Question About Proof: Is the Explanation from Paint Document Complete?

[Miike012]

The explanation given in the paint document that I copied from a book does not seem complete.

They are saying that the sum of two numbers is least when the two numbers are equal.

Here is my explanation for why this is not always true.

Let a and b be two positive numbers.

I will denote the sum S₁ as 2a and the sum S₂ as a + b

Then I can say that S₁>S₂ if a>b and therefore S₁ = S₂ for b = a which is greater than S₂ for a>b.

Second explanation: Let a and b be a positive number and I will define the sum S₁ to be the sum of a finite number of n terms therefore
S₁ = na = M = M + (b-b) = M + 0
If I allow M = na to equal the second term in S₁ then I have na = 0 or a = 0, however I defined n to be positive and therefore a must equal itself. and therefore I can conclude that
na>0 or S₁> n*0 = 0 where RHS is equal to the sum of n zeros.


Am I misinterpreting what they are saying in the paint document?

 

[Ray Vickson]

The explanation given in the paint document that I copied from a book does not seem complete.

They are saying that the sum of two numbers is least when the two numbers are equal.

Here is my explanation for why this is not always true.

Let a and b be two positive numbers.

I will denote the sum S₁ as 2a and the sum S₂ as a + b

Then I can say that S₁>S₂ if a>b and therefore S₁ = S₂ for b = a which is greater than S₂ for a>b.

Second explanation: Let a and b be a positive number and I will define the sum S₁ to be the sum of a finite number of n terms therefore
S₁ = na = M = M + (b-b) = M + 0
If I allow M = na to equal the second term in S₁ then I have na = 0 or a = 0, however I defined n to be positive and therefore a must equal itself. and therefore I can conclude that
na>0 or S₁> n*0 = 0 where RHS is equal to the sum of n zeros.


Am I misinterpreting what they are saying in the paint document?

It never said anything like what you claim. It says that if the *sum is given*, the *product* is maximized when the two numbers are equal, and if the *product is given* the *sum* is minimized when the two numbers are equal.

 

[Miike012]

Give me an example

 

[Miike012]

Wouldn't it be easier to explain with an example. You could say the sum is equal to 6
S = 6 = 1+5 = 2+4 = 3+3
1*5 = 5
2*4 = 8
3*3 = 9

 

[Tsunoyukami]

Let's look at a concrete example.

Suppose a + b = 6. Here I will only consider a few possibilities. We could have (a, b) = (1, 5) or (2, 4) or (3, 3) or (3, 3), or (4, 2) or (5, 1). Then we could write a = 6 - b. The product of these two numbers can be written P = ab = (6-b)b = 6b - b^2 or 0 = b^2 - 6b + P (a quadratic in b). To show that there is a max (or min) when b = a is easy if we can differentiate with respect to b and set the derivative to 0 and solve for b (we should find b = 3 for this case). Indeed, then P = 9 which is, indeed the max (you could check this for a few values for yourself in order to "convince" yourself of its validity without rigour. That being said, you should try to prove it with more rigour than I've provided in this one example of its validity.

A similar but "reverse" idea can be applied to show that if P is given S is least when a = b.

EDIT: As you said, it's easier to 'explain' by using an example. However, the example has only shown to be true for that specific value (ie. a+b = 6). We would prefer to show it is true in general for any positive a and b. Another thing is to note that you only considered whole numbers - what about every other combination of numbers - 1.1 and 4.9, does it hold true for them? (Yes.)

But what if I asked you to show it for every combination of positive numbers that sums to 6? You would have to explicitly show me an infinite number of calculations before I could truly be 100% convinced of its validity.

 

[Miike012]

Deleted

 

[Ray Vickson]

Let's look at a concrete example.

Suppose a + b = 6. Here I will only consider a few possibilities. We could have (a, b) = (1, 5) or (2, 4) or (3, 3) or (3, 3), or (4, 2) or (5, 1). Then we could write a = 6 - b. The product of these two numbers can be written P = ab = (6-b)b = 6b - b^2 or 0 = b^2 - 6b + P (a quadratic in b). To show that there is a max (or min) when b = a is easy if we can differentiate with respect to b and set the derivative to 0 and solve for b (we should find b = 3 for this case). Indeed, then P = 9 which is, indeed the max (you could check this for a few values for yourself in order to "convince" yourself of its validity without rigour. That being said, you should try to prove it with more rigour than I've provided in this one example of its validity.

A similar but "reverse" idea can be applied to show that if P is given S is least when a = b.

EDIT: As you said, it's easier to 'explain' by using an example. However, the example has only shown to be true for that specific value (ie. a+b = 6). We would prefer to show it is true in general for any positive a and b. Another thing is to note that you only considered whole numbers - what about every other combination of numbers - 1.1 and 4.9, does it hold true for them? (Yes.)

But what if I asked you to show it for every combination of positive numbers that sums to 6? You would have to explicitly show me an infinite number of calculations before I could truly be 100% convinced of its validity.

No need; the paint document he attached proved it in general.

 

[Tsunoyukami]

Oops, I guess I completely misread that. However, if you were to try to prove this yourself it would not suffice to provide one example in which it is true.