If we define real numbers, as is sometimes done, with field axioms, and order axioms, and completeness (or continuity) axiom, then, rational numbers fulfill field axioms and order axioms, but they do not fulfill completeness axiom.
So, completeness axiom helps us, in a sense, to distinguish between rational and real numbers.
Moreover, axioms for the real numbers characterize real numbers up to isomorphism, so, in a sense, there is only one structure that fulfills axioms for the real numbers, and that structure is a system of real numbers.
Because both real numbers and rational numbers fulfill field and order axioms, the field and order axioms do not characterize only one structure (up to isomorphism).
It is written somewhere that, for example, we can prove existence of $n$-th (for every $n$) roots of nonnegative real numbers in the set of real numbers, by taking as our starting point the set of axioms for the real numbers.
Also, there is a set-theoretic result that there is no bijection between the set of real numbers and the set of rational numbers, and the one which states that there is a bijection from naturals onto rationals, so, rationals are countable and reals are uncountable.
But, as axioms for the real number system characterize reals up to isomorphism, I started to think that we could be able to prove that there is no bijection between rationals and reals by using only these two premises:
1) rationals and reals satisfy field and order axioms
2) reals satisfy the completeness axiom but rationals do not
and by using the concept of bijection.
What I mean is, that representation of reals and rationals as infinite sums in some base would not be allowed in such a proof.
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