Chapters

Minimum and Maximum

So, let's say you are writing a mathematical program, and you reason 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{\frac{\left|x-y\right|}{x-y}+1}{2}\Rightarrow\left\{+\rightarrow1, -\rightarrow 0 \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!

Complex Functions Divisibility

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Minimum and Maximum

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