I stumbled on your blog because I am trying to prove mathematically that 100% solvency is impossible as long as we ‘all’ treat human needs as commodities.

I am trying to prove it mathematically because for some reason its the only way to prove something (apparently looking out the window and seeing reality is not enough).

I was wondering if you would be willing to take up the challenge?

When I say solvent: I mean one has the means to pay all expenses and debts.

When I say 100% solvency: I mean that every entity in society is able to be solvent all at the same time.

Intuitively I know its not possible, and history has demonstrated it’s not possible, and the continual need to treat more and more resources as commodities proves it’s not possible, however, I can’t seem to put it into a mathematical formula.

Do you think it’s possible to put this into a formula?

Regards

Dean

P.S. I have found that the majority of economists treat wealth as income. I do not understand why they do this as one’s mans income is another man’s expense. If everyone paid all their debts all at the same time there would be no money (Bank of England 2014 – Introduction to Money) and therefore no wealth. Therefore wealth is something projected into the future (i.e. debt) and not something in the present (i.e. income)

]]>Dear Paul Couvrey,

I am an elementary teacher. I majored in English. I am certified to teach regular ed K-5, the arts K-5, and English to speakers of other languages K-12. So…no math degree. I want to understand the issue with the following because so many mathematicians and high school math teachers are saying one of the 3 items I’ve posted below. Please help!

I did not have Facebook when this problem was posted back in 2011.

6/2(1+2)

Note: sometimes displayed with an obelus. Other problem on Facebook: 48/2(9+3)

Some people argued that…

1. The answer is 9: (6/2)*(1+2)=?, (6/2)*(3)=?, (3)*(3)=9. These people followed PEMDAS based on MD having left to right precedence depending on which comes first. I had never heard of PEMDAS until this problem and another like it in July 2017. I graduated high school in 1999. These people said that once the 1+2 was solved inside the parenthesis, that the 2 before the parenthesis is not to be multiplied by the 3. That the student must now start from the beginning (6/2)*(3)=?

2a. The answer is 1: 6/[2(1+2)]=?, *Some simplified inside the parenthesis first and multiplied by the 2 before the parenthesis. In doing so, these people followed PEMDAS literally and also saw 2(1+2) as a unit. 6/[2(3)]=?, 6/[2*3]=?, 6/6=1, I at first, solved it that way until I started to see all of the arguments. People calling others idiots, etc. I was amazed by the anger, yet it was funny to me no matter how wrong, correct, or confused I was by my own participation in trying to solve this problem.

2b. The answer is 1: 6/[2(1+2)]=?, *Some used the distributive property and in doing so…followed PEMDAS literally giving multiplication precedence over division. So, in doing so, they saw 2(1+2) as a unit.

6/[2(3)]=?, 6/[(2*1)+(2*2)]=?, 6/[2+4]=?, 6/6=1

3. That the problem is ambiguous and that mathematicians do not write problems in this way. That they would have added parenthesis to the original problem to avoid the two different interpretations. That if a teacher wanted students to have an answer of 9, then they should add parentheses to show (6/2)*(1+2). That if a teacher wanted students to have an answer of 1, then they should add parenthesis and brackets to show 6/[2(1+2)].

On a YouTube video (can’t recall which), I followed the comments section argument between 2 engineers. They argued over the course of a year. They were insulting each other, but I was still able to see how each dealt with the 2 before the parenthesis.

*The engineer who saw 9 as the answer saw 2 by itself and said that once the inside of the parentheses were added, then the person solving the problem must start at the beginning of the problem and work left to right based on PEMDAS.

*The engineer who saw 1 as the answer saw 2(1+2) as a unit and said that the distributive property must be used. [(2*1)+(2*2)], [2+4], 6. I will say that when I solved the problem, that I first solved for 1+2 then multiplied by the 2 before the parenthesis. [2(3)], 6.

Also, I read through the comments section of the article and found a comment and a reply.

September 20, 2016 at 11:44 pm

“You seem to have missed that the distributive property means that implied multiplication must be done before regular multiplication and division.”–Allen Helmer (I am not sure if Allen was replying to you or someone else in the comments)

“That is not true. Distribution is a necessity of Algebra because order of operations cannot be performed in order if there are variables that are unknown numbers. Distribution never has to be performed in order of operations if there are only numbers.”–Paul Couvrey (This is the reply about the distributive property and whether or not one must use the distributive property if the problem is all numbers)

Because the entire problem is made of numbers, does that mean that the distributive property was not necessary in the way the engineer who arrived at 1 did? Since the problem uses all numbers? Also, I did not use the distributive property, but I still multiplied the 3 by the 2 before the parenthesis.

Anyway, I hope you understand my writing. Please let me know any response to what I have written.

Thank you,

Shandra Hart

7.20.17

]]>Thank you,

Shandra Hart

]]>Example: 2y-3

This means that if a number replaces y then 2y is “2 multiplied to y.”

If 4 replaces y, then the example turns to 2×4-3.

Here by order of operations, 2×4 would be first, which equals to 8, then the next operation would be 8-3. Result: 5

As a side note, the 2 is a coefficient of the linear term 2y and -3 is the coefficient of the constant term (in Algebra, terms are separated by addition or subtraction). “But -3 has no variable, how can it be a coefficient?” you may ask. -3 technically has a variable of y to the zero power. The variable (or any number) to the zero power is equal to 1 and -3 multiplied to 1 is still -3. And if you are wondering why something to the zero power is equal to 1, that is a completely different topic. Search it up on the internet!

]]>1. Operations within Grouping (parenthesis, brackets, fraction bar, square root… etc.)

2. Exponents

3. Division

4. Multiplication

5. Subtraction

6. Addition ]]>