Find Equivalence Classes and the quotient set defined by the relationship in Reals.
$begingroup$
$xTyiff |x^{2}-2|=|y^{2}-2|$
I'd love to solve it but we don't deal in absolutes. Joke aside I have no clue where to start with this one, it has me stumped.
If I had been provided the answer to the problem I could just bump around until I got there but its not the case.
$x^{2}-y^{2}=0 rightarrow (x+y)(x-y)=0 rightarrow x=y lor x=-y$
Is the only thing that I can scavenge from this, but I doubt it covers all the classes, thus I can't even begin to partition the set.
$cl(x)={xin R:x,-x,x-2,2-x}$ But I'm not sure x-2 and 2-x counts in general I think it's only valid if x=0.
discrete-mathematics equivalence-relations set-partition
asked Dec 20 '18 at 17:55
AgustinAgustin
33
$endgroup$
add a comment |
$begingroup$
$xTyiff |x^{2}-2|=|y^{2}-2|$
I'd love to solve it but we don't deal in absolutes. Joke aside I have no clue where to start with this one, it has me stumped.
If I had been provided the answer to the problem I could just bump around until I got there but its not the case.
$x^{2}-y^{2}=0 rightarrow (x+y)(x-y)=0 rightarrow x=y lor x=-y$
Is the only thing that I can scavenge from this, but I doubt it covers all the classes, thus I can't even begin to partition the set.
$cl(x)={xin R:x,-x,x-2,2-x}$ But I'm not sure x-2 and 2-x counts in general I think it's only valid if x=0.
discrete-mathematics equivalence-relations set-partition
asked Dec 20 '18 at 17:55
AgustinAgustin
33
$endgroup$
$begingroup$
Work out the cases when expressions in absolutes are positive and negative. For each, you'll get one class, $2$ in total, if I'm not mistaken. You can draw the graphs of each to get the visual representation of $x$ being related to $y$.
$endgroup$
– user626177
Dec 20 '18 at 20:38
$begingroup$
Bear in mind a class $[k] = {xin mathbb R| xTk} = {x in mathbb R:|x^2 -2| = |k^2 - 2|}$. Just solve that.
$endgroup$
– fleablood
Dec 20 '18 at 21:38
add a comment |
$begingroup$
$xTyiff |x^{2}-2|=|y^{2}-2|$
I'd love to solve it but we don't deal in absolutes. Joke aside I have no clue where to start with this one, it has me stumped.
If I had been provided the answer to the problem I could just bump around until I got there but its not the case.
$x^{2}-y^{2}=0 rightarrow (x+y)(x-y)=0 rightarrow x=y lor x=-y$
Is the only thing that I can scavenge from this, but I doubt it covers all the classes, thus I can't even begin to partition the set.
$cl(x)={xin R:x,-x,x-2,2-x}$ But I'm not sure x-2 and 2-x counts in general I think it's only valid if x=0.
discrete-mathematics equivalence-relations set-partition
asked Dec 20 '18 at 17:55
AgustinAgustin
33
$endgroup$
$xTyiff |x^{2}-2|=|y^{2}-2|$
I'd love to solve it but we don't deal in absolutes. Joke aside I have no clue where to start with this one, it has me stumped.
If I had been provided the answer to the problem I could just bump around until I got there but its not the case.
$x^{2}-y^{2}=0 rightarrow (x+y)(x-y)=0 rightarrow x=y lor x=-y$
Is the only thing that I can scavenge from this, but I doubt it covers all the classes, thus I can't even begin to partition the set.
$cl(x)={xin R:x,-x,x-2,2-x}$ But I'm not sure x-2 and 2-x counts in general I think it's only valid if x=0.
discrete-mathematics equivalence-relations set-partition
discrete-mathematics equivalence-relations set-partition
asked Dec 20 '18 at 17:55
AgustinAgustin
33
asked Dec 20 '18 at 17:55
AgustinAgustin
33
edited Dec 20 '18 at 19:36
Agustin
asked Dec 20 '18 at 17:55
AgustinAgustin
33
asked Dec 20 '18 at 17:55
AgustinAgustin
33
asked Dec 20 '18 at 17:55
AgustinAgustin
33
33
$begingroup$
Work out the cases when expressions in absolutes are positive and negative. For each, you'll get one class, $2$ in total, if I'm not mistaken. You can draw the graphs of each to get the visual representation of $x$ being related to $y$.
$endgroup$
– user626177
Dec 20 '18 at 20:38
$begingroup$
Bear in mind a class $[k] = {xin mathbb R| xTk} = {x in mathbb R:|x^2 -2| = |k^2 - 2|}$. Just solve that.
$endgroup$
– fleablood
Dec 20 '18 at 21:38
add a comment |
$begingroup$
Work out the cases when expressions in absolutes are positive and negative. For each, you'll get one class, $2$ in total, if I'm not mistaken. You can draw the graphs of each to get the visual representation of $x$ being related to $y$.
$endgroup$
– user626177
Dec 20 '18 at 20:38
$begingroup$
Bear in mind a class $[k] = {xin mathbb R| xTk} = {x in mathbb R:|x^2 -2| = |k^2 - 2|}$. Just solve that.
$endgroup$
– fleablood
Dec 20 '18 at 21:38
$begingroup$
Work out the cases when expressions in absolutes are positive and negative. For each, you'll get one class, $2$ in total, if I'm not mistaken. You can draw the graphs of each to get the visual representation of $x$ being related to $y$.
$endgroup$
– user626177
Dec 20 '18 at 20:38
$begingroup$
Work out the cases when expressions in absolutes are positive and negative. For each, you'll get one class, $2$ in total, if I'm not mistaken. You can draw the graphs of each to get the visual representation of $x$ being related to $y$.
$endgroup$
– user626177
Dec 20 '18 at 20:38
$begingroup$
Bear in mind a class $[k] = {xin mathbb R| xTk} = {x in mathbb R:|x^2 -2| = |k^2 - 2|}$. Just solve that.
$endgroup$
– fleablood
Dec 20 '18 at 21:38
$begingroup$
Bear in mind a class $[k] = {xin mathbb R| xTk} = {x in mathbb R:|x^2 -2| = |k^2 - 2|}$. Just solve that.
$endgroup$
– fleablood
Dec 20 '18 at 21:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Well if we fix $k in mathbb R$ then
the equivalence class $[k]$ is by definition ${xin mathbb R: |x^2 -2| = |k^2 - 2|}$.
Now if $|x^2 - 2| = |k^2 - 2|$ then a) $x^2 -2 = k^2-2$ or b) $x^2 - 2 = -(k^2 -2) = 2-k^2$.
a) $x^2 - 2 = k^2 -2implies x^2 = k^2 implies x =pm k$.
b) $x^2 - 2 = 2-k^2 implies x^2 = 4 -k^2implies |k| le 2$ and $x=pm sqrt{4-k^2}$.
So $[k]={xin mathbb R: |x^2 -2| = |k^2 - 2|}=$
$={k, -k}$ if $|k| > 2$ and $= {k,-k, sqrt{4-k^2}, -sqrt{4-k^2}}$ if $|k| le 2$
=====
Note: All of this assumes $T$ is an equivalence relation in the first place.
answered Dec 20 '18 at 21:36
fleabloodfleablood
71.5k22686
$endgroup$
$begingroup$
Doesn't tansitivity just follows from relation $=$ being transitive ?
$endgroup$
– user626177
Dec 20 '18 at 21:40
1
$begingroup$
Oh... $|x^2 - 2| = |y^2 - 2|$ and $|y^2-2| =|z^2 - 2|$ so .... Um, How hard a dope slap do you think a human can apply to himself? Well, double that and that's what i just gave to myself.
$endgroup$
– fleablood
Dec 20 '18 at 21:42
$begingroup$
And there it is, thanks, I know where I got lost now.
$endgroup$
– Agustin
Dec 20 '18 at 21:51
$begingroup$
Hahah + But it's not bad to see how to do it in non-trivial way too, for sure!
$endgroup$
– user626177
Dec 20 '18 at 21:51
1
$begingroup$
Yeah, I also did it the hard way, because it looked too similar to a distance relation and those aren't always transitive thus not equivalent relations, only bothered to ask the second part of the question.
$endgroup$
– Agustin
Dec 20 '18 at 21:54
add a comment |
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1 Answer
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$begingroup$
Well if we fix $k in mathbb R$ then
the equivalence class $[k]$ is by definition ${xin mathbb R: |x^2 -2| = |k^2 - 2|}$.
Now if $|x^2 - 2| = |k^2 - 2|$ then a) $x^2 -2 = k^2-2$ or b) $x^2 - 2 = -(k^2 -2) = 2-k^2$.
a) $x^2 - 2 = k^2 -2implies x^2 = k^2 implies x =pm k$.
b) $x^2 - 2 = 2-k^2 implies x^2 = 4 -k^2implies |k| le 2$ and $x=pm sqrt{4-k^2}$.
So $[k]={xin mathbb R: |x^2 -2| = |k^2 - 2|}=$
$={k, -k}$ if $|k| > 2$ and $= {k,-k, sqrt{4-k^2}, -sqrt{4-k^2}}$ if $|k| le 2$
=====
Note: All of this assumes $T$ is an equivalence relation in the first place.
answered Dec 20 '18 at 21:36
fleabloodfleablood
71.5k22686
$endgroup$
$begingroup$
Doesn't tansitivity just follows from relation $=$ being transitive ?
$endgroup$
– user626177
Dec 20 '18 at 21:40
1
$begingroup$
Oh... $|x^2 - 2| = |y^2 - 2|$ and $|y^2-2| =|z^2 - 2|$ so .... Um, How hard a dope slap do you think a human can apply to himself? Well, double that and that's what i just gave to myself.
$endgroup$
– fleablood
Dec 20 '18 at 21:42
$begingroup$
And there it is, thanks, I know where I got lost now.
$endgroup$
– Agustin
Dec 20 '18 at 21:51
$begingroup$
Hahah + But it's not bad to see how to do it in non-trivial way too, for sure!
$endgroup$
– user626177
Dec 20 '18 at 21:51
1
$begingroup$
Yeah, I also did it the hard way, because it looked too similar to a distance relation and those aren't always transitive thus not equivalent relations, only bothered to ask the second part of the question.
$endgroup$
– Agustin
Dec 20 '18 at 21:54
add a comment |
$begingroup$
Well if we fix $k in mathbb R$ then
the equivalence class $[k]$ is by definition ${xin mathbb R: |x^2 -2| = |k^2 - 2|}$.
Now if $|x^2 - 2| = |k^2 - 2|$ then a) $x^2 -2 = k^2-2$ or b) $x^2 - 2 = -(k^2 -2) = 2-k^2$.
a) $x^2 - 2 = k^2 -2implies x^2 = k^2 implies x =pm k$.
b) $x^2 - 2 = 2-k^2 implies x^2 = 4 -k^2implies |k| le 2$ and $x=pm sqrt{4-k^2}$.
So $[k]={xin mathbb R: |x^2 -2| = |k^2 - 2|}=$
$={k, -k}$ if $|k| > 2$ and $= {k,-k, sqrt{4-k^2}, -sqrt{4-k^2}}$ if $|k| le 2$
=====
Note: All of this assumes $T$ is an equivalence relation in the first place.
answered Dec 20 '18 at 21:36
fleabloodfleablood
71.5k22686
$endgroup$
$begingroup$
Doesn't tansitivity just follows from relation $=$ being transitive ?
$endgroup$
– user626177
Dec 20 '18 at 21:40
1
$begingroup$
Oh... $|x^2 - 2| = |y^2 - 2|$ and $|y^2-2| =|z^2 - 2|$ so .... Um, How hard a dope slap do you think a human can apply to himself? Well, double that and that's what i just gave to myself.
$endgroup$
– fleablood
Dec 20 '18 at 21:42
$begingroup$
And there it is, thanks, I know where I got lost now.
$endgroup$
– Agustin
Dec 20 '18 at 21:51
$begingroup$
Hahah + But it's not bad to see how to do it in non-trivial way too, for sure!
$endgroup$
– user626177
Dec 20 '18 at 21:51
1
$begingroup$
Yeah, I also did it the hard way, because it looked too similar to a distance relation and those aren't always transitive thus not equivalent relations, only bothered to ask the second part of the question.
$endgroup$
– Agustin
Dec 20 '18 at 21:54
add a comment |
$begingroup$
Well if we fix $k in mathbb R$ then
the equivalence class $[k]$ is by definition ${xin mathbb R: |x^2 -2| = |k^2 - 2|}$.
Now if $|x^2 - 2| = |k^2 - 2|$ then a) $x^2 -2 = k^2-2$ or b) $x^2 - 2 = -(k^2 -2) = 2-k^2$.
a) $x^2 - 2 = k^2 -2implies x^2 = k^2 implies x =pm k$.
b) $x^2 - 2 = 2-k^2 implies x^2 = 4 -k^2implies |k| le 2$ and $x=pm sqrt{4-k^2}$.
So $[k]={xin mathbb R: |x^2 -2| = |k^2 - 2|}=$
$={k, -k}$ if $|k| > 2$ and $= {k,-k, sqrt{4-k^2}, -sqrt{4-k^2}}$ if $|k| le 2$
=====
Note: All of this assumes $T$ is an equivalence relation in the first place.
answered Dec 20 '18 at 21:36
fleabloodfleablood
71.5k22686
$endgroup$
Well if we fix $k in mathbb R$ then
the equivalence class $[k]$ is by definition ${xin mathbb R: |x^2 -2| = |k^2 - 2|}$.
Now if $|x^2 - 2| = |k^2 - 2|$ then a) $x^2 -2 = k^2-2$ or b) $x^2 - 2 = -(k^2 -2) = 2-k^2$.
a) $x^2 - 2 = k^2 -2implies x^2 = k^2 implies x =pm k$.
b) $x^2 - 2 = 2-k^2 implies x^2 = 4 -k^2implies |k| le 2$ and $x=pm sqrt{4-k^2}$.
So $[k]={xin mathbb R: |x^2 -2| = |k^2 - 2|}=$
$={k, -k}$ if $|k| > 2$ and $= {k,-k, sqrt{4-k^2}, -sqrt{4-k^2}}$ if $|k| le 2$
=====
Note: All of this assumes $T$ is an equivalence relation in the first place.
answered Dec 20 '18 at 21:36
fleabloodfleablood
71.5k22686
edited Dec 20 '18 at 21:43
answered Dec 20 '18 at 21:36
fleabloodfleablood
71.5k22686
answered Dec 20 '18 at 21:36
fleabloodfleablood
71.5k22686
answered Dec 20 '18 at 21:36
fleabloodfleablood
71.5k22686
71.5k22686
$begingroup$
Doesn't tansitivity just follows from relation $=$ being transitive ?
$endgroup$
– user626177
Dec 20 '18 at 21:40
1
$begingroup$
Oh... $|x^2 - 2| = |y^2 - 2|$ and $|y^2-2| =|z^2 - 2|$ so .... Um, How hard a dope slap do you think a human can apply to himself? Well, double that and that's what i just gave to myself.
$endgroup$
– fleablood
Dec 20 '18 at 21:42
$begingroup$
And there it is, thanks, I know where I got lost now.
$endgroup$
– Agustin
Dec 20 '18 at 21:51
$begingroup$
Hahah + But it's not bad to see how to do it in non-trivial way too, for sure!
$endgroup$
– user626177
Dec 20 '18 at 21:51
1
$begingroup$
Yeah, I also did it the hard way, because it looked too similar to a distance relation and those aren't always transitive thus not equivalent relations, only bothered to ask the second part of the question.
$endgroup$
– Agustin
Dec 20 '18 at 21:54
add a comment |
$begingroup$
Doesn't tansitivity just follows from relation $=$ being transitive ?
$endgroup$
– user626177
Dec 20 '18 at 21:40
1
$begingroup$
Oh... $|x^2 - 2| = |y^2 - 2|$ and $|y^2-2| =|z^2 - 2|$ so .... Um, How hard a dope slap do you think a human can apply to himself? Well, double that and that's what i just gave to myself.
$endgroup$
– fleablood
Dec 20 '18 at 21:42
$begingroup$
And there it is, thanks, I know where I got lost now.
$endgroup$
– Agustin
Dec 20 '18 at 21:51
$begingroup$
Hahah + But it's not bad to see how to do it in non-trivial way too, for sure!
$endgroup$
– user626177
Dec 20 '18 at 21:51
1
$begingroup$
Yeah, I also did it the hard way, because it looked too similar to a distance relation and those aren't always transitive thus not equivalent relations, only bothered to ask the second part of the question.
$endgroup$
– Agustin
Dec 20 '18 at 21:54
$begingroup$
Doesn't tansitivity just follows from relation $=$ being transitive ?
$endgroup$
– user626177
Dec 20 '18 at 21:40
$begingroup$
Doesn't tansitivity just follows from relation $=$ being transitive ?
$endgroup$
– user626177
Dec 20 '18 at 21:40
1
1
$begingroup$
Oh... $|x^2 - 2| = |y^2 - 2|$ and $|y^2-2| =|z^2 - 2|$ so .... Um, How hard a dope slap do you think a human can apply to himself? Well, double that and that's what i just gave to myself.
$endgroup$
– fleablood
Dec 20 '18 at 21:42
$begingroup$
Oh... $|x^2 - 2| = |y^2 - 2|$ and $|y^2-2| =|z^2 - 2|$ so .... Um, How hard a dope slap do you think a human can apply to himself? Well, double that and that's what i just gave to myself.
$endgroup$
– fleablood
Dec 20 '18 at 21:42
$begingroup$
And there it is, thanks, I know where I got lost now.
$endgroup$
– Agustin
Dec 20 '18 at 21:51
$begingroup$
And there it is, thanks, I know where I got lost now.
$endgroup$
– Agustin
Dec 20 '18 at 21:51
$begingroup$
Hahah + But it's not bad to see how to do it in non-trivial way too, for sure!
$endgroup$
– user626177
Dec 20 '18 at 21:51
$begingroup$
Hahah + But it's not bad to see how to do it in non-trivial way too, for sure!
$endgroup$
– user626177
Dec 20 '18 at 21:51
1
1
$begingroup$
Yeah, I also did it the hard way, because it looked too similar to a distance relation and those aren't always transitive thus not equivalent relations, only bothered to ask the second part of the question.
$endgroup$
– Agustin
Dec 20 '18 at 21:54
$begingroup$
Yeah, I also did it the hard way, because it looked too similar to a distance relation and those aren't always transitive thus not equivalent relations, only bothered to ask the second part of the question.
$endgroup$
– Agustin
Dec 20 '18 at 21:54
add a comment |
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$begingroup$
Work out the cases when expressions in absolutes are positive and negative. For each, you'll get one class, $2$ in total, if I'm not mistaken. You can draw the graphs of each to get the visual representation of $x$ being related to $y$.
$endgroup$
– user626177
Dec 20 '18 at 20:38
$begingroup$
Bear in mind a class $[k] = {xin mathbb R| xTk} = {x in mathbb R:|x^2 -2| = |k^2 - 2|}$. Just solve that.
$endgroup$
– fleablood
Dec 20 '18 at 21:38