Definition or Axiom for less than for real numbers

Question

I was curious as to what the definition for $<$ would be concerning real numbers. I have looked at Peanos axioms which discusses inequalities for every $a, b\in\mathbb{N}$, $a<b$ iff there is some $c\in \mathbb{N}$ such that $a+c=b$. I am curious as to what it would be for real numbers. I could try to guess what it might be, but I am wanting to be exact. Any help or resources would be appreciated.

Answer 1

One can define it as follows: $x\leq y \iff (\exists z)(z^2 = y-x)$

Since someone voted to delete my answer, but did not leave a comment as to why this answer is incorrect, I will explain my answer more thoroughly.

My answer consider's the reals as a field in the ring language $\mathcal{L} =\{1,0,\times,+, -\}$. The reals as a field are naturally a structure in this language. My answer shows that $\leq$ is definable (in the model theory sense) in this structure in this particular language.

Answer 2

It depends on which definition of the real numbers you're using.

If you define real numbers as Dedekind cuts, then each real number $x$ contains a set $A_x$ of rational numbers (which are intended to represent the rational numbers which are less than $x$). With this definition, the definition of $<$ is that $x < y$ whenever $A_x$ is a proper subset of $A_y$.

If you define real numbers as equivalence classes of Cauchy sequences, then each real number $x$ is considered to be a class of Cauchy sequences $\{x_n\}$. With this definition, the definition of $<$ is that $x < y$ whenever, for all sufficiently large $n$, $x_n < y_n$, where the $<$ operator here is the one defined on the rational numbers. (But in order for this definition to be valid, it's necessary to prove that it gives the same answer regardless of which Cauchy sequence is chosen from each equivalence class.)

You can define the real numbers by specifying a list of axioms that they must obey and then proving that there is exactly one structure which satisfies all of the axioms. If you do this, then the meaning of $<$ is not explicitly defined; it's just specified by the axioms.

Finally, there's a definition of $<$ which is independent of the underlying definition. You can define $\le$ by saying that $x \le y$ whenever there is a number $z$ such that $z^2 = y - x$. Then you can say that $x < y$ whenever $x \le y$ and $x \ne y$; or, alternatively, you can say that $x < y$ whenever it is not the case that $y \le x$.