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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. –  Cameron Buie May 30 '12 at 15:59
    
for max(x,y) it's f: R^2 -> R, for min(x,y) it's f: N^2 -> R –  c1571333 May 30 '12 at 16:00
    
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 $(x,y)\mapsto\max(x,y)$. –  anon May 30 '12 at 16:08
    
Sorry, yes, this is what I meant. Was not familiar how to use notation on this site. Now I know :) –  c1571333 May 30 '12 at 16:16
    
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. –  Cameron Buie Jan 26 '13 at 21:44

1 Answer 1

Let's define $f:\mathbb{R}^2\to\mathbb{R}$ by $f(x,y)=\max(x,y)$ and $g:\mathbb{N}^2\to\mathbb{R}$ by $g(x,y)=\min(x,y)$. I claim that the range of $f$ is all of $\mathbb{R}$ and the range of $g$ is all of $\mathbb{N}$. The first claim and half of the second come from considering the following: $f(x,x)=?$, $g(x,x)=?$.

The other half of the second claim comes by justifying that if $x,y\in\mathbb{N}$, then $g(x,y)\in\mathbb{N}$.

Does that get you unstuck? Work with it a bit, and let me know.

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