How Can We Prove the Max and Min Formulas Using Basic Algebraic Operations?

[Seydlitz]

Homework Statement

The function max will return which number is the largest between the two parameters x and y. The function min will return the smallest number between the two. If the numbers are the same they will give out that same value.

Prove that

<br /> max(x,y) = \frac{x+y+|y-x|}{2}<br /> \\<br /> min(x,y) =\frac{x+y-|y-x|}{2}<br />

Homework Equations

None

The Attempt at a Solution

Suppose $x=y=0$
<br /> max(0,0) = <br /> \\<br /> \frac{0+0+|0-0|}{2} = 0<br /> <br />
Suppose $x = a+1$ and $y = a$
<br /> \begin{align*}<br /> max((a+1),a) &amp;= \frac{(a+1)+a+|a-(a+1)|}{2}\\<br /> &amp;=\frac{a+1+a+1}{2}\\<br /> &amp;=\frac{2a+2}{2} \\<br /> &amp;=(a+1) \\<br /> \end{align*}<br />
Where $x>y$

The similar reasoning can be applied to the case where $y>x$ and to $min(x,y)$ function.

Is this done or enough? I'm not sure if this is already enough for the purpose, to prove a formula. To what extent should we go so that it's correct?

Thank You

 

[jbunniii]

Your special cases, $x = y = 0$ and $x = a+1, y = a$ aren't going to do much for you.

Instead, try considering these two cases: $x \geq y$ and $x < y$.

 

[Seydlitz]

Your special cases, $x = y = 0$ and $x = a+1, y = a$ aren't going to do much for you.

Instead, try considering these two cases: $x \geq y$ and $x < y$.

Ok considering the given cases, then it is certain that the prove rest on the inequality $|y-x|$ and what happens with it when $x\geq y$ and $x<y$. I think I can do the symbolic manipulation to prove that it is indeed true.

Though my question is why I've to prove the formula using those cases? What is the difference logically with my a+1 case? (It rests with the fact that $x>y$) Is it that because they are more general, rather than when a+1?

 

[jbunniii]

Ok considering the given cases, then it is certain that the prove rest on the inequality $|y-x|$ and what happens with it when $x\geq y$ and $x<y$. I think I can do the symbolic manipulation to prove that it is indeed true.

Note that I suggested considering these two cases because (1) they are the only possibilities, and (2) the expression $|y-x|$ can be simplified in both cases (try rewriting it without the absolute value).

Though my question is why I've to prove the formula using those cases? What is the difference logically with my a+1 case? (It rests with the fact that $x>y$) Is it that because they are more general, rather than when a+1?

Yes. Your solution does not cover, for example, the case where $x = 1$ and $y = 1/2$.

 

[Seydlitz]

Note that I suggested considering these two cases because (1) they are the only possibilities, and (2) the expression $|y-x|$ can be simplified in both cases (try rewriting it without the absolute value).

Yes. Your solution does not cover, for example, the case where $x = 1$ and $y = 1/2$.

Yes the final simplified statement match with the original case once the inequality is changed, say $|y-x|$ to $(y-x)$ if $x<y$. Thanks for your help.

Ah yes! I really didn't think of that.