Mixing
![$sin(x)+cos(x)+sin(2x)>1$ [closed]](https://lh3.googleusercontent.com/-yoXg4bOFDf4/AAAAAAAAAAI/AAAAAAAAAFA/SaQb6f8L_yg/s72-c/photo.jpg?sz=32)
Need help finding the equivalence relation!
$begingroup$
"Let A be the set of all people in the world, R a relation on A defined by (a, b) ∈ R, if and only if a is a twin of b. Is this relation an equivalence relation? Explain!"
An equivalence relation is a reflexive, symmetric and transitive relation.
A = set of all people in the world
R = {(a,b)| a and b are twins}
Okay so i thought this:
This relation is not reflexive (a,a) or (b,b) because I am not my twin.
But, it is symmetric because (a,b) and (b,a), that means I am my twin's twin and he/she is also my twin.
Transitive in this case would basically mean reflexive, but like I said that is false.
Therefore my relation is not an equivalence relation...
Is my argument correct?
relations equivalence-relations
asked Jan 24 at 10:25
Sofia Sofia
62
$endgroup$
add a comment |
$begingroup$
"Let A be the set of all people in the world, R a relation on A defined by (a, b) ∈ R, if and only if a is a twin of b. Is this relation an equivalence relation? Explain!"
An equivalence relation is a reflexive, symmetric and transitive relation.
A = set of all people in the world
R = {(a,b)| a and b are twins}
Okay so i thought this:
This relation is not reflexive (a,a) or (b,b) because I am not my twin.
But, it is symmetric because (a,b) and (b,a), that means I am my twin's twin and he/she is also my twin.
Transitive in this case would basically mean reflexive, but like I said that is false.
Therefore my relation is not an equivalence relation...
Is my argument correct?
relations equivalence-relations
asked Jan 24 at 10:25
Sofia Sofia
62
$endgroup$
$begingroup$
Related to your point about transitivity. You are right in this case, the relationship is neither reflexive nor transitive.
$endgroup$
– RcnSc
Jan 24 at 10:31
add a comment |
$begingroup$
"Let A be the set of all people in the world, R a relation on A defined by (a, b) ∈ R, if and only if a is a twin of b. Is this relation an equivalence relation? Explain!"
An equivalence relation is a reflexive, symmetric and transitive relation.
A = set of all people in the world
R = {(a,b)| a and b are twins}
Okay so i thought this:
This relation is not reflexive (a,a) or (b,b) because I am not my twin.
But, it is symmetric because (a,b) and (b,a), that means I am my twin's twin and he/she is also my twin.
Transitive in this case would basically mean reflexive, but like I said that is false.
Therefore my relation is not an equivalence relation...
Is my argument correct?
relations equivalence-relations
asked Jan 24 at 10:25
Sofia Sofia
62
$endgroup$
"Let A be the set of all people in the world, R a relation on A defined by (a, b) ∈ R, if and only if a is a twin of b. Is this relation an equivalence relation? Explain!"
An equivalence relation is a reflexive, symmetric and transitive relation.
A = set of all people in the world
R = {(a,b)| a and b are twins}
Okay so i thought this:
This relation is not reflexive (a,a) or (b,b) because I am not my twin.
But, it is symmetric because (a,b) and (b,a), that means I am my twin's twin and he/she is also my twin.
Transitive in this case would basically mean reflexive, but like I said that is false.
Therefore my relation is not an equivalence relation...
Is my argument correct?
relations equivalence-relations
relations equivalence-relations
asked Jan 24 at 10:25
Sofia Sofia
62
asked Jan 24 at 10:25
Sofia Sofia
62
asked Jan 24 at 10:25
Sofia Sofia
62
asked Jan 24 at 10:25
Sofia Sofia
62
asked Jan 24 at 10:25
Sofia Sofia
62
62
$begingroup$
Related to your point about transitivity. You are right in this case, the relationship is neither reflexive nor transitive.
$endgroup$
– RcnSc
Jan 24 at 10:31
add a comment |
$begingroup$
Related to your point about transitivity. You are right in this case, the relationship is neither reflexive nor transitive.
$endgroup$
– RcnSc
Jan 24 at 10:31
$begingroup$
Related to your point about transitivity. You are right in this case, the relationship is neither reflexive nor transitive.
$endgroup$
– RcnSc
Jan 24 at 10:31
$begingroup$
Related to your point about transitivity. You are right in this case, the relationship is neither reflexive nor transitive.
$endgroup$
– RcnSc
Jan 24 at 10:31
add a comment |
1 Answer
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$begingroup$
Yes, your reasoning is correct.
BTW, with the alternative definition of $(a, b) ∈ R$ if and only if $a$ is a twin of $b$ or $a$ and $b$ are the same person, the relation $R$ becomes an equivalence relation.
answered Jan 24 at 10:45
Reiner MartinReiner Martin
3,509414
$endgroup$
add a comment |
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$begingroup$
Yes, your reasoning is correct.
BTW, with the alternative definition of $(a, b) ∈ R$ if and only if $a$ is a twin of $b$ or $a$ and $b$ are the same person, the relation $R$ becomes an equivalence relation.
answered Jan 24 at 10:45
Reiner MartinReiner Martin
3,509414
$endgroup$
add a comment |
$begingroup$
Yes, your reasoning is correct.
BTW, with the alternative definition of $(a, b) ∈ R$ if and only if $a$ is a twin of $b$ or $a$ and $b$ are the same person, the relation $R$ becomes an equivalence relation.
answered Jan 24 at 10:45
Reiner MartinReiner Martin
3,509414
$endgroup$
add a comment |
$begingroup$
Yes, your reasoning is correct.
BTW, with the alternative definition of $(a, b) ∈ R$ if and only if $a$ is a twin of $b$ or $a$ and $b$ are the same person, the relation $R$ becomes an equivalence relation.
answered Jan 24 at 10:45
Reiner MartinReiner Martin
3,509414
$endgroup$
Yes, your reasoning is correct.
BTW, with the alternative definition of $(a, b) ∈ R$ if and only if $a$ is a twin of $b$ or $a$ and $b$ are the same person, the relation $R$ becomes an equivalence relation.
answered Jan 24 at 10:45
Reiner MartinReiner Martin
3,509414
edited Jan 24 at 23:16
answered Jan 24 at 10:45
Reiner MartinReiner Martin
3,509414
answered Jan 24 at 10:45
Reiner MartinReiner Martin
3,509414
answered Jan 24 at 10:45
Reiner MartinReiner Martin
3,509414
3,509414
add a comment |
add a comment |
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$begingroup$
Related to your point about transitivity. You are right in this case, the relationship is neither reflexive nor transitive.
$endgroup$
– RcnSc
Jan 24 at 10:31