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

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 \(x,y\) such that \(\displaystyle\underline{x\lt y}\), which is a very important assumption for what will come. If we graph those into the real number line then we will see that \(x\) is in the left in regards of \(y\) as shown in this image:

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

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 \((x+y)/2\) to the smallest number, that is \(x\), or to the biggest number \(y\). To see what I mean look carefully at this image:

So by adding the distance from \((x+y)/2\) to \(y\) to \((x+y)/2\) itself we will obtain \(y\). And if we substract the distance from \((x+y)/2\) to \(x\) to \((x+y)/2\) itself we will obtain \(x\). So we actually found a way to get \(x\) and \(y\) 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 \(a\) to \(b\)? We use absolute values! To get: \(\text{distance from }a\text{ to }b=|a-b|.\) 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:

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

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

## Exercises:

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

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

I hope this helps. Best wishes, \(\cal H\)akim.

## Comments

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TopNewestWell done Champ – Ayush Mohare · 2 years, 11 months ago

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Thank you! – Ayesha Nasir · 2 years, 11 months ago

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Great note!!! Thanks.... :) – Krishna Ramesh · 2 years, 11 months ago

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Nice approach .. :) – Ramesh Goenka · 2 years, 11 months ago

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Out of topic, are you Hakim on MSE? – Tunk-Fey Ariawan · 2 years, 9 months ago

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Tunk-Fey on MSE. – حكيم الفيلسوف الضائع · 2 years, 8 months ago

Yeah, and you areLog in to reply