Prove the formula for the maximum of two numbers

[nietzsche]

Homework Statement

The maximum of two numbers x and y is denoted by max(x,y). Thus max(-1,3) = max (3,3) = 3. Prove that:

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

Homework Equations

N/A

The Attempt at a Solution

I have no idea where to begin. I've thought about it for a long time, I swear! Any hints on how to get started?

 

[Brian-san]

You'll need to consider two cases: x>y and y>x. Then use that with the definition of absolute value. That should show the right results.

 

[nietzsche]

Here's an attempt:

Assume x > y .

<br /> \begin{align*}<br /> \mathrm{max}(x,y) &amp;= \frac{x+y+x-y}{2}\\<br /> \mathrm{max}(x,y) &amp;= \frac{x+y+|x-y|}{2}\\<br /> \mathrm{max}(x,y) &amp;= \frac{x+y+|y-x|}{2}<br /> \end{align*}<br />

But I haven't proved it yet...

 

[nietzsche]

You'll need to consider two cases: x>y and y>x. Then use that with the definition of absolute value. That should show the right results.

Okay, so I'm on the right track with that last post.

Assume y > x .
<br /> \begin{align*}<br /> \mathrm{max}(x,y) &amp;= \frac{x+y-x+y}{2}\\<br /> &amp;= \frac{x+y+|y-x|}{2}<br /> \end{align*}<br />

 

[nietzsche]

So is that a sufficient proof? Because it could just as well be

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

couldn't it?

 

[Brian-san]

With each case, you can show that the formula leaves you with one value x, or y, that should be enough for the proof.

 

[nietzsche]

With each case, you can show that the formula leaves you with one value x, or y, that should be enough for the proof.

Ah, thank you very much.

 

[nietzsche]

(1) Assume x > y .

<br /> \begin{align*}<br /> \mathrm{max}(x,y) &amp;= \frac{x+y+|y-x|}{2}\\<br /> &amp;= \frac{x+y+(x-y)}{2}\\<br /> &amp;= \frac{2x}{2}\\<br /> &amp;= x<br /> \end{align*}<br />(2) Now assume y > x .

<br /> \begin{align*}<br /> \mathrm{max}(x,y) &amp;= \frac{x+y+|y-x|}{2}\\<br /> &amp;= \frac{x+y+(y-x)}{2}\\<br /> &amp;= \frac{2y}{2}\\<br /> &amp;= y<br /> \end{align*}<br />Therefore:
<br /> \mathrm{max}(x,y) &amp;= \frac{x+y+|y-x|}{2}<br />

is true for all values of x,y provided x,y are real numbers.

(Although I'm still not thoroughly convinced that it's sufficient...)

 

[Brian-san]

Since x and y are arbitrary real numbers, I don't see why it wouldn't be considered a strong enough proof. You can make an argument about how you come to that particular formula, but I don't think it's necessary.

 

[ehild]

There is the third possibility that y=x. Include and your proof is complete.

ehild

 

[nietzsche]

There is the third possibility that y=x. Include and your proof is complete.

ehild

Thanks very much.

 

[emyt]

you're reading from the spivak textbook are you? your post on x^2 + xy + y^2 > 0 was also an exercise from the textbook

 

[nietzsche]

yes i am. it's such an awful textbook, it gives practically no examples.

or maybe i should say that it's a textbook that require a bit more thinking...