The metaphor of the lemon seller(or why social inequality causes poverty)

This week, I was just scrolling down through my facebook page when I saw a post claiming that social inequality causes poverty. In my "free-market" mind, I instantly disagreed with that statement and started to see the comments flooding that post and I´m not gonna lie to you, I was agreeing with every single complaint exposed. Well, when I was just about to leave Facebook and do sth decent in my life I saw my "anarcho-capitalist" friend replying to a comment in agreement with that post:
                          "Prove that social inequality causes poverty, take your time"
And then, all of a sudden, I started to think about proofs that could support that statement, that´s when I came up with a metaphor (It was kinda dumb at the beggining, but when I showed to my friend we had this discussion and it got improved).
So initially, picture an isolated country, that has no commercial partners, no foreign income (investments) and all the commercial trades occur only inside this country. Also consider that NO ONE is gonna throw their money into the sea.

So you might as well agree with me that the amount of money flowing is constant. Let´s assign a value to that: X. If you are familiar with thermodynamics, just imagine that the remote island is an isolated system and the money is the whole energy, and the internal trades are just one molecule giving energy to another (if A has energy of E  andB has E1, if there is a trade,  A becomes E-delE and B becomes B+delE, therefore the total sum is CONSTANT).
Ok, geekness apart, just picture now that we are separating the population of this island into 2 groups (the richest and the poorest), it´s just a mental separation (not an apartheid), they coexist in the same space. Let´s assign the poorest to the red dots and the richest to the orange dots. So now we have two subsystems (the richest and the poorest) included in one system (the island), so should we assume that they behave the same as the system? the answer is NOOOO. Note that now there is an income and an outcome related to that groups. So now the reader can predict clearly my point here. Let´s start from the cause of this whole situation here. What is the meaning of social inequality? A group has more money than other group and this situation is constant, it´s not like the most privilegied(or privileged?) group is gonna be the most privilegied(or privileged?) for only a couple of weeks or even a couple or years. It means that they are gonna be in that situation for a LOONNNGG time, and what does that mean in macroeconomics 101? THAT THEY ARE MAKING PROFIT. Therefore, the inequality rises constantly.
Ok, now we are going to analyse what is going on with each group. The total amount of money changes due to the incomes or outcomes, but the total sum of the whole system is constant, X. Now, as written on the last  paragraph,  Let´s assume that the red guys(poorest) give dA to the richest. If they are giving, the brown guys (richest) are earning sth. But which value?... dA!!! Remember: there are no external incomes or outcomes related to the whole system.
So in the instant t we have:
                                          RED MONEY: R
                                          BROWN MONEY: B
                                          TOTAL: R+B=X
in the instant t +dt:
                                          RED MONEY: R-dA
                                          BROWN MONEY: B+dA
                                          TOTAL: R-dA+B+dA=X
Now, compare the values, in the instant t+dt, the richest became richer and the poorest, poorer due to the inequality (which is our cause). THEREFORE,as consequence of the cause, inequality generates poverty.
Yeah, I know, you might be asking to yourself now: "Ok, but what if this "isolated system that you are talking about opens to the real world,in other words, what if exists foreign commercial partners,  foreign income (investments) and foreign commercial trades? ".
To answer this question I'm gonna have to draw another map (on paint btw):

    Imagine now a land that has partners and they trade goods. The population is the same as the previous one. Now, let's say that (for the sake of the poors) the island has an economic superavit (income>outcome) so you gotta agree with me that X is not a constant anymore and it's increasing. It means that the money flow is higher than the previous situation. More money circulating causes inflation. You may ask as my friend did: "Why?". And NOW when the metaphor of the lemon seller shows up. Travel with me through the power of imagination to a microscale. And by that I mean: imagine a lemon seller and yourself being a system. First of all, imagine this system a isolated system. You got $ 5 and the lemon seller is wondering how much he is gonna charge you for his tasty lemons. Well, if he's clever enough, he's gonna charge you a value such that you can afford buy at least 2 lemons(Yeah, he's kind). For instance, if he charges you $ 3 for a lemon, he's gonna earn only  $ 3, and he's not gonna be able to earn more money. But, if he $ 2,50 for a lemon, he's gonna earn all of your money. Ok, now imagine a situation that exists income. Imagine an amount of $ 80000 going directly to your hand. You are going to buy the whole s*** if he mantain the price. But he's clever, don't forget that, and he's gonna increase the lemon value, which is inflation. Law of supply and demand. Inflation. Now, return to our macroeconomical open land, if there's a huge amount of income, prices are gonna go up. "Does it mean that the rich guy is gonna get less richer?", you may ask, and I answer, "YES". And then you reply "wtf".  Come on, by now you know that prices going up affect mainly the poorest, because the price is equal to both, therefore, if you have less money, the percentage of this rise is higher.
In sum, if you overlap the first situation (remote island) and the second one (not so remote land). You get that inequality is in fact one of the many causes of poverty.

A brief math problem

MATHEMATICS

Q1. Suppose f is a function from positive integers to positive integers satisfying (I) f(1)=1,(II) f(2n)=f(n), and (III) f(2n+1)=f(2n)+1, for all positives integers n. Find the maximum of f(n) when n is grater than or equal to 1 and less than or equal to 1994. 

Ans.: 

First of all, let’s organize the things a little bit. Instead of 

f(2n+1)=f(2n)+1, we’re gonna change it into (IV) f(2n+1)=f(n)+1 (II on I)  Furthermore, let’s see if we find out a pattern on the f(n) value by testing 10 possibilities (1≤𝑛 ≤5).

Note: You’re gonna notice that we can find f(n) for every n in the domain, once we already have f(1), and all the values to n even and n odd.

•      f(1)=1

•      For n=1: f(2)=f(1)=1 (comes from II)               

                   f(3)=f(2)+1=2 (comes from III)   

•      For n=2: f(4)=f(2)=1

                    f(5)=f(2)+1=2

•      For n=3: f(6)=f(3)=2                

                   f(7)=3

•      For n=4: f(8)=f(4)=1

                       f(9)=2

•      For n=5: f(10)=f(5)=2                

                            f(11)=3

   What we can extract from the topics above is that the maximum in a series of numbers always shows up first when you pick an odd number in the domain. For instance, in crescent order, the first maximum is f(3)=2, the second maximum is f(7)=3. Note that f(11) IS NOT THE THIRD MAXIMUM once 3 has already appeared. It happens because if your number in the domain is an even number, f(e), where e is an even number, equals to f(e/2), so you’ll never have a brand new maximum in this situation. Whereas, if you pick an odd number, f(o), where o is an odd number, equals to f(e/2) +1, so it could be a brand new maximum. 

   

So, let’s see if we find out an algorithm that show us when there is a brand new maximum. Note that we can find other b.n.m (brand new maximum) when we have (V) f(2(l.b.n.d.)+1)=f(l.b.n.d.)+1 (from IV), where l.b.n.d. is the last brand new domain. Let’s analyze: 

First b.n.m: f(3)=2

Second b.n.m: f(7)=3, “l.b.n.d = 3”

Third b.n.m: f(2(7) + 1) = f(15) = f(7) + 1 = 4, “l.b.n.d = 7”

Fourth b.n.m: f(2(15) + 1) = f(31) = 5, “l.b.n.d=15”

So what we can see is that, 𝑓(2^𝑎−1)=𝑎, where a is a b.n.m.

If n goes up to 1994, if we want to see what’s the maximum f(n), we must find 2^𝑎−1 value, which must be the closest to 2(1994) +1. So a, in this case, equals to 11.

So f(2047)=11, is the maximum.