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[parent] rational numbers are real numbers (Result)

Let us first show that the natural numbers $ 0,1,2,\ldots$ are contained in the real numbers as constructed above. Heuristically, this should be clear. We start with 0. By adding $ 1$ repeatedly we obtain the natural numbers

$\displaystyle 0, \quad 0+1, \quad (0+1)+1, \quad ((0+1)+1)+1, \ldots, $
To make this precise, let $ \mathbbmss{N}$ be the natural numbers. (We assume that these exist. For example, all the usual constructions of $ \mathbbmss{R}$ rely on the existence of the natural numbers.) Then we can define a map $ f\colon \mathbbmss{N}\to \mathbbmss{R}$ as
  1. $ f(0)=0$, or more precisely, $ f(0_\mathbbmss{N})=0_\mathbbmss{R}$,
  2. $ f(a+1)=f(a)+1$ for $ a\in \mathbbmss{N}$.
By induction on $ a$ one can prove that
$\displaystyle f(a + b)$ $\displaystyle =$ $\displaystyle f(a) + f(b),$  
$\displaystyle f(a b)$ $\displaystyle =$ $\displaystyle f(a) f(b), \quad a,b\in \mathbbmss{N}$  

and
$\displaystyle f(a)$ $\displaystyle \ge$ $\displaystyle 0, \ a\in \mathbbmss{N}\ $   with equality only when$\displaystyle \ a=0.$  

The last claim follows since $ f(a)>0$ for $ a=1,2,\ldots$ (by induction), and $ f(0)=0$. It follows that $ f$ is an injection: If $ a\le b$, then $ f(a)=f(b)$ implies that $ f(a)=f(a)+f(b-a)$, so $ a=b$.

To conclude, let us show that $ f(\mathbbmss{N})\subset\mathbbmss{R}$ satisfies the Peano axioms with zero element $ f(0)$ and sucessor operator

$\displaystyle S\colon f(\mathbbmss{N})$ $\displaystyle \to$ $\displaystyle f(\mathbbmss{N})$  
$\displaystyle k$ $\displaystyle \mapsto$ $\displaystyle f(f^{-1}(k)+1)$  

First, as $ f$ is a bijection, $ x=y$ if and only if $ S(x)=S(y)$ is clear. Second, if $ S(k)=0$ for some $ k=f(a)\in f(\mathbbmss{N})$, then $ a+1=0$; a contradiction. Lastly, the axiom of induction follows since $ \mathbbmss{N}$ satisfies this axiom. We have shown that $ f(\mathbbmss{N})$ are a subset of the real numbers that behave as the natural numbers.

From the natural numbers, the integers and rationals can be defined as

$\displaystyle \mathbbmss{Z}$ $\displaystyle =$ $\displaystyle \mathbbmss{N}\cup \{-z\in \mathbbmss{R}: z\in \mathbbmss{N}\},$  
$\displaystyle \mathbbmss{Q}$ $\displaystyle =$ $\displaystyle \left\{ \frac{a}{b} : a\in \mathbbmss{Z}, b\in \mathbbmss{N}\setminus\{0\} \right\}.$  

Mathematically, $ \mathbbmss{Z}$ and $ \mathbbmss{Q}$ are subrings of $ \mathbbmss{R}$ that are ring isomorphic to the integers and rationals, respectively.

Other constructions

The above construction follows [1]. However, there are also other constructions. For example, in [2], natural numbers in $ \mathbbmss{R}$ are defined as follows. First, a set $ L\subseteq \mathbbmss{R}$ is inductive if
  1. $ 0\in L$,
  2. if $ a\in L$, then $ a+1\in L$.
Then the natural numbers are defined as real numbers that are contained in all inductive sets. A third approach is to explicitly exhibit the natural numbers when constructing the real numbers. For example, in [3], it is shown that the rational numbers form a subfield of $ \mathbbmss{R}$ using explicit Dedekind cuts.

Bibliography

1
H.L. Royden, Real analysis, Prentice Hall, 1988.
2
M. Spivak, Calculus, Publish or Perish.
3
W. Rudin, Principles of mathematical analysis, McGraw-Hill, 1976.



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See Also: proof that Q is the prime subfield of any field of characteristic 0


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Cross-references: Dedekind cuts, subfield, rational numbers, isomorphic, ring, subrings, rationals, integers, subset, axiom, axiom of induction, contradiction, bijection, operator, zero element, Peano axioms, implies, injection, induction, map, clear, real numbers, contained, natural numbers
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This is version 3 of rational numbers are real numbers, born on 2006-03-13, modified 2006-04-18.
Object id is 7719, canonical name is RationalNumbersAreRealNumbers.
Accessed 1694 times total.

Classification:
AMS MSC12D99 (Field theory and polynomials :: Real and complex fields :: Miscellaneous)
 26-00 (Real functions :: General reference works )
 54C30 (General topology :: Maps and general types of spaces defined by maps :: Real-valued functions)

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