The max and min functions: what are they? A response to Milly Choochoo.

2 months and 3 weeks ago the user Milly Choochoo asked in this thread a question about how to deal with expressions involving max and min functions. Unfortunately, he didn't receive much attention. But I wrote to him an answer, that I will share with you if it can help anyone.


Dear Milly;

What is max{a,b}\operatorname{max}\{a,b\}? It's just a function that gives us the biggest number from a,ba,b. The same thing for min{a,b}\operatorname{min}\{a,b\}, except that this time, it gives us the smallest number from a,ba,b. Let's begin with some few examples to wrap our heads around those two functions. From the definitions we got max{3,8}\operatorname{max}\{-3,8\} is equal to 88. Since out of 3-3 and 88 it's the latter that is the biggest. And for min{7,1}\operatorname{min}\{-7,1\} we can see that it is equal to 7-7 since this last one is the smallest number out of 7-7 and 11.

Now that we have understood those functions, we need to determine a way that will reveal to us a formula to find the min and max of any two numbers. Let's take two arbitrary numbers as an example like x,yx,y such that x<y\displaystyle\underline{x\lt y}, which is a very important assumption for what will come. If we graph those into the real number line then we will see that xx is in the left in regards of yy as shown in this image:

http://i.stack.imgur.com/OpjU2.png http://i.stack.imgur.com/OpjU2.png

One of the numbers we can define based on xx and yy is the their average, (x+y)/2(x+y)/2. What this average graphically means, is that it represent the number whose corresponding point on the number line is a midpoint of the segment [x,y][x,y], as this image show:

http://i.stack.imgur.com/x10g3.png http://i.stack.imgur.com/x10g3.png

We've made some neat progress till now, but we'll get to the next point only by seeing that this midpoint unlocks to us a new possibility. This midpoint makes an axis of symmetry, and so with the rightly chosen way we can go from the quantity (x+y)/2(x+y)/2 to the smallest number, that is xx, or to the biggest number yy. To see what I mean look carefully at this image:

http://i.stack.imgur.com/ozkyR.png http://i.stack.imgur.com/ozkyR.png

So by adding the distance from (x+y)/2(x+y)/2 to yy to (x+y)/2(x+y)/2 itself we will obtain yy. And if we substract the distance from (x+y)/2(x+y)/2 to xx to (x+y)/2(x+y)/2 itself we will obtain xx. So we actually found a way to get xx and yy out of some formula that will soon discover. Note however that those two distances are the same. And how is distance represented? I mean how to find the distance from aa to bb? We use absolute values! To get: distance from a to b=ab.\text{distance from }a\text{ to }b=|a-b|. So if we use that property we will get that:

\eqalign{ \operatorname{max}\{x,y\}&=\dfrac{x+y}{2}+\left|\dfrac{x+y}{2}-y\right|\\ &=\dfrac{x+y}{2}+\left|\dfrac{x+y-2y}{2}\right| \\ &=\dfrac{x+y}{2}+\left|\dfrac{x-y}{2}\right| \\ &=\dfrac{x+y}{2}+\left|\dfrac{1}2(x-y)\right| \\ &=\dfrac{x+y}{2}+\dfrac12\left|(x-y)\right| \quad\text{using properties of the }|\,\cdot\,| \\ &=\dfrac{x+y}{2}+\dfrac{\left|(x-y)\right|}2 \\ &=\dfrac{x+y+\left|x-y\right|}2\quad\blacksquare \\ }

Now I'm sure you can do the same yourself to prove that:

min{x,y}=x+yxy2.\operatorname{min}\{x,y\}=\dfrac{x+y-\left|x-y\right|}2.

So now whenever you see a problem involving min{x,y}\operatorname{min}\{x,y\} or max{x,y}\operatorname{max}\{x,y\} just replace it with the formulas we found above, and the rest would be easy.


Exercises:

1) Prove that min{x,x}=max{x,x}=x.\operatorname{min}\{x,x\}=\operatorname{max}\{x,x\}=x.

2) Derive a formula for min{x,y,z}\operatorname{min}\{x,y,z\} and max{x,y,z}.\operatorname{max}\{x,y,z\}.


I hope this helps. Best wishes, \calH\cal Hakim.

Note by حكيم الفيلسوف الضائع
5 years, 5 months ago

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Well done Champ

Ayush Mohare - 5 years, 5 months ago

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Nice approach .. :)

Ramesh Goenka - 5 years, 5 months ago

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Great note!!! Thanks.... :)

Krishna Ramesh - 5 years, 5 months ago

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Thank you!

Ayesha Nasir - 5 years, 5 months ago

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Out of topic, are you Hakim on MSE?

Tunk-Fey Ariawan - 5 years, 3 months ago

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Yeah, and you are Tunk-Fey on MSE.

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