Max & Min

So, let's say you are writing your nice little conjecture, and you write that $$x\ne y\ \Rightarrow\ \frac{x}{y}>1.$$ Right? The ratio must be more than one if they are not equal. But then, someone tells you, it's only true if $$x>y.$$ Argh! There goes your wonderful conjecture. You could write that $$x\ne y\Rightarrow\frac{\max\left(x,y\right)}{\min\left(x,y\right)}>1.$$ But, those are non-standard functions!

Let's try to make a $\max$ function using only standard functions.

What we'll do is figure out whether $x-y$ is positive or negative. Then, we'll return $x$ if it is positive and $y$ if it is negative.  To figure out whether it is positive or negative, we compute $$\text{isPositive}\left(x-y\right)\ =\ \frac{\left|x-y\right|}{x-y}\Rightarrow\left\{+\rightarrow1,\ -\rightarrow-1\right\}$$ If positive, we'll get $1,$ and if negative, we will get $-1.$ Great -- how do we put this to use? First,  we'll configure the positive-negative function to output binary. This means outputting $1$ if it is positive, and $0$ if it is negative.   $$\text{isPositive}\left(x-y\right)\ =\ \frac{\left|x-y\right|}{x-y}+1\Rightarrow\left\{+\rightarrow2,\ -\rightarrow0\right\}$$ $$\text{isPositive}\left(x-y\right)\ =\ \frac{\frac{\left|x-y\right|}{x-y}+1}{2}\Rightarrow\left\{+\rightarrow1,\ -\rightarrow0\right\}$$ Okay, cool. Now, when the function returns positive, we want to return $x.$ We can do that by multiplying by $x:$ $$\max\left(x,y\right)\ =\ x\cdot\frac{\frac{\left|x-y\right|}{x-y}+1}{2}\Rightarrow\left\{x\rightarrow x,\ y\rightarrow0\right\}$$ Oh, but now we get $0$ when the maximum is $y!$ We can add the same function, but comparing against $y:$ $$\max\left(x,y\right)\ =\ x\cdot\frac{\frac{\left|x-y\right|}{x-y}+1}{2}+y\cdot\frac{\frac{\left|x-y\right|}{y-x}+1}{2}\Rightarrow\left\{x\rightarrow x,\ y\rightarrow y\right\}$$

Great! Now we just need a minimum function. All we need to do is switch the outputs: $$\min\left(x,y\right)\ =\ y\cdot\frac{\frac{\left|x-y\right|}{x-y}+1}{2}+x\cdot\frac{\frac{\left|x-y\right|}{y-x}+1}{2}\Rightarrow\left\{x\rightarrow x,\ y\rightarrow y\right\}$$ Can you fix my conjecture for me now? Thanks!