Rassah: Do you believe that the equation x^2 = 2 have at least one solution?
Yes? Should be two solutions.
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?
Yes, but then why can't I match that number in that infinity set with any other number, like 1, in another infinity set?
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.
