The basic axioms defined on $\R$ (we assume here the existence of the elements in $\R$, since the rigorous derivation and proof of their existence from the intuitive assumption of the existence of $\N$ is very exhausting and treated elsewhere):

1.1) Commutativity: $$\forall x, y \in \R, \qquad \left\{ \begin{array}{l} x + y = y +x \\ x \cdot y = y \cdot x \end{array} \right. $$ 1.2) Associativity: $$\forall x, y, z \in \R, \qquad \left\{ \begin{array}{l} (x + y) + z = x + (y + z) \\ (x \cdot y) \cdot z = x \cdot (y \cdot z) \end{array} \right. $$ 1.3) Existence of an identity element (for each operation): $$ \exists 1, 0 \in \R, \quad \forall x \in R \qquad \left\{ \begin{array}{l} x + 0 = x \\ x \cdot 1 = x \end{array} \right. $$ 1.4) Existence of the negative: $$ \forall x \in \R, \quad \exists y \in \R, \qquad x + y = 0, \quad [\; y \; \text{denoted as} \; (-x) \;] $$ 1.5) Existence of the reciprocal: $$ \forall x \in \R \; (x \ne 0), \quad \exists y \in \R, \qquad x \cdot y = 1, \quad [\; y \; \text{denoted as} \; (x^{-1}) \;] $$ 1.6) Distributivity: $$ \forall x, y, z \in \R \qquad x \cdot (y + z) = x \cdot y + x \cdot z $$

Using the above defined axioms we can derive the basic rules of algebra.

2.1) Cancellation rules:

$\underline{\text{2.1.1}} \\ \forall x, y, z \in \R, \quad if \quad x+y = x+z, \quad then \quad y = z $

$\underline{\text{2.1.2}} \\ \forall x, y, z \in \R \;(x \ne 0), \quad if \quad x \cdot y = x \cdot z, \quad then \quad y = z $

$$ \underline{\text{Proof}} $$ $\underline{\text{2.1.1}} \\ x + y = x + z \qquad / + a \quad (\exists a \in \R, \quad x + a = 0) \\ $

$(x + y) + a = (x + z) + a \quad \rightarrow \quad x+(y+a)=x+(z+a) \quad \rightarrow \quad \\ x+(a+y)=x+(a+z) \quad \rightarrow \quad (x+a)+y = (x+a)+z \quad \rightarrow \quad \\ 0+y=0+z \quad \rightarrow \quad y+0=z+0\quad \quad \rightarrow \quad y=z \\ [\; \text{Using 1.2, 1.4.} \;] \quad \Box \\ $

$\underline{\text{2.1.2}} \\ x \cdot y = x \cdot z \qquad / \cdot a \quad (\exists a \in \R, \quad x \cdot a = 1, \; because \quad x \ne 0) \\ $ $(x \cdot y) \cdot a = (x \cdot z) \cdot a \quad \rightarrow \quad x \cdot (y \cdot a)=x \cdot(z \cdot a) \quad \rightarrow \quad \\ x\cdot (a\cdot y)=x\cdot (a\cdot z) \quad \rightarrow \quad (x\cdot a)\cdot y = (x\cdot a)\cdot z \quad \rightarrow \quad \\ 1\cdot y=1\cdot z \quad \rightarrow \quad y\cdot 1=z\cdot 1\quad \quad \rightarrow \quad y=z \\ [\; \text{Using 1.1, 1.2, 1.4.} \;] \quad \Box \\ $

2.2) Uniqueness of negative and reciprocal elements:

$\underline{\text{2.2.1}} \\ \forall x \in \R, \quad \text{there is only one} \quad a \in \R, \\ \text{such that} \quad x + a = 0$

$ \underline{\text{2.2.2}} \\ \forall x \in \R \; (x \ne 0), \quad \text{there is only one} \quad a \in \R, \\ \text{such that} \quad x \cdot a = 1$

$$ \underline{\text{Proof (by contradiction)}} $$

$ \underline{\text{2.2.1}} \\ \text{Assume} \; \exists a,b \in \R,\quad a \ne b, \; \text{such that} \; x + a = 0, \; x + b = 0. \\ \text{Then, by the transitivity of equality,} \; x + a = x + b, \text{using 2.1.1 we get } \\ a=b, \; \text{leading to a contradiction!} \\ \text{Thus, the negative must be a unique element for each} \; x \in \R. \quad \Box$

$ \underline{\text{2.2.2}} \\ \text{Assume} \; \exists a,b \in \R,\quad a \ne b, \; \text{such that} \; x \cdot a = 1, \; x \cdot b = 1. \\ \text{Then, by the transitivity of equality,} \; x \cdot a = x \cdot b, \text{using 2.1.2 we get } \\ a=b, \; \text{leading to a contradiction!} \\ \text{Thus, the reciprocal must be a unique element for each} \; x \in \R. \quad \Box \\$

2.3) The sign manipulation theorems:

$ \qquad 2.3.1 \qquad \forall x \in \R, \quad -(-x) = x \\ $ $ \qquad 2.3.2 \qquad \forall x \in \R, \quad (-1) \cdot x = -x \\$ $ \qquad 2.3.3 \qquad \forall x,y \in \R, \quad (-1) \cdot (-1) = 1 \\$ $ \qquad 2.3.4 \qquad \forall x,y \in \R, \quad -(x + y) = (-x) + (-y) \\$ $ \qquad 2.3.5 \qquad \forall x,y \in \R, \quad (-x) \cdot (-y) = x \cdot y $
$$ \underline{\text{Proof}} $$
$\underline{\text{2.3.1}} \\ \text{Using the transitivity property of equality and axiom 1.4,} \\ \exists a \in \R, \; x + a = 0. \\ \text{Also,} \; \exists b \in \R, \; (-x) + b = 0. \; \text{Using transitivity,} \; x + a = (-x) + b. \\ \text{According to 1.4,} \; a = (-x), \; b = -(-x). \\ \text{Thus,} \\ \\ x + a = (-x) + b \quad \rightarrow \quad x + (-x) = (-x) +(-(-x)) \quad \rightarrow \quad \\ (-x) + x = (-x) + (-(-x)) \quad \rightarrow \quad x = -(-x) \\ [\; \text{using} \; 2.1.1 \; \text{and} \; 1.1 \;] \quad \Box$

$\underline{\text{2.3.2}} \\ \text{In order to prove this algebra rule, we need to first prove:}$ $$ \forall x \in \R, \; 0 \cdot x = 0. $$$ \text{Using 1.6,} \quad x \cdot (0+0) = x \cdot 0 + x \cdot 0. \\ \text{Also, by 1.4,} \; \exists a \in \R, \; \text{such that} \; x \cdot 0 + a = 0. \\ x \cdot (0+0) = x \cdot 0 + x \cdot 0 \qquad / + a \quad \rightarrow \quad \\ x \cdot 0 + a = (x \cdot 0 + x \cdot 0) + a \quad \rightarrow \quad \\ x \cdot 0 + a = x \cdot 0 + (x \cdot 0 + a) \quad \rightarrow \quad 0 = x \cdot 0 + 0 \quad \rightarrow \quad \\ 0 = x \cdot 0 \\ [\; \text{Using 1.4 and 1.2.} \;] \quad \Box \\ \text{Now, we move on to the actual property proof:} \\ x \cdot (1 + (-1)) = x \cdot 1 + x \cdot (-1) = x + (-1) \cdot x \\ x \cdot (1 + (-1)) = x \cdot 0 = 0 \cdot x = 0 \qquad [\; \text{Using 1.1, 1.3, 1.4, 1.6} \;] \\ \text{By the transitivity of equality,} \; x + (-1) \cdot x = 0. \\ \text{Using 1.4,} \; \exists a \in \R, \; x + a = 0 \; [ \; a \; \text{denoted} \; (-x) \; ]. \\ x + a = x + (-1)x \quad \rightarrow \quad a = (-x) = (-1)x \\ [\; \text{Using} \; 2.1.1 \;] \quad \Box$

$ \underline{\text{2.3.3}} \\ \text{Using 2.3.2,} \\ (-1) \cdot x = -x, \; \text{substitute } \; x = -1 \quad \rightarrow \\ (-1) \cdot (-1) = -(-1) = 1 \\ [\; \text{Using}\; 2.3.1 \; \text{in last step} \;] \quad \Box$

$ \underline{\text{2.3.4}} \\ \text{Using 2.3.2 and 1.6 ,} \\ -(x + y) = (-1)\cdot (x + y) = (-1)\cdot x +(-1)\cdot y = (-x) + (-y) \quad \Box$

$ \underline{\text{2.3.5}} \\ \text{Using 2.3.2, 2.3.3, 1.1, 1.2, 1.3, } \\ (-x) \cdot (-y) = ((-1)\cdot x) \cdot ((-1) \cdot y) = (x\cdot (-1)) \cdot ((-1) \cdot y) \\ = x\cdot ((-1) \cdot (-1)) \cdot y = x \cdot 1 \cdot y = x \cdot y \quad \Box \\$

2.4) The reciprocal theorems:

$ 2.4.1 \qquad \forall x \in \R \; (x \ne 0), \quad (x^{-1})^{-1} = x \\ $ $ 2.4.2 \qquad \forall x,y \in \R \; (x \ne 0, \; y \ne 0), \quad (xy)^{-1} = x^{-1}\cdot y^{-1}\\$
$$ \underline{\text{Proof}} $$
$\underline{\text{2.4.1}} \\ \text{Using 1.5,} \\ \exists a \in \R, \; x \cdot a = 1 \qquad [\; a \; \text{denoted} \; as \; x^{-1} \;] \\ \text{Also,} \; \exists b \in \R, \; x^{-1} \cdot b = 1 \qquad [\; b \; \text{denoted} \; as \; (x^{-1})^{-1} \;] \\ \text{Using transitivity of equality, } \\ x\cdot a = x^{-1} \cdot b \quad \rightarrow \quad x\cdot x^{-1} = x^{-1} \cdot b \quad \rightarrow \quad \\ x^{-1}\cdot x = x^{-1} \cdot b \quad \rightarrow \quad x = b = (x^{-1})^{-1} \qquad [\; \text{Using 2.1.2} \;] \quad \Box$

$\underline{ \text{2.4.2}} \\ \text{Using 1.5,} \\ \exists a \in \R, \; x \cdot a = 1 \qquad [\; a \; \text{denoted} \; as \; x^{-1} \;] \\ \exists b \in \R, \; y \cdot b = 1 \qquad [\; b \; \text{denoted} \; as \; y^{-1} \;] \\ \text{Also,} \; \exists c \in \R, \; xy \cdot c = 1 \qquad [\; c \; \text{denoted} \; as \; (xy)^{-1} \;] \\ \text{Multiplication of the two first equations gives: } \; xy \cdot ab = 1 \\ [\; \text{Using 1.1, 1.2 , 1.3} \;] \\ \text{Then, by transitivity of equality, } \\ xy \cdot ab = xy \cdot c \quad \rightarrow \quad ab = c \\ \rightarrow \quad x^{-1}\cdot y^{-1} = (xy)^{-1} \qquad [\; \text{Using 2.1.2} \;] \quad \Box$