How to define relation of "bigger" and "less" and not run into a circularity?

The axioms for the real numbers provide one way to settle this issue. The axioms are broken into two portions: the field axioms concerning algebraic properties of the binary operations $+$ and $\times$; and the order axioms which require the existence of a total order $x<y$ satisfying three axioms:

1. if $x < y$ then $x+z < y+z$ for all $z$;
2. if $0 < x$ and $0 < y$ then $0 < x \times y$.
3. The completeness axiom.

You can then define the positive numbers to be all numbers $x$ such that $0<x$.

On the other hand you don't have to be satisfied with the axiomatic approach to the real numbers themselves, because the real numbers can be constructed from the rational numbers, which can be constructed from the integers, which can be constructed from the natural numbers, which satisfy Peano's Axioms. This is a long drawn out story, but the salient point for your question is that in Peano's Axioms for the natural numbers there is no need to even mention the concept of positive or negative numbers, but on the other hand the order relation on natural numbers can be defined like this (addition and multiplication having already been defined):

• Given natural numbers $m,n$ define $m < n$ if there exists a nonzero natural number $k$ such that $m+k=n$.

Once that task is accomplished, one embeds the natural numbers into the integers and defines order similarly:

• Given integers $m,n$, define $m<n$ if there exists a nonzero natural number $k$ such that $m+k=n$.

Then one embeds the integers into the rationals and defines order on the rationals:

• Given rationals $r=\frac{a}{b}$, $s=\frac{c}{d}$ where $a,c$ are integers and $b,d$ are nonzero natural numbers, define $r < s$ if $ad<bc$.

Finally, one embeds the rationals into the reals and defines order on the reals:

• Given reals $x \ne y$, define $x < y$ if there exist Cauchy sequences of rational numbers $(r_n)$, $(s_n)$ converging to $x,y$ respectively, such that $r_n < s_n$ for all $n$.

It all depends on how you define the real numbers. The important thing is that you can define "positive" without any reference to $<$.

In the Dedekind cuts approach, where a real is a pair $(A,B)$ of subsets of $\mathbb{Q}$ such that $A\sqcup B = \mathbb{Q}$ and $\forall a\in A, b \in B, a<b$, then a real $(A,B)$ is said to be positive if and only if there exists $a\in A, a>0$ (note that this $>$ is the ordering on $\mathbb{Q}$, so it's not circular).

In the Cauchy sequences approach, where a real is an equivalence class of Cauchy sequences in $\mathbb{Q}$ (the equivalence relation is induced by the ideal of sequences converging to $0$), then a real $[(q_n)]$ is said to be positive if there is $\epsilon >0$ (again, this is in $\mathbb{Q}$ so no circularity here) and $n$ such that $\forall k \geq n, q_k \geq \epsilon$ (again, in $\mathbb{Q}$). One easily checks that this doesn't depend on the choice of representative of the class $[(q_n)]$, and hence yields a sound definition.

There are of course other approaches to defining the real numbers, and they all come with different ways of defining positivity.

What matters is that the ordering always comes from the ordering of $\mathbb{Q}$, which has been defined before, so it's never circular. Now one might ask how the ordering on $\mathbb{Q}$ is defined. And once again, this depends on the definition of $\mathbb{Q}$.

For instance, defining $\mathbb{Q}$ as $\mathbb{Z}\times \mathbb{N}^*$ modulo the equivalence relation $(a,b) \sim (c,d) \iff ad= bc$, then you can say that the class of $(a,b)$ is positive if and only if $a>0$ (this is in $\mathbb{Z}$, so no circularity here). One again checks that this doesn't depend on the representative, and so this is indeed well defined. Once you have the definition of positivity you can define an ordering as you mentioned in the beginning of your post.

It remains to define an ordering on $\mathbb{Z}$, and similarly this can be done without circularity by using the ordering on $\mathbb{N}$ (of course it also depends on the definition of $\mathbb{Z}$ one is using).

The last step is defining an ordering on $\mathbb{N}$. Obviously this also depends on the definition of $\mathbb{N}$ , but it's most usually defined as the first infinite ordinal (existing because of the axiom of infinity) and so it comes with a "natural" ordering : $\in$. Indeed, in this definition, $0 := \emptyset, 1:= \{0\}$ and more generally, $n+1 := n\cup \{n\}$ and so $n=\{0,1,...,n-1\}$, so $<$ coincides with $\in$. Now $\in$ (membership) is a primitive notion in set theory and so we don't define it.

$a\ge b$ if there is a number $c$ with $a=b+c^2$