Can I really factor a constant into the $\min$ function?

Say I have $\min(5x_1,x_2)$ and I multiply the whole function by $10$, i.e. $10\min(5x_1,x_2)$. Does that simplify to $\min(50x_1,10x_1)$? In one of my classes I think my professor did this but I'm not sure (he makes very hard to read and seemingly bad notes), and I'm just trying to put these notes together. Thanks!

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It may be useful (later) to know that $\min(a,b)=\frac12 (a+b-|a-b|)$. –  ՃՃՃ Feb 7 at 3:54
Wow thanks everyone. Everyone had good answers, too bad we aren't allowed to pick more than one correct answer. –  TheHopefulActuary Feb 7 at 4:02
That is not simplification, it is a functional identity when the factor is positive, for negative factors the identity switches from max to min or min to max, good exercise for you to show that. –  Arjang Feb 7 at 5:04

Let's try it out. There's three cases:

Case 1:

$5x_1 < x_2$, therefore $10\min(5x_1, x_2)=50x_1$

If $5x_1 < x_2$, then $50x_1 <10x_2$. Therefore, $\min(50x_1 ,10x_2) = 50x_1$.

Case 2:

$5x_1 > x_2$, therefore $10\min(5x_1, x_2)=10x_2$

If $5x_1 > x_2$, then $50x_1 > 10x_2$.

Therefore, $\min(50x_1 ,10x_2) = 10x_2$.

Case 3:

$5x_1 = x_2$, therefore $10\min(5x_1, x_2) = 50x_1 = 10x_2$

If $5x_1 = x_2$, then $50x_1 = 10x_2$.

Therefore, $\min(50x_1 ,10x_2) = 50x_1 = 10x_2$.

Summary

We get the same value from both $10\min(5x_1, x_2)$ and $\min(50x_1, 10x_2)$, therefore the two expressions are equal. However, my intuition warns me against extending this to say $c\min(x_1, x_2) = \min(cx_1, cx_2)$. (My thought is that, if $c < 0$, we would have $c\min(x_1, x_2) = \max(cx_1, cx_2)$, but I haven't checked.)

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Your thoughts are correct, instead of this special case, do the general case for c>0 and another for c<0. –  Arjang Feb 7 at 5:16
@Arjang I believe I will leave that as an exercise to the reader. It isn't that hard, and the process is the same as I've outlined above (except that I did it with an actual number, rather than with an arbitrary constant.) To the reader: If you can't figure out how to do it for a negative constant, leave a note! –  anorton Feb 7 at 13:02

Yes, that will work, assuming the constant is positive. Consider any numbers $a, b, c$, where $c > 0$. If $a \leq b$, then $\min(a,b) = a$ and $ca \leq cb$, so $c\min(a,b) = c(a) = \min(ca,cb)$. If $b < a$, then $cb < ca$, so $c\min(a,b) = cb = \min(ca,cb)$. So in either case $c\min(a,b) = \min(ca,cb)$. Hope that helps.

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Ys, that is legal as long as the constant is not negative. I.e., $10 \cdot \max(3, 5) = 10 \cdot 5 = 50$ is the same as $\max(10 \cdot 3, 10 \cdot 5) = 50$, but try multiplying by $-10$...

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Sure. Here's why: $\min(a, b)$ is equal to $a$, if it is smaller than $b$, and otherwise to $b$.

$\min(10a, 10b)$ is equal to $10a$, if $10a$ is smaller than $10b$, and otherwise to $10b$.

$10\cdot\min(a, b)$ is equal to $10a$, if $a$ is smaller than $b$, and otherwise to $10b$.

But $a$ is smaller than $b$ exactly when $10a$ is smaller than $10b$, so $\min(10a, 10b)$ and $10\cdot\min(a, b)$ always have the same value.

In general $k\cdot\min(x_1, x_2, \ldots, x_n) = \min(k\cdot x_1, k\cdot x_2, \ldots, k\cdot x_n)$ by the same argument.

The key thing to learn here is that this is not some game where you push symbols around on the paper according to arbitrary rules, and some pushing around is allowed and some isn't. These symbols mean things, and if you think about what they mean you can figure out what is correct and what isn't.

A comment correctly points out that my "in general" statement actually fails when $k\lt 0$. It might be a good exercise to try to find the place in the argument that fails when $k\lt 0$.

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Your "in general" statement might need to be qualified with $k \ge 0$ –  anorton Feb 7 at 3:48
Very useful, thanks. –  MJD Feb 7 at 3:50