oh like PPC.

... has not been trustworthy in a long long time.

No its not safe/It depends on your definition of "safe".

yeah i understand this a hell lot better than you, but it was you who asked the question: "why can't I match that number in that infinity set with any other number, like 1, in another infinity set?" and im trying to explain.

You don't have to(and you can't infinity does not exist in the natural numbers)...
The algebraic numbers(all roots of all rational polynomials) have the same cardinality as the natural numbers(N).

This is easily(lol no!) seen by that N and N*N(all pairs of natural numbers, (0,0), (1,2), (237319313,5), ...) have the same cardinality(do you need proof?), as there exists bijective functions between them. Consider that for a moment: their is as many natural numbers as pairs of natural numbers.

we can of course repete the above to say that N ~ N*N ~ N*(N*N) ~ ...

lets now consider the following:
Given a countable family(set of sets) of countable sets A_n for n in N. The the result(B) of the union of all these sets(U A_n), is then again countable.
the proof of that is quite simple:
as every A_n is countable there exists by definition a surjective function from N onto A_n, we call that function g_n : N -> A_n.
Now because we know that N*N is countable we have a function f: N -> N*N.
We can now construct a surjection(h: N*N -> B) onto the union of the A_n's: h(a,b) = g_a(b). which is surjective. and the proof is done.

to see that the algebraic number are countable, i will just spew out some facts that you should think about:
Z(the set of all positive and negative integers) is countable.
Q(the set of all quotients between integers) also countable.
Pol_n(Q) (the set of all polynomials, that have a degree of n, with quotients in Q) is countable.
and finally a polynomial that have degree n have at most n roots.

These facts together with the above theorem about unions of countable sets, is sufficient to proof that the algebraic numbers are countable.

But then we have to consider numbers like e and pi, which are not algebraic, and are not roots in any polynomial. The set of non-algebraic/transcendent numbers, are the ones that are not countable, by the diagonal argument.

and you really really want R to not have "holes" (R is complete) of "nonexistent" numbers, for the intermediate value theorem to work...

If you denies the existence of R(and only accept the existence of algebraic numbers), you accept that there exist continuous functions that gives positive and negative values, but the equation f(x) = 0 is false for all x.

The positive root of that polynomial(the square root of 2) is not a rational number(Q)(do you require a proof? please say so).
So at least there exist a number that cannot be written as a quotient of two integers(Z), right?
That means, as a consequence, that there must exist another set of numbers(hint: R) which contains the numbers that are not rational.

Do you see where this is leading?

don't try to assign meaning to what dank says, you can't. dank is the one and only enlightened god, and we are too. its all love. just love.

Rassah: Do you believe that the equation x^2 = 2 have at least one solution?

i don't think so. but around 250 - 300 is likely, if we have a crash again.
After a new all time high, bitcoin's history tells us that we will likely go back down to the last all time high, and trade around that point for a few months.

No, you cannot count all the real numbers(R), you cannot create a one-to-one mapping between N and R, as there are not enough natural numbers, by cantors diagonal argument. If you give me a any mapping between N and R, that you say is surjective, I can always constructively find a number in R that you miss with your surjection, and your surjection is not surjective.

Cardinality(roughly equvivalent of size) of set are defined by the existence of injections and bijections(and surjections, but that require AC) between them. As we can construct a bijection(f(x) = 2x) between the natural numbers(N) and the even natural numbers(2,4,6,8,...), we say that the even natural numbers have the same cardinality as all of the natural numbers.
On the other hand any injection from the natural numbers can't be surjective, and can't therefor be a bijection, we then say that R have a greater cardinality then N. But we can show that the set of all subsets of N have the same cardinality as R.


(You should really go read some elementary set theory, your posts about them seems that you are unable to comprehend what the fuck is going on. This is in no way criticism, set theory is at some times really counter intuitive, and most have, like you, difficulty of understanding it.)

and yet be easily verifiable for a computer.

There have no mentioned of cryptocurrencies in the OP. But the short answer would be no, or simple barter.

oh! its simple, its called selling stuff you made, for some other stuff someone other made that you considered of equal value.

It have exists is the last tens of thousand years, and probably before that too.

Do your believe that for every set A there exists a set, called the power set of A, P(A), that contains every subset of the set A? (The axiom of power sets)?

don't you see? dank is enlightened.

it depends on what ever you talk about natural numbers(1,2,3,4,...) or ordinal numbers(1,2,3,4,...,infinity,2 infinity,...infinity^2,...infinity^infinity), or real numbers(1,2,3,pi, 5,7878787,...) or extended real numbers(-infinity,...-1,0,1,...+infinity).
The problem with ordinal numbers is that plus does no longer commute.

WTF!!  Do you even know what you are talking about, do you even know what compiling means?

fuck you!

true, and all the network code is written in bash with nc.

no only APL and perl.