What does $frac{x+y}{x-y}$ represent? [on hold]












-2














If you have two positive variables like $x$ and $y$ such that $x>y$, how could the following correlation be interpreted?



$$frac{x+y}{x-y}$$



What are the minimum and maximum possible values?










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put on hold as off-topic by Adrian Keister, zipirovich, Austin Mohr, Mark Bennet, amWhy 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, zipirovich, Mark Bennet, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    What does this have to do with linear algebra?
    – Omnomnomnom
    Jan 3 at 20:36






  • 1




    Just take $x=y+1$ and you can make it arbitrarily large by making $y$ large. You can do something similar to find the smallest values.
    – MAX
    Jan 3 at 20:38








  • 1




    We could rewrite your expression as $$ frac{x+y}{x-y} = 1 + frac{2y}{x-y} $$ I don't know if this is a particularly useful thing to do though
    – Omnomnomnom
    Jan 3 at 20:38






  • 2




    It might also be helpful to think of your expression as $$ frac{x+y}{x-y} = frac{1 + frac yx}{1 - frac yx} $$
    – Omnomnomnom
    Jan 3 at 20:40










  • What do you mean "how could the following correlation be interpreted"? It's x +y divided by x-y. It's exactly what it says it is.
    – fleablood
    Jan 3 at 21:26
















-2














If you have two positive variables like $x$ and $y$ such that $x>y$, how could the following correlation be interpreted?



$$frac{x+y}{x-y}$$



What are the minimum and maximum possible values?










share|cite|improve this question









New contributor




AHBagheri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











put on hold as off-topic by Adrian Keister, zipirovich, Austin Mohr, Mark Bennet, amWhy 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, zipirovich, Mark Bennet, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    What does this have to do with linear algebra?
    – Omnomnomnom
    Jan 3 at 20:36






  • 1




    Just take $x=y+1$ and you can make it arbitrarily large by making $y$ large. You can do something similar to find the smallest values.
    – MAX
    Jan 3 at 20:38








  • 1




    We could rewrite your expression as $$ frac{x+y}{x-y} = 1 + frac{2y}{x-y} $$ I don't know if this is a particularly useful thing to do though
    – Omnomnomnom
    Jan 3 at 20:38






  • 2




    It might also be helpful to think of your expression as $$ frac{x+y}{x-y} = frac{1 + frac yx}{1 - frac yx} $$
    – Omnomnomnom
    Jan 3 at 20:40










  • What do you mean "how could the following correlation be interpreted"? It's x +y divided by x-y. It's exactly what it says it is.
    – fleablood
    Jan 3 at 21:26














-2












-2








-2







If you have two positive variables like $x$ and $y$ such that $x>y$, how could the following correlation be interpreted?



$$frac{x+y}{x-y}$$



What are the minimum and maximum possible values?










share|cite|improve this question









New contributor




AHBagheri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











If you have two positive variables like $x$ and $y$ such that $x>y$, how could the following correlation be interpreted?



$$frac{x+y}{x-y}$$



What are the minimum and maximum possible values?







algebra-precalculus






share|cite|improve this question









New contributor




AHBagheri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




AHBagheri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 20:46









amWhy

192k28224439




192k28224439






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asked Jan 3 at 20:33









AHBagheri

1022




1022




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New contributor





AHBagheri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






AHBagheri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




put on hold as off-topic by Adrian Keister, zipirovich, Austin Mohr, Mark Bennet, amWhy 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, zipirovich, Mark Bennet, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by Adrian Keister, zipirovich, Austin Mohr, Mark Bennet, amWhy 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, zipirovich, Mark Bennet, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    What does this have to do with linear algebra?
    – Omnomnomnom
    Jan 3 at 20:36






  • 1




    Just take $x=y+1$ and you can make it arbitrarily large by making $y$ large. You can do something similar to find the smallest values.
    – MAX
    Jan 3 at 20:38








  • 1




    We could rewrite your expression as $$ frac{x+y}{x-y} = 1 + frac{2y}{x-y} $$ I don't know if this is a particularly useful thing to do though
    – Omnomnomnom
    Jan 3 at 20:38






  • 2




    It might also be helpful to think of your expression as $$ frac{x+y}{x-y} = frac{1 + frac yx}{1 - frac yx} $$
    – Omnomnomnom
    Jan 3 at 20:40










  • What do you mean "how could the following correlation be interpreted"? It's x +y divided by x-y. It's exactly what it says it is.
    – fleablood
    Jan 3 at 21:26














  • 1




    What does this have to do with linear algebra?
    – Omnomnomnom
    Jan 3 at 20:36






  • 1




    Just take $x=y+1$ and you can make it arbitrarily large by making $y$ large. You can do something similar to find the smallest values.
    – MAX
    Jan 3 at 20:38








  • 1




    We could rewrite your expression as $$ frac{x+y}{x-y} = 1 + frac{2y}{x-y} $$ I don't know if this is a particularly useful thing to do though
    – Omnomnomnom
    Jan 3 at 20:38






  • 2




    It might also be helpful to think of your expression as $$ frac{x+y}{x-y} = frac{1 + frac yx}{1 - frac yx} $$
    – Omnomnomnom
    Jan 3 at 20:40










  • What do you mean "how could the following correlation be interpreted"? It's x +y divided by x-y. It's exactly what it says it is.
    – fleablood
    Jan 3 at 21:26








1




1




What does this have to do with linear algebra?
– Omnomnomnom
Jan 3 at 20:36




What does this have to do with linear algebra?
– Omnomnomnom
Jan 3 at 20:36




1




1




Just take $x=y+1$ and you can make it arbitrarily large by making $y$ large. You can do something similar to find the smallest values.
– MAX
Jan 3 at 20:38






Just take $x=y+1$ and you can make it arbitrarily large by making $y$ large. You can do something similar to find the smallest values.
– MAX
Jan 3 at 20:38






1




1




We could rewrite your expression as $$ frac{x+y}{x-y} = 1 + frac{2y}{x-y} $$ I don't know if this is a particularly useful thing to do though
– Omnomnomnom
Jan 3 at 20:38




We could rewrite your expression as $$ frac{x+y}{x-y} = 1 + frac{2y}{x-y} $$ I don't know if this is a particularly useful thing to do though
– Omnomnomnom
Jan 3 at 20:38




2




2




It might also be helpful to think of your expression as $$ frac{x+y}{x-y} = frac{1 + frac yx}{1 - frac yx} $$
– Omnomnomnom
Jan 3 at 20:40




It might also be helpful to think of your expression as $$ frac{x+y}{x-y} = frac{1 + frac yx}{1 - frac yx} $$
– Omnomnomnom
Jan 3 at 20:40












What do you mean "how could the following correlation be interpreted"? It's x +y divided by x-y. It's exactly what it says it is.
– fleablood
Jan 3 at 21:26




What do you mean "how could the following correlation be interpreted"? It's x +y divided by x-y. It's exactly what it says it is.
– fleablood
Jan 3 at 21:26










3 Answers
3






active

oldest

votes


















1














If $x > y$ then $x - y>0$ and $x + y > 0$ so $frac {x+y}{x-y} > 0$.



$x+y > x-y$ so $frac {x+y}{x-y} > 1$.



Other than that can make it as big or as small as we like.



To solve for $frac {x+y}{x-y} = M > 1$ we need $x+y = (x-y)M$.



If we let $x+y = M$ and $x-y = 1$ we can solve:



$x = y+1$



$x + y = y+1 + y = 2y+1 = M$



$y = frac {M-1}2$ (which is larger than $0$ because $M> 1$)



$x = frac {M-1}2 + 1$.



$M$ can be as close to $1$ as we like or as large as we like. We will always be able to find $x,y > 0$ where $frac {x+y}{x-y} = M$ for any possible value larger than $1$.



So there is no minimum and there is not maximum.



Although there is no minimum, we can't get as small as $1$ but we can get as close to $1$ as we like. So this means $1$ is the greatest lower bound of values or an infimum possible values.



You shouldn't confuse infimum, an absolute lowest limit that may or may not be possible, with minimum, the lowest value that IS reached.






share|cite|improve this answer





























    0














    Assuming that $,x > y > 0,,$ then let $,x = a+b,, y = a-b ,$ where $,a > b > 0.,$The ratio $, r = (x+y)/(x-y) = a/b.,$ Since $, x = a+b, $ and $, a>b, ,$ then $,x,$ is partitioned into two parts $,a,$ and $,b,$ where $,a,$ is the larger part. The ratio $,r,$ of $,a,$ to $,b,$ is at least $,1,$ and unboundedly large as $,b,$ gets small. Thus, $, 1 < r < infty ,$ is the best bounds.






    share|cite|improve this answer





























      0














      It is a function of two variables so it represents a surface that sits above and below the $x,y$ plane. Very much like $f(x)$ sits above and below the number line.

      Here, $xgt y$ so only that part of the $(x,y)$ plane is involved - to the righr of $x=y$.

      In the region $x+ygt0$, the surface is above the plane; it cuts the plane at $x+y=0$ and is below the plane in the region $x+ylt0$.






      share|cite|improve this answer




























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1














        If $x > y$ then $x - y>0$ and $x + y > 0$ so $frac {x+y}{x-y} > 0$.



        $x+y > x-y$ so $frac {x+y}{x-y} > 1$.



        Other than that can make it as big or as small as we like.



        To solve for $frac {x+y}{x-y} = M > 1$ we need $x+y = (x-y)M$.



        If we let $x+y = M$ and $x-y = 1$ we can solve:



        $x = y+1$



        $x + y = y+1 + y = 2y+1 = M$



        $y = frac {M-1}2$ (which is larger than $0$ because $M> 1$)



        $x = frac {M-1}2 + 1$.



        $M$ can be as close to $1$ as we like or as large as we like. We will always be able to find $x,y > 0$ where $frac {x+y}{x-y} = M$ for any possible value larger than $1$.



        So there is no minimum and there is not maximum.



        Although there is no minimum, we can't get as small as $1$ but we can get as close to $1$ as we like. So this means $1$ is the greatest lower bound of values or an infimum possible values.



        You shouldn't confuse infimum, an absolute lowest limit that may or may not be possible, with minimum, the lowest value that IS reached.






        share|cite|improve this answer


























          1














          If $x > y$ then $x - y>0$ and $x + y > 0$ so $frac {x+y}{x-y} > 0$.



          $x+y > x-y$ so $frac {x+y}{x-y} > 1$.



          Other than that can make it as big or as small as we like.



          To solve for $frac {x+y}{x-y} = M > 1$ we need $x+y = (x-y)M$.



          If we let $x+y = M$ and $x-y = 1$ we can solve:



          $x = y+1$



          $x + y = y+1 + y = 2y+1 = M$



          $y = frac {M-1}2$ (which is larger than $0$ because $M> 1$)



          $x = frac {M-1}2 + 1$.



          $M$ can be as close to $1$ as we like or as large as we like. We will always be able to find $x,y > 0$ where $frac {x+y}{x-y} = M$ for any possible value larger than $1$.



          So there is no minimum and there is not maximum.



          Although there is no minimum, we can't get as small as $1$ but we can get as close to $1$ as we like. So this means $1$ is the greatest lower bound of values or an infimum possible values.



          You shouldn't confuse infimum, an absolute lowest limit that may or may not be possible, with minimum, the lowest value that IS reached.






          share|cite|improve this answer
























            1












            1








            1






            If $x > y$ then $x - y>0$ and $x + y > 0$ so $frac {x+y}{x-y} > 0$.



            $x+y > x-y$ so $frac {x+y}{x-y} > 1$.



            Other than that can make it as big or as small as we like.



            To solve for $frac {x+y}{x-y} = M > 1$ we need $x+y = (x-y)M$.



            If we let $x+y = M$ and $x-y = 1$ we can solve:



            $x = y+1$



            $x + y = y+1 + y = 2y+1 = M$



            $y = frac {M-1}2$ (which is larger than $0$ because $M> 1$)



            $x = frac {M-1}2 + 1$.



            $M$ can be as close to $1$ as we like or as large as we like. We will always be able to find $x,y > 0$ where $frac {x+y}{x-y} = M$ for any possible value larger than $1$.



            So there is no minimum and there is not maximum.



            Although there is no minimum, we can't get as small as $1$ but we can get as close to $1$ as we like. So this means $1$ is the greatest lower bound of values or an infimum possible values.



            You shouldn't confuse infimum, an absolute lowest limit that may or may not be possible, with minimum, the lowest value that IS reached.






            share|cite|improve this answer












            If $x > y$ then $x - y>0$ and $x + y > 0$ so $frac {x+y}{x-y} > 0$.



            $x+y > x-y$ so $frac {x+y}{x-y} > 1$.



            Other than that can make it as big or as small as we like.



            To solve for $frac {x+y}{x-y} = M > 1$ we need $x+y = (x-y)M$.



            If we let $x+y = M$ and $x-y = 1$ we can solve:



            $x = y+1$



            $x + y = y+1 + y = 2y+1 = M$



            $y = frac {M-1}2$ (which is larger than $0$ because $M> 1$)



            $x = frac {M-1}2 + 1$.



            $M$ can be as close to $1$ as we like or as large as we like. We will always be able to find $x,y > 0$ where $frac {x+y}{x-y} = M$ for any possible value larger than $1$.



            So there is no minimum and there is not maximum.



            Although there is no minimum, we can't get as small as $1$ but we can get as close to $1$ as we like. So this means $1$ is the greatest lower bound of values or an infimum possible values.



            You shouldn't confuse infimum, an absolute lowest limit that may or may not be possible, with minimum, the lowest value that IS reached.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 3 at 21:43









            fleablood

            68.4k22685




            68.4k22685























                0














                Assuming that $,x > y > 0,,$ then let $,x = a+b,, y = a-b ,$ where $,a > b > 0.,$The ratio $, r = (x+y)/(x-y) = a/b.,$ Since $, x = a+b, $ and $, a>b, ,$ then $,x,$ is partitioned into two parts $,a,$ and $,b,$ where $,a,$ is the larger part. The ratio $,r,$ of $,a,$ to $,b,$ is at least $,1,$ and unboundedly large as $,b,$ gets small. Thus, $, 1 < r < infty ,$ is the best bounds.






                share|cite|improve this answer


























                  0














                  Assuming that $,x > y > 0,,$ then let $,x = a+b,, y = a-b ,$ where $,a > b > 0.,$The ratio $, r = (x+y)/(x-y) = a/b.,$ Since $, x = a+b, $ and $, a>b, ,$ then $,x,$ is partitioned into two parts $,a,$ and $,b,$ where $,a,$ is the larger part. The ratio $,r,$ of $,a,$ to $,b,$ is at least $,1,$ and unboundedly large as $,b,$ gets small. Thus, $, 1 < r < infty ,$ is the best bounds.






                  share|cite|improve this answer
























                    0












                    0








                    0






                    Assuming that $,x > y > 0,,$ then let $,x = a+b,, y = a-b ,$ where $,a > b > 0.,$The ratio $, r = (x+y)/(x-y) = a/b.,$ Since $, x = a+b, $ and $, a>b, ,$ then $,x,$ is partitioned into two parts $,a,$ and $,b,$ where $,a,$ is the larger part. The ratio $,r,$ of $,a,$ to $,b,$ is at least $,1,$ and unboundedly large as $,b,$ gets small. Thus, $, 1 < r < infty ,$ is the best bounds.






                    share|cite|improve this answer












                    Assuming that $,x > y > 0,,$ then let $,x = a+b,, y = a-b ,$ where $,a > b > 0.,$The ratio $, r = (x+y)/(x-y) = a/b.,$ Since $, x = a+b, $ and $, a>b, ,$ then $,x,$ is partitioned into two parts $,a,$ and $,b,$ where $,a,$ is the larger part. The ratio $,r,$ of $,a,$ to $,b,$ is at least $,1,$ and unboundedly large as $,b,$ gets small. Thus, $, 1 < r < infty ,$ is the best bounds.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 3 at 21:49









                    Somos

                    13.1k11034




                    13.1k11034























                        0














                        It is a function of two variables so it represents a surface that sits above and below the $x,y$ plane. Very much like $f(x)$ sits above and below the number line.

                        Here, $xgt y$ so only that part of the $(x,y)$ plane is involved - to the righr of $x=y$.

                        In the region $x+ygt0$, the surface is above the plane; it cuts the plane at $x+y=0$ and is below the plane in the region $x+ylt0$.






                        share|cite|improve this answer


























                          0














                          It is a function of two variables so it represents a surface that sits above and below the $x,y$ plane. Very much like $f(x)$ sits above and below the number line.

                          Here, $xgt y$ so only that part of the $(x,y)$ plane is involved - to the righr of $x=y$.

                          In the region $x+ygt0$, the surface is above the plane; it cuts the plane at $x+y=0$ and is below the plane in the region $x+ylt0$.






                          share|cite|improve this answer
























                            0












                            0








                            0






                            It is a function of two variables so it represents a surface that sits above and below the $x,y$ plane. Very much like $f(x)$ sits above and below the number line.

                            Here, $xgt y$ so only that part of the $(x,y)$ plane is involved - to the righr of $x=y$.

                            In the region $x+ygt0$, the surface is above the plane; it cuts the plane at $x+y=0$ and is below the plane in the region $x+ylt0$.






                            share|cite|improve this answer












                            It is a function of two variables so it represents a surface that sits above and below the $x,y$ plane. Very much like $f(x)$ sits above and below the number line.

                            Here, $xgt y$ so only that part of the $(x,y)$ plane is involved - to the righr of $x=y$.

                            In the region $x+ygt0$, the surface is above the plane; it cuts the plane at $x+y=0$ and is below the plane in the region $x+ylt0$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 3 at 21:55









                            Empy2

                            33.5k12261




                            33.5k12261















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