $(6,18)$ and of course not $(6,7)$. (Writing down a few examples is always a good way to test your understanding.)
answered Nov 28 '18 at 1:33
Ethan Bolker
41.4k547108
2
I would then say R is NOT reflexive as $(a,a) notin R$ for all $a in A$. If $(a,a)$ existed it would divide itself.
– Arthur Green
Nov 28 '18 at 1:37
1
You're right - it's not reflexive.
– Ethan Bolker
Nov 28 '18 at 1:38
Would it make sense to say that the relation is: Not anti-symmetric as as $(a,b) wedge (b,a) exists in R$ I have only seen the $exists$ symbol in front of the pairs before. Usually with a $forall$
– Arthur Green
Nov 29 '18 at 14:47
add a comment |
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1 Answer
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Almost. I wouldn't say "$(a,b)$ exists".
This relation is the set of pairs with gcd greater than $1$ where neither divides the other. So it contains $(6,15)$ but not $(6,18)$ and of course not $(6,7)$. (Writing down a few examples is always a good way to test your understanding.)
answered Nov 28 '18 at 1:33
Ethan Bolker
41.4k547108
2
I would then say R is NOT reflexive as $(a,a) notin R$ for all $a in A$. If $(a,a)$ existed it would divide itself.
– Arthur Green
Nov 28 '18 at 1:37
1
You're right - it's not reflexive.
– Ethan Bolker
Nov 28 '18 at 1:38
Would it make sense to say that the relation is: Not anti-symmetric as as $(a,b) wedge (b,a) exists in R$ I have only seen the $exists$ symbol in front of the pairs before. Usually with a $forall$
– Arthur Green
Nov 29 '18 at 14:47
add a comment |
Almost. I wouldn't say "$(a,b)$ exists".
This relation is the set of pairs with gcd greater than $1$ where neither divides the other. So it contains $(6,15)$ but not $(6,18)$ and of course not $(6,7)$. (Writing down a few examples is always a good way to test your understanding.)
answered Nov 28 '18 at 1:33
Ethan Bolker
41.4k547108
2
I would then say R is NOT reflexive as $(a,a) notin R$ for all $a in A$. If $(a,a)$ existed it would divide itself.
– Arthur Green
Nov 28 '18 at 1:37
1
You're right - it's not reflexive.
– Ethan Bolker
Nov 28 '18 at 1:38
Would it make sense to say that the relation is: Not anti-symmetric as as $(a,b) wedge (b,a) exists in R$ I have only seen the $exists$ symbol in front of the pairs before. Usually with a $forall$
– Arthur Green
Nov 29 '18 at 14:47
add a comment |
Almost. I wouldn't say "$(a,b)$ exists".
This relation is the set of pairs with gcd greater than $1$ where neither divides the other. So it contains $(6,15)$ but not $(6,18)$ and of course not $(6,7)$. (Writing down a few examples is always a good way to test your understanding.)
answered Nov 28 '18 at 1:33
Ethan Bolker
41.4k547108
Almost. I wouldn't say "$(a,b)$ exists".
This relation is the set of pairs with gcd greater than $1$ where neither divides the other. So it contains $(6,15)$ but not $(6,18)$ and of course not $(6,7)$. (Writing down a few examples is always a good way to test your understanding.)
answered Nov 28 '18 at 1:33
Ethan Bolker
41.4k547108
answered Nov 28 '18 at 1:33
Ethan Bolker
41.4k547108
answered Nov 28 '18 at 1:33
Ethan Bolker
41.4k547108
answered Nov 28 '18 at 1:33
Ethan Bolker
41.4k547108
41.4k547108
2
I would then say R is NOT reflexive as $(a,a) notin R$ for all $a in A$. If $(a,a)$ existed it would divide itself.
– Arthur Green
Nov 28 '18 at 1:37
1
You're right - it's not reflexive.
– Ethan Bolker
Nov 28 '18 at 1:38
Would it make sense to say that the relation is: Not anti-symmetric as as $(a,b) wedge (b,a) exists in R$ I have only seen the $exists$ symbol in front of the pairs before. Usually with a $forall$
– Arthur Green
Nov 29 '18 at 14:47
add a comment |
2
I would then say R is NOT reflexive as $(a,a) notin R$ for all $a in A$. If $(a,a)$ existed it would divide itself.
– Arthur Green
Nov 28 '18 at 1:37
1
You're right - it's not reflexive.
– Ethan Bolker
Nov 28 '18 at 1:38
Would it make sense to say that the relation is: Not anti-symmetric as as $(a,b) wedge (b,a) exists in R$ I have only seen the $exists$ symbol in front of the pairs before. Usually with a $forall$
– Arthur Green
Nov 29 '18 at 14:47
2
2
I would then say R is NOT reflexive as $(a,a) notin R$ for all $a in A$. If $(a,a)$ existed it would divide itself.
– Arthur Green
Nov 28 '18 at 1:37
I would then say R is NOT reflexive as $(a,a) notin R$ for all $a in A$. If $(a,a)$ existed it would divide itself.
– Arthur Green
Nov 28 '18 at 1:37
1
1
You're right - it's not reflexive.
– Ethan Bolker
Nov 28 '18 at 1:38
You're right - it's not reflexive.
– Ethan Bolker
Nov 28 '18 at 1:38
Would it make sense to say that the relation is: Not anti-symmetric as as $(a,b) wedge (b,a) exists in R$ I have only seen the $exists$ symbol in front of the pairs before. Usually with a $forall$
– Arthur Green
Nov 29 '18 at 14:47
Would it make sense to say that the relation is: Not anti-symmetric as as $(a,b) wedge (b,a) exists in R$ I have only seen the $exists$ symbol in front of the pairs before. Usually with a $forall$
– Arthur Green
Nov 29 '18 at 14:47
add a comment |
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What‘s the question?
– Lukas Kofler
Nov 28 '18 at 1:28
1
That's a correct interpretation yes.
– Ovi
Nov 28 '18 at 1:28
1
$R$ is a relation defined as the set of all ordered pairs of positive integers $(a,b)$ such that $a,b$ satisfy the property that the greatest common divisor of $a$ and $b$ is greater than $1$ while also satisfying the property that $a$ is not a multiple of $b$ as well as $b$ is not a multiple of $a$.
– JMoravitz
Nov 28 '18 at 1:28
@LukasKofler I posted the question, but I haven't given it a shot as I wasn't sure if my interpretation of the relation statement was correct.
– Arthur Green
Nov 28 '18 at 1:31
2
Examples of pairs in the relation would be things such as $(6,9)$, $(15,10)$ and $(20,15)$ etc... while examples of pairs not in the relation would be things such as $(1,5)$, $(2,7)$, $(5,5)$ and $(10,20)$
– JMoravitz
Nov 28 '18 at 1:31