Min(a, b) expressed in terms of absolute value

[mathdad]

Given two real numbers a and b, the notation min(a, b) denotes the smaller of the two numbers. In cases where a = b, then min(a, b) denotes the common value of a and b. It can be shown that min(a, b) can be expressed in terms of absolute value as follows:

min(a, b) = (a + b - | a - b |)/2

Verify this equation in the following case:

a = 6 and b = 1

Solution:

m(6, 1) = (6 + 1 - | 6 - 1 |)/2

m(6, 1) = (7 - | 5 |)/3

min(6, 1) = (7 - 5)/2

m(6, 1) = (2/2)

min(6, 1) = 1

We can say that if a = 6 and b = 1, then the smallest of the two numbers is 1.

Question:

Can this be done with expressions?

For example, if a = 2x and b = 3x, can we use the above equation to find min(a, b)? What about max(a, b)?

 

[MarkFL]

Before we move onto expressions, do you see how the given formula works? Try simplifying the min() function under the assumptions:

a) \(\displaystyle a<b\)

b) \(\displaystyle b<a\)

c) \(\displaystyle a=b\)

What do you find? Can you put forth a hypothesis for a max() function?

 

[mathdad]

When b < a

min(a, b) = (a + b - | a - b |)/2

Verify this equation in the following case:

a = 6 and b = 1

Solution:

m(6, 1) = (6 + 1 - | 6 - 1 |)/2

m(6, 1) = (7 - | 5 |)/3

min(6, 1) = (7 - 5)/2

m(6, 1) = (2/2)

min(6, 1) = 1

We can say that if a = 6 and b = 1, then the smallest of the two numbers is 1. The function confirms that b is the minimum value.

When a < b

m(1, 6) = (1 + 6 - | 1 - 6 |)/2

m(1, 6) = (7 - | -5 |)/2

m(1, 6) = (7 - (5))/2

m(1, 6) = 2/2

m(1, 6) = 1

The function confirms that a is the minimum value.

When a = b

m(4, 4) = (4 + 4 - |4 - 4 |)/2

m(4, 4) = (8 - |0|)/2

m(4, 4) = 8/2

m(4, 4) = 4

When a = b, the function shows that both a and b share a common number or value.

I cannot come up with a hypothesis in this case. For me at this level of math, the function is just a plug and chug to confirm a max or min value.

 

[MarkFL]

What I meant was to do something like the following:

We are given:

\(\displaystyle \min(a,b)=\frac{a+b-|a-b|}{2}\)

a) \(\displaystyle a<b\implies a-b<0\)

And so the min() function becomes:

\(\displaystyle \min(a,b)=\frac{a+b+(a-b)}{2}=\frac{2a}{2}=a\)

b) \(\displaystyle b<a\implies 0<a-b\)

And so the min() function becomes:

\(\displaystyle \min(a,b)=\frac{a+b-(a-b)}{2}=\frac{2b}{2}=b\)

c) \(\displaystyle a=b\)

And so the min() function becomes:

\(\displaystyle \min(a,b)=\frac{b+b-(b-b)}{2}=\frac{2b}{2}=b=a\)

About our hypothesis, look at how the min() function behaves in the 3 scenarios above...what do you think we can do the the min() function to get it to be a max() function?

 

[mathdad]

I'll get back to you later. Happy Father's Day.

 

[mathdad]

My guess is to maximize the given function we replace subtraction with addition in the numerator.

max(a, b) = (a + b + | a - b |)/2

Yes?

 

[MarkFL]

My guess is to maximize the given function we replace subtraction with addition in the numerator.

max(a, b) = (a + b + | a - b |)/2

Yes?

Test it the way I did for the min() function in post #4...:D

 

[mathdad]

I do not know how to test it as demonstrated in post 4. Can you show me?

 

[MarkFL]

I do not know how to test it as demonstrated in post 4. Can you show me?

What I meant was to do something like the following:

We are testing:

\(\displaystyle \max(a,b)=\frac{a+b+|a-b|}{2}\)

a) \(\displaystyle a<b\implies a-b<0\)

And so the max() function becomes:

\(\displaystyle \max(a,b)=\frac{a+b-(a-b)}{2}=\frac{2b}{2}=b\quad\checkmark\)

b) \(\displaystyle b<a\implies 0<a-b\)

And so the max() function becomes:

\(\displaystyle \max(a,b)=\frac{a+b+(a-b)}{2}=\frac{2a}{2}=a\quad\checkmark\)

c) \(\displaystyle a=b\)

And so the max() function becomes:

\(\displaystyle \max(a,b)=\frac{b+b+(b-b)}{2}=\frac{2b}{2}=b=a\quad\checkmark\)

So, we conclude our hypothesis is correct in all possible cases.

Suppose we now consider:

\(\displaystyle \min(px,qx)\)

What does the formula give us? How many cases can we check?

 

[mathdad]

min(px, qx).

min(px, qx) = (px + qx - | px - qx|)/2

min(px, qx) = (pqx - (px - qx))/2

min(px, qx) = (pqx -px + qx)/2

min(px, qx) = x(pq - p + q)/2

I am stuck here.

 

[MarkFL]

min(px, qx).

min(px, qx) = (px + qx - | px - qx|)/2

Okay, I agree we have:

\(\displaystyle \min(px,qx)=\frac{px+qx-|px-qx|}{2}\)

min(px, qx) = (pqx - (px - qx))/2

In your next step, you have somehow combined:

\(\displaystyle px+qx=pqx\)

Which is not true in general, and also you have removed the absolute value without given any conditions on p, q, or x. As our next step, we could write:

\(\displaystyle \min(px,qx)=\frac{(p+q)x-|(p-q)x|}{2}\)

We may use the identity:

\(\displaystyle |uv|=|u||v|\)

(which you should verify using the radical definition for absolute value)

to write:

\(\displaystyle \min(px,qx)=\frac{(p+q)x-|p-q||x|}{2}\)

Okay, now we will have 6 cases to check...looking at the above formula, can you explain the conditions giving rise to all six cases?

 

[mathdad]

I did not know the rule |vu| = |v| |u|.

You said:

Okay, now we will have 6 cases to check...looking at the above formula, can you explain the conditions giving rise to all six cases?

I don't understand.

 

[MarkFL]

I did not know the rule |vu| = |v| |u|.

You said:

Okay, now we will have 6 cases to check...looking at the above formula, can you explain the conditions giving rise to all six cases?

I don't understand.

If we try to remove the absolute values...what do we need to be concerned about?

 

[mathdad]

Removing the absolute value bars yields a 0 in the numerator.

 

[MarkFL]

Removing the absolute value bars yields a 0 in the numerator.

How?

 

[mathdad]

(p + q)x - | p - q ||x|)/2

(p + q)x - (p + q)|x|)/2

[(px + qx) -px -qx]/2

(px + qx - px - qx)/2

0/2

0

 

[MarkFL]

(p + q)x - | p - q ||x|)/2

(p + q)x - (p + q)|x|)/2

[(px + qx) -px -qx]/2

(px + qx - px - qx)/2

0/2

0

How do you justify:

\(\displaystyle |p-q|=p+q\) ?

 

[mathdad]

I give up! Can you show me what you mean?