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In the book Principles of Mathematical Analysis by Rudin, I read that "a < b" is defined this way: if b - a is positive, then a < b or b > a. Then some questions arose to me: we know that "minus" is a reverse operation of "plus" since a - b = a + (-b), but how is the elementary arithmetic "a + b" is defined, and how about "a * b"? Will there be any differences between the definitions with regard to the real number system, rational number system, and natural number system?

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    $\begingroup$ The Peano axioms can be used to define the natural numbers, as well as $+$ and $*$ for the natural numbers. $\endgroup$
    – Regret
    Nov 3 '14 at 11:48
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    $\begingroup$ See my answer here. $\endgroup$
    – Asaf Karagila
    Nov 3 '14 at 11:57
  • $\begingroup$ If $b-a \gt 0$, then $b \gt a$. You know $b \not \gt a$ $\endgroup$ Jan 5 '15 at 4:55
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"God gave us the integers.
All else is the work of man."
Leopold Kronecker

A sketch of the ideas:

Addition and multiplication for $a, b \in \mathbb{R} \backslash \mathbb{Q}$ can be defined by using sequences $(a_n)_{n \in \mathbb{N}}$ and $(b_n)_{n \in \mathbb{N}}$ from $\mathbb{Q}$. $$ \begin{array}[llllll]\\ a+b&=\lim\limits_{n \to \infty} (a_n+b_n), & a=\lim\limits_{n \to \infty} a_n, & a_n \in \mathbb{Q} & n \in \mathbb{N}\\ a \cdot b&=\lim\limits_{n \to \infty} (a_n \cdot b_n), & b=\lim\limits_{n \to \infty} b_n, & b_n \in \mathbb{Q} \\ \tag{1} \end{array} $$

This works because the limes of the sum is independent of the sequences selected. So addition in $\mathbb{R}$ can be reduced to addition in $\mathbb{Q}$.

But addition and multiplication in $\mathbb{Q} \backslash \mathbb{Z}$ can be reduced to addition in $\mathbb{Z}$ $$ \begin{array}[llllll]\\ a+b&=\frac{a_1b_2+b_1a_2}{a_2b_2}, & a=\frac{a_1}{a_2}, & a_n \in \mathbb{Z} & n \in \{1,2\} \\ a \cdot b&=\frac{a_1 \cdot b_1}{a_2 \cdot b_2}, & b=\frac{b_1}{b_2}, & b_n \in \mathbb{Z} \end{array} \\ \tag{2} $$

And addition and multiplication in $\mathbb{Z}$ can be reduced to multiplication and addition in $\mathbb{N}$. $$ \begin{array}[llllll]\\ a+b&=(a_1+b_1)-(a_2+b_2), & a=a_1-a_2, & a_n \in \mathbb{N} & n \in \{1,2\} \\ a \cdot b&=(a_1 b_1 + a_2 b_2) - (a_1 b_2 + a_2 b_1), & b=b_1-b_2, & b_n \in \mathbb{N} \end{array} \\ \tag{3} $$

All these definitions are independent of the selected $a_n$ and $b_n$ as far as they meet the required specifications in regard of $a$ and $b$.

Multiplication in $\mathbb{N}$ can be reduced to addition in $\mathbb{N}$ $$ \begin{array}[llllll]\\ 1 \cdot b&=b, & a, b \in \mathbb{N} & \\ a \cdot b&=(a-1)b +b & \\ \end{array} \\ \tag{4} $$

Addition in $\mathbb{N}$ can be reduced to counting ($\text{successor}(n)$ of a number $n \in \mathbb{N}$ is the number following $n$):

$$ \begin{array}[llllll]\\0+b&=b & a, b \in \mathbb{N} & \\ a+b&=\text{successor}((a-1)+b) \end{array} \\ \tag{5} $$

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  • $\begingroup$ Nice answer, really reduced it to the essence of the question $\endgroup$ Dec 19 '14 at 20:35

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