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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?

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.

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.

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 $x+ylt0$.