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I came across the following problem in my self-study of real analysis:

For any real numbers $a$ and $b$, show that $$\max \{a,b \} = \frac{1}{2}(a+b+|a-b|)$$ and $$\min\{a,b \} = \frac{1}{2}(a+b-|a-b|)$$

So $a \geq b$ iff $a-b \ge0$ and $b \ge a$ iff $b-a \ge 0$. At first glance, it seems like an average of distances. For the first case, go to the point $a+b$, add $|a-b|$ and divide by $2$. Similarly with the second case.

Would you just break it up in cases and verify the formulas? Or do you actually need to come up with the formulas?

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One way different from any posted thus far (that I am aware of) is given in the following paper: Dan Kalman, The maximum and minimum of two numbers using the quadratic formula, College Mathematics Journal 15 #4 (September 1984), 329-330. –  Dave L. Renfro Mar 26 at 16:07

2 Answers 2

up vote 13 down vote accepted

How can you come up with the formulas if they're already given? :)

Yes, you're right, it is a good idea to look at this formula as strongly related to an average. I find it more intuitive to look at it as $$\max{\{a,b\}} = \frac{a+b}{2} + \frac{|a-b|}{2},$$ so go to the midpoint of $a$ and $b$ and add half the distance $|a-b|$ between them to get to the larger among them. Similarly for the second formula.

Yes, the proof is probably easiest if you break it into the two cases you suggest and use the definition of the absolute value.

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So we wan modify this for any $n$. In particular, for $n=3$, $\max \{a,b \} = \frac{a+b}{3} + |a-b|$ and $\min \{a,b \} = \frac{a+b}{3} - |a-b|$? –  Damien Jun 22 '11 at 17:38
    
Bueler: Can we instead start at $\frac{a+b}{3}$ and get to $\max \{a,b \}$? In other words, do we have to choose the midpoint? –  Damien Jun 22 '11 at 17:47
    
@Damien: (this was addressing the previous version of your comment) No, the formula you're suggesting is wrong: Take $a = 0$ and $b=6$, for instance. Your formula gives $8$. In my answer I just rewrote your formula $$\frac{1}{2}(a+b+|a-b|) = \frac{1}{2}(a+b) + \frac{1}{2}|a-b|.$$ –  t.b. Jun 22 '11 at 17:47
    
In other words, suppose someone told me to come up with a formula for $\max \{a,b \}$. Can I start at $\frac{a+b}{3}$ and get to the maximum also? Or is $\frac{a+b}{2}$ the only option. –  Damien Jun 22 '11 at 17:50
    
@Damien: But what meaning does $\frac{a+b}{3}$ have? The formula you doesn't work so simply, consider what happens when you replace $a$ by $-a$ and $b$ by $-b$? The case "$n=2$" is simpler becuase the midpoint of $-a,-b$ is $-$ the midpoint of $a,b$. –  t.b. Jun 22 '11 at 17:51

The formulas can be thought of in geometric terms. First you know that the distance between the points $a, b$ is just the absolute value of their difference, that is $d = |a - b|$. Now, the midpoint between $a, b$ is just $\frac{a+b}{2}$. Then each formula just tells you the following:

  1. To get the maximum $\max{(a, b)}$ you step on the midpoint $\frac{a+b}{2}$ and then "walk" half the distance between them on the positive direction on real line, that is, you add $\frac{|a-b|}{2}$ so you get $$\max{(a, b)} = \frac{a+b + |a-b|}{2}$$
  2. To get the minimum $\min{(a, b)}$ you would do the same thing, but you would "walk" now in the negative direction, so you would subtract $\frac{|a-b|}{2}$ to get now $$\min{(a, b)} = \frac{a+b - |a-b|}{2}$$
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