Trying to make an argument that irrational numbers are countable

The explaination is huge so stick to the end., Also sorry for occasional bad English, because I am non native English speaker

So the way I want to start is by defining only 2 numbers in the entire number system at first, 1 and -1( Calling this tiny set of numbers A) , and basic operations being addition, multiplication and power( addition alone won't be enough)

Now let's define natural numbers as the sum of all positive A numbers with repetition allowed. So first number 1. Next number 1+1, then next 1+1+1........ And so on. That would now continue to infinity giving 1+1+1+1+1........ And that's where we get all our natural numbers

x=n+n+n+n+n..........{n€A,n≠-1}

(I don't know how to add suffix and prefix on keyboard so everytime I write n, you have to assume that each n represents a different number here, same goes for any other variable I use)

Next would be the set of integers. Instead of only 1, let's define it as sum of all positive and negative( only -1 and 1) that would be 1-1(0),1,-1,1+1(2),-1+1(-2), and we know that they are countable too

x=n+n+n+n+n.........{n€A}

Note:since 1+1-1+1-1+1-1..... is 1/2, we are including one more statement to the definition that |I| should be a whole number

Now as we have defined natural and integers, we can safely use N and I to define next numbers.

Now comes rational numbers. We can define them as the repeated summation and multiplication of all Integers to the power of 1 or -1. Meaning we can multiply or add any number to get a rational number.

Eg 10^-1*9^2+2= a rational number

And again, we know that all rationals are countables.

x=n^h(+*)n^h(+*)n^h(+*)n..........{n€I,h€A}

Note:(+*) means any of + or * could be placed there, irrespective of what was there in the previous place

Note: considering that the sums could go to infinity, the transcendental numbers like π and logs would also fall under rational number, hence we are adding one more statement that all summations and multiplications couldn't have infinite numbers.

Now comming to the final man of the match, the irrationals. (This is gonna be a little tricky but stick with me for a sec) we can define irrationals as the repeated summation or multiplication of all rational numbers to the power of rational numbers. Most of you know this definition is incomplete for now, but let's first write what this interprets

x=n^n(+*)n^n(+*)n^n(+*)n.....{n€R}

To rectify the error, we need to define one more term, I would call it, let's say complexity value. Now complexity value of a number would be the number of n(or let's call it terms) in addition or multiplication, used to define a number. So if the n is in power, it wouldn't count. So for √2, complexity value is 1, since there is only 1 term. With that lets set some rules for this number.

Rule 1: c value should be the least value possible for a number, so, you cannot write √2 as 2¼*2¼ and say it's c value is 2, it will still be 1

Rule 2: for a power of a term, the complexity of the term should always be less than that of the term, this rule also applies for the power of power of it's term and so on.....

(For c=1, power will be 1)

The second rule, is what changes everything.

So now, all the numbers with c value 1 are rationals, since there is no number in their power. But then what about √2? It has a number in power. So here is the answer, √2 can also be written as 0+2½, where 0 is a rational number, so √2 now has one digit ahead of it, ie:0, and then we are writing √2, so now the c value of √2 is actually 2.

Another eg, 2^½^(√3+3√4) now what's the complexity value of this?, Let's start with the uppermost power(√3+3√4). Since first term is in power, we add a 0 there first,(0+√3+3√4) giving us c value 3. Now considering the second rule, that c value of power cannot be greater than the base, we need to add some zeros there making it 0+0+0+½, giving it c value of 4. Now finally the base, we add 4 zeros, and now the entire term has a c value of 5, meaning you atleast need 5 terms to define this number.

Well, why add such a rule then? Cause this rule now arranges the irrationals in an orderly way. So now, we have a new definition for irrational, it is the repetitive addition and multiplication of rationals and irrationals with irrational powers. Sounds confusing, but I don't know how to frame it better, and you know what I mean by now. So

X=n^n(+*)n^n(+*)n^n(+*).........{n belongs to Q, with complexity value of each term more than its previous term}

Giving that irrational numbers are countable

Again, with one extra rule that no roots of negative numbers allowed, otherwise it would become a complex set.

This is my argument, I may be wrong but let's hear it from yall