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I am trying to find image of $(x,y)\mapsto \max(x,y)$ and $(x,y)\mapsto\min(x,y)$, defined as follows:

$$\Bbb R^2\to\Bbb R:\quad \max(x,y) = \begin{cases} x \text{ if } x \ge y, \\ y \text{ otherwise};\end{cases}$$

$$\Bbb N^2\to\Bbb R :\quad \min(x,y) = \begin{cases} y \text{ if } x \ge y, \\ x \text{ otherwise}.\end{cases}$$

I am not quite sure where to start.

Basically, if I take $\max(x,y)$, I am trying to prove that $S=T$, where $S$ is the respective image, and $T$ for $\max(x,y)$ is: $z = x$ for $z\in T$, if $x \ge y$ and $z = y$ for $z\in T$, if $x < y$.

So, $S\subseteq T$. But how do I prove that all elements in $T$ are in $S$? Thanks for any pointers!

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    What is the domain you are working with? $\mathbb{R}^2$? Without knowing that, it will be impossible to answer your question.2012-05-30
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    for max(x,y) it's f: R^2 -> R, for min(x,y) it's f: N^2 -> R2012-05-30
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    I thought you wanted the image of the ratio $\frac{\max}{\min}$. Ambiguous usage of the forward slash! Also, equations like `(x,y)=max(x,y)` don't make sense; you perhaps wanted to write something like $(x,y)\mapsto\max(x,y)$.2012-05-30
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    Sorry, yes, this is what I meant. Was not familiar how to use notation on this site. Now I know :)2012-05-30
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    I know this is well after the fact, but if you wish to give explicit formulae for these functions and avoid the piecewise notation, we can define $$\max(x,y)=\frac{x+y+|x-y|}{2}$$ and $$\min(x,y)=\frac{x+y-|x-y|}{2},$$ with appropriate domains.2013-01-26

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