Proof

Proof of Rule 1) For any $$x\neq 0$$ it is $$x \cdot x >0$$. This is a consequence from the rules of multiplying positive and negative real numbers. Proof of Rule 2) $$1 > 0$$. From the uniqueness of 1 and the uniqueness of 0, it follows that $$1 \neq 0$$. Therefore, according to the definition of order relation, it must be either $$1 > 0$$ or $$-1 > 0$$. Since $$1\cdot 1 =1$$ , it must be $$1 > 0$$, according to the rule Rule 1).

Proof of Rule 3)  From $$x > y$$ it follows that $$x+a > y+a$$ for any $$a\in\mathbb R$$. (The rule is analogously valid and proven for any inequalities with "$$<$$" in between). $$x > y$$ means, by definition, that $$x - y > 0$$. Therefore, it is $$x - y + a - a > 0$$, and so it follows that $$x + a - (y + a) > 0$$, and finally that $$x + a > y + a$$.

Proof of Rule 4)  If $$x > y$$ and $$a > b$$, then it follows that $$x+a > y+b$$. (The rule is analogously valid and proven for any inequalities with "$$<$$" in between). By definition, it is $$x - y > 0$$ and $$a - b > 0$$. From the definition of order relation, it follows that $$(x - y) + (a -b) > 0$$, which by definition results in $$x + a > y + b$$.

Proof of Rule 5) Transitivity of the "$$<$$" (analogously the "$$>$$" relation) If $$x < y$$ and $$y < z$$, then by definition it is $$0 < y - x$$ and $$0 < z - y$$. From the definition of order relation, it follows that $$0 < y - x + z - y= z - x$$, which by definition results in $$x < z$$.

Proof of Rule 6) The inequality $$x > y$$ does not change, if it is multiplied by any number $$a > 0$$, i.e. it is $$ax > ay$$. (The rule is analogously valid and proven for an inequalities with "$$<$$" in between). By definition, it is $$x -y > 0$$. From the definition of order relation, it follows that $$a(x-y) = ax - ay > 0$$, which, by definition, results in $$ax > ay$$.

Proof of Rule 7) The inequality $$x > y$$ changes, if it is multiplied by any number $$a < 0$$, i.e. it is $$ax < ay$$. (The rule is analogously valid and proven for an inequality with "$$<$$" in between, which then changes into "$$>$$"). By definition, it is $$x - y > 0$$ and $$-a > 0$$. From the definition of order relation, it follows that $$(-a)(x-y) = -ax + ay > 0$$, which, by definition, results in $$ay > ax$$, and finally in $$ax < ay$$.

Proof of Rule 8)  If $$x > 0$$, then $$x^{-1} > 0$$. It is $$x^{-1}=x^{-1}\cdot 1=x^{-1}\cdot(x^{-1}\cdot x) =(x^{-1}\cdot x^{-1})\cdot x$$, which is greater then $$0$$, following the precondition, the Rule 1) and the definition of order relation.

Proof of Rule 9)  If $$x < 0$$, then $$x^{-1} < 0$$. It is $$x^{-1}=x^{-1}\cdot 1=x^{-1}\cdot(x^{-1}\cdot x) =(x^{-1}\cdot x^{-1})\cdot x$$, which is smaller then $$0$$, following the precondition, the Rule 1) and the Rule 7).

Proof of Rule 10) If $$y > x > 0$$, then $$x^{-1} > y^{-1}$$. (The rule is analogously valid and proven for the inequality $$0 < x < y$$, resulting in $$y^{-1} < x^{-1}$$). According to the precondition $$x$$ and $$y$$ are both positive. Therefore, it follows from the definition of order relation that $$xy > 0$$, and according to Rule 8) $$(xy)^{-1} > 0$$. Multiplying the inequality $$y > x$$ by the number $$(xy)^{-1}$$ does not change the inequality by Rule 6), therefore $$y(xy)^{-1} > x (xy)^{-1}$$. This is equivalent to $$x^{-1} > y^{-1}$$.

Proof of Rule 11) If $$0 \le x < y$$ and $$0 \le a < b$$  then $$ax < by$$. (The rule is analogously valid and proven for $$by > ax$$, following from $$y > x \ge 0$$ and $$b > a \ge 0$$). The argument is valid, if $$x=0$$ or $$a=0$$, since then $$0 < by$$ follows from the definition of order relation. So assume $$0 < x < y$$ and $$0 < a < b$$. In this case we have $$ax < ay$$ and $$ay < by$$, following in both cases from the Rule 6. The argument follows now from the transitivity (Rule 5)).

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