Mixing
Proof of equivalence relations
$begingroup$
I have just started my math class and I think I might not be completely understanding the 3 properties of a equivalence relation: reflexivity, symmetry, transitivity.
I have these 2 examples and I would like to know if my steps are good or what I am missing.
What would be the main difference by the bigger or equal to 0 and bigger than 0?
Knowing $a, b in mathbb{Z}$:
A. $a sim b$ if $ab geq 0$
for reflexivity: $asim a$ if $a^2 geq 0$ then yes
for symmetry: $a sim b$ if $b sim a$ then $ab geq 0$ if $ba geq 0$ then yes
for transitivity: $a sim b$ if $b sim a$ then $a sim c$ then $ab geq 0$ if $bc geq 0$ if $ac geq 0$ then no
B. $a sim b$ if $ab > 0$
for reflexivity: $a sim a$ if $a-a=0$ so yes
for symmetry: $a sim b$ if $b sim a$ then $a - b = c in mathbb{Z}$ so $b - a = -c in mathbb{Z}$ then yes
for transitivity: $a sim b$ if $b sim a$ then $a sim c$ Here I have no idea how to prove it
Thank you for your feedback!
elementary-set-theory equivalence-relations
edited Jan 8 at 22:07
T. Fo
466311
asked Jan 8 at 20:55
Goose LookingGoose Looking
11
$endgroup$
add a comment |
$begingroup$
I have just started my math class and I think I might not be completely understanding the 3 properties of a equivalence relation: reflexivity, symmetry, transitivity.
I have these 2 examples and I would like to know if my steps are good or what I am missing.
What would be the main difference by the bigger or equal to 0 and bigger than 0?
Knowing $a, b in mathbb{Z}$:
A. $a sim b$ if $ab geq 0$
for reflexivity: $asim a$ if $a^2 geq 0$ then yes
for symmetry: $a sim b$ if $b sim a$ then $ab geq 0$ if $ba geq 0$ then yes
for transitivity: $a sim b$ if $b sim a$ then $a sim c$ then $ab geq 0$ if $bc geq 0$ if $ac geq 0$ then no
B. $a sim b$ if $ab > 0$
for reflexivity: $a sim a$ if $a-a=0$ so yes
for symmetry: $a sim b$ if $b sim a$ then $a - b = c in mathbb{Z}$ so $b - a = -c in mathbb{Z}$ then yes
for transitivity: $a sim b$ if $b sim a$ then $a sim c$ Here I have no idea how to prove it
Thank you for your feedback!
elementary-set-theory equivalence-relations
edited Jan 8 at 22:07
T. Fo
466311
asked Jan 8 at 20:55
Goose LookingGoose Looking
11
$endgroup$
$begingroup$
Well, for the second one, $0not sim 0$ so it isn't reflexive.
$endgroup$
– lulu
Jan 8 at 20:57
$begingroup$
when does the 0 ~ O come from?
$endgroup$
– Goose Looking
Jan 8 at 20:59
$begingroup$
Sorry? under the second relation $0$ is not equivalent to itself, since $0^2$ is not greater than $0$.
$endgroup$
– lulu
Jan 8 at 21:21
$begingroup$
I just edited, but @GooseLooking in the future please write your questions using the MathJax typesetting. Thank you and welcome to the site!
$endgroup$
– T. Fo
Jan 8 at 21:45
add a comment |
$begingroup$
I have just started my math class and I think I might not be completely understanding the 3 properties of a equivalence relation: reflexivity, symmetry, transitivity.
I have these 2 examples and I would like to know if my steps are good or what I am missing.
What would be the main difference by the bigger or equal to 0 and bigger than 0?
Knowing $a, b in mathbb{Z}$:
A. $a sim b$ if $ab geq 0$
for reflexivity: $asim a$ if $a^2 geq 0$ then yes
for symmetry: $a sim b$ if $b sim a$ then $ab geq 0$ if $ba geq 0$ then yes
for transitivity: $a sim b$ if $b sim a$ then $a sim c$ then $ab geq 0$ if $bc geq 0$ if $ac geq 0$ then no
B. $a sim b$ if $ab > 0$
for reflexivity: $a sim a$ if $a-a=0$ so yes
for symmetry: $a sim b$ if $b sim a$ then $a - b = c in mathbb{Z}$ so $b - a = -c in mathbb{Z}$ then yes
for transitivity: $a sim b$ if $b sim a$ then $a sim c$ Here I have no idea how to prove it
Thank you for your feedback!
elementary-set-theory equivalence-relations
edited Jan 8 at 22:07
T. Fo
466311
asked Jan 8 at 20:55
Goose LookingGoose Looking
11
$endgroup$
I have just started my math class and I think I might not be completely understanding the 3 properties of a equivalence relation: reflexivity, symmetry, transitivity.
I have these 2 examples and I would like to know if my steps are good or what I am missing.
What would be the main difference by the bigger or equal to 0 and bigger than 0?
Knowing $a, b in mathbb{Z}$:
A. $a sim b$ if $ab geq 0$
for reflexivity: $asim a$ if $a^2 geq 0$ then yes
for symmetry: $a sim b$ if $b sim a$ then $ab geq 0$ if $ba geq 0$ then yes
for transitivity: $a sim b$ if $b sim a$ then $a sim c$ then $ab geq 0$ if $bc geq 0$ if $ac geq 0$ then no
B. $a sim b$ if $ab > 0$
for reflexivity: $a sim a$ if $a-a=0$ so yes
for symmetry: $a sim b$ if $b sim a$ then $a - b = c in mathbb{Z}$ so $b - a = -c in mathbb{Z}$ then yes
for transitivity: $a sim b$ if $b sim a$ then $a sim c$ Here I have no idea how to prove it
Thank you for your feedback!
elementary-set-theory equivalence-relations
elementary-set-theory equivalence-relations
edited Jan 8 at 22:07
T. Fo
466311
asked Jan 8 at 20:55
Goose LookingGoose Looking
11
edited Jan 8 at 22:07
T. Fo
466311
asked Jan 8 at 20:55
Goose LookingGoose Looking
11
edited Jan 8 at 22:07
T. Fo
466311
edited Jan 8 at 22:07
T. Fo
466311
edited Jan 8 at 22:07
T. Fo
466311
466311
asked Jan 8 at 20:55
Goose LookingGoose Looking
11
asked Jan 8 at 20:55
Goose LookingGoose Looking
11
asked Jan 8 at 20:55
Goose LookingGoose Looking
11
11
$begingroup$
Well, for the second one, $0not sim 0$ so it isn't reflexive.
$endgroup$
– lulu
Jan 8 at 20:57
$begingroup$
when does the 0 ~ O come from?
$endgroup$
– Goose Looking
Jan 8 at 20:59
$begingroup$
Sorry? under the second relation $0$ is not equivalent to itself, since $0^2$ is not greater than $0$.
$endgroup$
– lulu
Jan 8 at 21:21
$begingroup$
I just edited, but @GooseLooking in the future please write your questions using the MathJax typesetting. Thank you and welcome to the site!
$endgroup$
– T. Fo
Jan 8 at 21:45
add a comment |
$begingroup$
Well, for the second one, $0not sim 0$ so it isn't reflexive.
$endgroup$
– lulu
Jan 8 at 20:57
$begingroup$
when does the 0 ~ O come from?
$endgroup$
– Goose Looking
Jan 8 at 20:59
$begingroup$
Sorry? under the second relation $0$ is not equivalent to itself, since $0^2$ is not greater than $0$.
$endgroup$
– lulu
Jan 8 at 21:21
$begingroup$
I just edited, but @GooseLooking in the future please write your questions using the MathJax typesetting. Thank you and welcome to the site!
$endgroup$
– T. Fo
Jan 8 at 21:45
$begingroup$
Well, for the second one, $0not sim 0$ so it isn't reflexive.
$endgroup$
– lulu
Jan 8 at 20:57
$begingroup$
Well, for the second one, $0not sim 0$ so it isn't reflexive.
$endgroup$
– lulu
Jan 8 at 20:57
$begingroup$
when does the 0 ~ O come from?
$endgroup$
– Goose Looking
Jan 8 at 20:59
$begingroup$
when does the 0 ~ O come from?
$endgroup$
– Goose Looking
Jan 8 at 20:59
$begingroup$
Sorry? under the second relation $0$ is not equivalent to itself, since $0^2$ is not greater than $0$.
$endgroup$
– lulu
Jan 8 at 21:21
$begingroup$
Sorry? under the second relation $0$ is not equivalent to itself, since $0^2$ is not greater than $0$.
$endgroup$
– lulu
Jan 8 at 21:21
$begingroup$
I just edited, but @GooseLooking in the future please write your questions using the MathJax typesetting. Thank you and welcome to the site!
$endgroup$
– T. Fo
Jan 8 at 21:45
$begingroup$
I just edited, but @GooseLooking in the future please write your questions using the MathJax typesetting. Thank you and welcome to the site!
$endgroup$
– T. Fo
Jan 8 at 21:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let's go through the second carefully, then you can do the first yourself. Define $a sim b$ if and only if $ab>0$.
Reflexivity means $a sim a$ for all $a in mathbb{Z}$. Let $a in mathbb{Z}$, then $a sim a iff a^2 > 0$ which is only true if $ane 0$, hence the relation is not reflexive
Symmetry means $asim b iff b sim a$ for all $a,b in mathbb{Z}$. Let $a,b in mathbb{Z}$ then $asim b iff ab > 0 iff ba > 0 iff b sim a$, so indeed the relation is symmetric
Transitivity means $asim b, bsim c iff a sim c$ for all $a,b,c in mathbb{Z}$. Let $a,b,c in mathbb{Z}$ then
$$
begin{split}
asim b,bsim c
&iff ab>0,bc>0\
&iff a,b text{ have same sign and } b,c text{ have same sign}\
&iff a,b,c text{ all have the same sign}\
&iff ac > 0 \
&iff a sim c,
end{split}
$$
so the
relation is transitive
answered Jan 8 at 21:05
gt6989bgt6989b
33.9k22455
$endgroup$
$begingroup$
So it isn't reflexive bc it can't be applied to all value of a? (since a != 0)
$endgroup$
– Goose Looking
Jan 8 at 21:10
$begingroup$
@GooseLooking correct, it must be true for all values of $a$
$endgroup$
– gt6989b
Jan 8 at 21:10
$begingroup$
Ok perfect. For symmetry, if a or b =0, then it can't be symmetric, correct? bc 0 is in ℤ, so how can we say for sure ab>0 ?
$endgroup$
– Goose Looking
Jan 8 at 21:12
$begingroup$
Symmetry means that for any $a,b in mathbb{Z}$, you have $a sim b iff b sim a$. Now note that $a sim b iff ab > 0 iff b sim a$, so it is symmetric, since if $a=0$ or $b=0$, then $a sim b$ does not hold...
$endgroup$
– gt6989b
Jan 9 at 4:53
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's go through the second carefully, then you can do the first yourself. Define $a sim b$ if and only if $ab>0$.
Reflexivity means $a sim a$ for all $a in mathbb{Z}$. Let $a in mathbb{Z}$, then $a sim a iff a^2 > 0$ which is only true if $ane 0$, hence the relation is not reflexive
Symmetry means $asim b iff b sim a$ for all $a,b in mathbb{Z}$. Let $a,b in mathbb{Z}$ then $asim b iff ab > 0 iff ba > 0 iff b sim a$, so indeed the relation is symmetric
Transitivity means $asim b, bsim c iff a sim c$ for all $a,b,c in mathbb{Z}$. Let $a,b,c in mathbb{Z}$ then
$$
begin{split}
asim b,bsim c
&iff ab>0,bc>0\
&iff a,b text{ have same sign and } b,c text{ have same sign}\
&iff a,b,c text{ all have the same sign}\
&iff ac > 0 \
&iff a sim c,
end{split}
$$
so the
relation is transitive
answered Jan 8 at 21:05
gt6989bgt6989b
33.9k22455
$endgroup$
$begingroup$
So it isn't reflexive bc it can't be applied to all value of a? (since a != 0)
$endgroup$
– Goose Looking
Jan 8 at 21:10
$begingroup$
@GooseLooking correct, it must be true for all values of $a$
$endgroup$
– gt6989b
Jan 8 at 21:10
$begingroup$
Ok perfect. For symmetry, if a or b =0, then it can't be symmetric, correct? bc 0 is in ℤ, so how can we say for sure ab>0 ?
$endgroup$
– Goose Looking
Jan 8 at 21:12
$begingroup$
Symmetry means that for any $a,b in mathbb{Z}$, you have $a sim b iff b sim a$. Now note that $a sim b iff ab > 0 iff b sim a$, so it is symmetric, since if $a=0$ or $b=0$, then $a sim b$ does not hold...
$endgroup$
– gt6989b
Jan 9 at 4:53
add a comment |
$begingroup$
Let's go through the second carefully, then you can do the first yourself. Define $a sim b$ if and only if $ab>0$.
Reflexivity means $a sim a$ for all $a in mathbb{Z}$. Let $a in mathbb{Z}$, then $a sim a iff a^2 > 0$ which is only true if $ane 0$, hence the relation is not reflexive
Symmetry means $asim b iff b sim a$ for all $a,b in mathbb{Z}$. Let $a,b in mathbb{Z}$ then $asim b iff ab > 0 iff ba > 0 iff b sim a$, so indeed the relation is symmetric
Transitivity means $asim b, bsim c iff a sim c$ for all $a,b,c in mathbb{Z}$. Let $a,b,c in mathbb{Z}$ then
$$
begin{split}
asim b,bsim c
&iff ab>0,bc>0\
&iff a,b text{ have same sign and } b,c text{ have same sign}\
&iff a,b,c text{ all have the same sign}\
&iff ac > 0 \
&iff a sim c,
end{split}
$$
so the
relation is transitive
answered Jan 8 at 21:05
gt6989bgt6989b
33.9k22455
$endgroup$
$begingroup$
So it isn't reflexive bc it can't be applied to all value of a? (since a != 0)
$endgroup$
– Goose Looking
Jan 8 at 21:10
$begingroup$
@GooseLooking correct, it must be true for all values of $a$
$endgroup$
– gt6989b
Jan 8 at 21:10
$begingroup$
Ok perfect. For symmetry, if a or b =0, then it can't be symmetric, correct? bc 0 is in ℤ, so how can we say for sure ab>0 ?
$endgroup$
– Goose Looking
Jan 8 at 21:12
$begingroup$
Symmetry means that for any $a,b in mathbb{Z}$, you have $a sim b iff b sim a$. Now note that $a sim b iff ab > 0 iff b sim a$, so it is symmetric, since if $a=0$ or $b=0$, then $a sim b$ does not hold...
$endgroup$
– gt6989b
Jan 9 at 4:53
add a comment |
$begingroup$
Let's go through the second carefully, then you can do the first yourself. Define $a sim b$ if and only if $ab>0$.
Reflexivity means $a sim a$ for all $a in mathbb{Z}$. Let $a in mathbb{Z}$, then $a sim a iff a^2 > 0$ which is only true if $ane 0$, hence the relation is not reflexive
Symmetry means $asim b iff b sim a$ for all $a,b in mathbb{Z}$. Let $a,b in mathbb{Z}$ then $asim b iff ab > 0 iff ba > 0 iff b sim a$, so indeed the relation is symmetric
Transitivity means $asim b, bsim c iff a sim c$ for all $a,b,c in mathbb{Z}$. Let $a,b,c in mathbb{Z}$ then
$$
begin{split}
asim b,bsim c
&iff ab>0,bc>0\
&iff a,b text{ have same sign and } b,c text{ have same sign}\
&iff a,b,c text{ all have the same sign}\
&iff ac > 0 \
&iff a sim c,
end{split}
$$
so the
relation is transitive
answered Jan 8 at 21:05
gt6989bgt6989b
33.9k22455
$endgroup$
Let's go through the second carefully, then you can do the first yourself. Define $a sim b$ if and only if $ab>0$.
Reflexivity means $a sim a$ for all $a in mathbb{Z}$. Let $a in mathbb{Z}$, then $a sim a iff a^2 > 0$ which is only true if $ane 0$, hence the relation is not reflexive
Symmetry means $asim b iff b sim a$ for all $a,b in mathbb{Z}$. Let $a,b in mathbb{Z}$ then $asim b iff ab > 0 iff ba > 0 iff b sim a$, so indeed the relation is symmetric
Transitivity means $asim b, bsim c iff a sim c$ for all $a,b,c in mathbb{Z}$. Let $a,b,c in mathbb{Z}$ then
$$
begin{split}
asim b,bsim c
&iff ab>0,bc>0\
&iff a,b text{ have same sign and } b,c text{ have same sign}\
&iff a,b,c text{ all have the same sign}\
&iff ac > 0 \
&iff a sim c,
end{split}
$$
so the
relation is transitive
answered Jan 8 at 21:05
gt6989bgt6989b
33.9k22455
answered Jan 8 at 21:05
gt6989bgt6989b
33.9k22455
answered Jan 8 at 21:05
gt6989bgt6989b
33.9k22455
answered Jan 8 at 21:05
gt6989bgt6989b
33.9k22455
33.9k22455
$begingroup$
So it isn't reflexive bc it can't be applied to all value of a? (since a != 0)
$endgroup$
– Goose Looking
Jan 8 at 21:10
$begingroup$
@GooseLooking correct, it must be true for all values of $a$
$endgroup$
– gt6989b
Jan 8 at 21:10
$begingroup$
Ok perfect. For symmetry, if a or b =0, then it can't be symmetric, correct? bc 0 is in ℤ, so how can we say for sure ab>0 ?
$endgroup$
– Goose Looking
Jan 8 at 21:12
$begingroup$
Symmetry means that for any $a,b in mathbb{Z}$, you have $a sim b iff b sim a$. Now note that $a sim b iff ab > 0 iff b sim a$, so it is symmetric, since if $a=0$ or $b=0$, then $a sim b$ does not hold...
$endgroup$
– gt6989b
Jan 9 at 4:53
add a comment |
$begingroup$
So it isn't reflexive bc it can't be applied to all value of a? (since a != 0)
$endgroup$
– Goose Looking
Jan 8 at 21:10
$begingroup$
@GooseLooking correct, it must be true for all values of $a$
$endgroup$
– gt6989b
Jan 8 at 21:10
$begingroup$
Ok perfect. For symmetry, if a or b =0, then it can't be symmetric, correct? bc 0 is in ℤ, so how can we say for sure ab>0 ?
$endgroup$
– Goose Looking
Jan 8 at 21:12
$begingroup$
Symmetry means that for any $a,b in mathbb{Z}$, you have $a sim b iff b sim a$. Now note that $a sim b iff ab > 0 iff b sim a$, so it is symmetric, since if $a=0$ or $b=0$, then $a sim b$ does not hold...
$endgroup$
– gt6989b
Jan 9 at 4:53
$begingroup$
So it isn't reflexive bc it can't be applied to all value of a? (since a != 0)
$endgroup$
– Goose Looking
Jan 8 at 21:10
$begingroup$
So it isn't reflexive bc it can't be applied to all value of a? (since a != 0)
$endgroup$
– Goose Looking
Jan 8 at 21:10
$begingroup$
@GooseLooking correct, it must be true for all values of $a$
$endgroup$
– gt6989b
Jan 8 at 21:10
$begingroup$
@GooseLooking correct, it must be true for all values of $a$
$endgroup$
– gt6989b
Jan 8 at 21:10
$begingroup$
Ok perfect. For symmetry, if a or b =0, then it can't be symmetric, correct? bc 0 is in ℤ, so how can we say for sure ab>0 ?
$endgroup$
– Goose Looking
Jan 8 at 21:12
$begingroup$
Ok perfect. For symmetry, if a or b =0, then it can't be symmetric, correct? bc 0 is in ℤ, so how can we say for sure ab>0 ?
$endgroup$
– Goose Looking
Jan 8 at 21:12
$begingroup$
Symmetry means that for any $a,b in mathbb{Z}$, you have $a sim b iff b sim a$. Now note that $a sim b iff ab > 0 iff b sim a$, so it is symmetric, since if $a=0$ or $b=0$, then $a sim b$ does not hold...
$endgroup$
– gt6989b
Jan 9 at 4:53
$begingroup$
Symmetry means that for any $a,b in mathbb{Z}$, you have $a sim b iff b sim a$. Now note that $a sim b iff ab > 0 iff b sim a$, so it is symmetric, since if $a=0$ or $b=0$, then $a sim b$ does not hold...
$endgroup$
– gt6989b
Jan 9 at 4:53
add a comment |
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$begingroup$
Well, for the second one, $0not sim 0$ so it isn't reflexive.
$endgroup$
– lulu
Jan 8 at 20:57
$begingroup$
when does the 0 ~ O come from?
$endgroup$
– Goose Looking
Jan 8 at 20:59
$begingroup$
Sorry? under the second relation $0$ is not equivalent to itself, since $0^2$ is not greater than $0$.
$endgroup$
– lulu
Jan 8 at 21:21
$begingroup$
I just edited, but @GooseLooking in the future please write your questions using the MathJax typesetting. Thank you and welcome to the site!
$endgroup$
– T. Fo
Jan 8 at 21:45