Let S be the set of nonzero integers. Define a relation R on S by letting aRb mean that b/a is an integer. Is R an antisymmetric relation on S?Equivalence Classes and such for the Relation: x and y have the same first and third letter.How to prove for each positive integer $n$, the sum of the first $n$ odd positive integers is $n^2$?Antisymmetric Relation: How can I use the formal definition?Binary Search 2Log(n)+1 steps?Questions on equivalence relation and functionsDeteremining whether the relation R on the set of all Real Numbers is Reflexive, Symmetric, Antisymmetric, Transitive, and/or IrreflexiveIf R is a relation, then what is $R^0$?how many equivalance classes are in the functionDetermining whether the relation R on the set of all web pages is reflexive, symmetric, antisymmetric or TransitiveRelation that is transitive, symmetric but not antisymmetric nor reflexive
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Let S be the set of nonzero integers. Define a relation R on S by letting aRb mean that b/a is an integer. Is R an antisymmetric relation on S?
Equivalence Classes and such for the Relation: x and y have the same first and third letter.How to prove for each positive integer $n$, the sum of the first $n$ odd positive integers is $n^2$?Antisymmetric Relation: How can I use the formal definition?Binary Search 2Log(n)+1 steps?Questions on equivalence relation and functionsDeteremining whether the relation R on the set of all Real Numbers is Reflexive, Symmetric, Antisymmetric, Transitive, and/or IrreflexiveIf R is a relation, then what is $R^0$?how many equivalance classes are in the functionDetermining whether the relation R on the set of all web pages is reflexive, symmetric, antisymmetric or TransitiveRelation that is transitive, symmetric but not antisymmetric nor reflexive
$begingroup$
A question from the back of my Discrete Math classes textbook. I cant think of any example where (a,b) belongs to R and (b,a) belongs to R but a $neq$ b.
The answer in the back literally just says "no".
This is my first question on here please be gentle.
discrete-mathematics relations
asked Mar 14 at 17:15
CheckMateSergeiCheckMateSergei
12
$endgroup$
add a comment |
$begingroup$
A question from the back of my Discrete Math classes textbook. I cant think of any example where (a,b) belongs to R and (b,a) belongs to R but a $neq$ b.
The answer in the back literally just says "no".
This is my first question on here please be gentle.
discrete-mathematics relations
asked Mar 14 at 17:15
CheckMateSergeiCheckMateSergei
12
$endgroup$
$begingroup$
Literally write out a few examples. What do you observe?
$endgroup$
– Robert Shore
Mar 14 at 17:18
add a comment |
$begingroup$
A question from the back of my Discrete Math classes textbook. I cant think of any example where (a,b) belongs to R and (b,a) belongs to R but a $neq$ b.
The answer in the back literally just says "no".
This is my first question on here please be gentle.
discrete-mathematics relations
asked Mar 14 at 17:15
CheckMateSergeiCheckMateSergei
12
$endgroup$
A question from the back of my Discrete Math classes textbook. I cant think of any example where (a,b) belongs to R and (b,a) belongs to R but a $neq$ b.
The answer in the back literally just says "no".
This is my first question on here please be gentle.
discrete-mathematics relations
discrete-mathematics relations
asked Mar 14 at 17:15
CheckMateSergeiCheckMateSergei
12
asked Mar 14 at 17:15
CheckMateSergeiCheckMateSergei
12
asked Mar 14 at 17:15
CheckMateSergeiCheckMateSergei
12
asked Mar 14 at 17:15
CheckMateSergeiCheckMateSergei
12
asked Mar 14 at 17:15
CheckMateSergeiCheckMateSergei
12
12
$begingroup$
Literally write out a few examples. What do you observe?
$endgroup$
– Robert Shore
Mar 14 at 17:18
add a comment |
$begingroup$
Literally write out a few examples. What do you observe?
$endgroup$
– Robert Shore
Mar 14 at 17:18
$begingroup$
Literally write out a few examples. What do you observe?
$endgroup$
– Robert Shore
Mar 14 at 17:18
$begingroup$
Literally write out a few examples. What do you observe?
$endgroup$
– Robert Shore
Mar 14 at 17:18
add a comment |
1 Answer
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$begingroup$
Welcome to MSE! The answer is ''depends''.
Define the relation $aRb$ iff $a$ divides $b$, i.e., there exists $c$ with $acdot c=b$.
If the set on which the relation is defined is the set of integers, then $-1$ divides $1$ and $1$ divides $-1$, but $1ne -1$.
If the set on which the relation is defined is the set of natural numbers, then the relation is antisymmetric.
answered Mar 14 at 17:20
WuestenfuxWuestenfux
5,2731513
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Welcome to MSE! The answer is ''depends''.
Define the relation $aRb$ iff $a$ divides $b$, i.e., there exists $c$ with $acdot c=b$.
If the set on which the relation is defined is the set of integers, then $-1$ divides $1$ and $1$ divides $-1$, but $1ne -1$.
If the set on which the relation is defined is the set of natural numbers, then the relation is antisymmetric.
answered Mar 14 at 17:20
WuestenfuxWuestenfux
5,2731513
$endgroup$
add a comment |
$begingroup$
Welcome to MSE! The answer is ''depends''.
Define the relation $aRb$ iff $a$ divides $b$, i.e., there exists $c$ with $acdot c=b$.
If the set on which the relation is defined is the set of integers, then $-1$ divides $1$ and $1$ divides $-1$, but $1ne -1$.
If the set on which the relation is defined is the set of natural numbers, then the relation is antisymmetric.
answered Mar 14 at 17:20
WuestenfuxWuestenfux
5,2731513
$endgroup$
add a comment |
$begingroup$
Welcome to MSE! The answer is ''depends''.
Define the relation $aRb$ iff $a$ divides $b$, i.e., there exists $c$ with $acdot c=b$.
If the set on which the relation is defined is the set of integers, then $-1$ divides $1$ and $1$ divides $-1$, but $1ne -1$.
If the set on which the relation is defined is the set of natural numbers, then the relation is antisymmetric.
answered Mar 14 at 17:20
WuestenfuxWuestenfux
5,2731513
$endgroup$
Welcome to MSE! The answer is ''depends''.
Define the relation $aRb$ iff $a$ divides $b$, i.e., there exists $c$ with $acdot c=b$.
If the set on which the relation is defined is the set of integers, then $-1$ divides $1$ and $1$ divides $-1$, but $1ne -1$.
If the set on which the relation is defined is the set of natural numbers, then the relation is antisymmetric.
answered Mar 14 at 17:20
WuestenfuxWuestenfux
5,2731513
answered Mar 14 at 17:20
WuestenfuxWuestenfux
5,2731513
answered Mar 14 at 17:20
WuestenfuxWuestenfux
5,2731513
answered Mar 14 at 17:20
WuestenfuxWuestenfux
5,2731513
5,2731513
add a comment |
add a comment |
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$begingroup$
Literally write out a few examples. What do you observe?
$endgroup$
– Robert Shore
Mar 14 at 17:18