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Formally the rational numbers are defined as $\mathbb Z \times \mathbb Z / \{0\}$, where $(m_1, n_1)$ and $(m_2, n_2)$ being equivalent if $m_1n_2 = m_2 n_1$.

This set equipped with $+$ and $\times$ as defined in the Wiki (http://en.wikipedia.org/wiki/Rational_number) form a field, that is the field of quotients of $\mathbb Z$.

How can one say for $n \in \mathbb Z$ that $n = (n,1)$ ? This equality is true only up to isomorphism ?

Also, how can one say $(m_1, n_1) = \text {some decimal expansion} : c_n c_{n-1} \ldots c_0. c_{-1}c_{n-2}\ldots$ ?

I mean $(m_1, n_1)$ is an element of a set but in normal use we set it equal to a real number ?

Also the result of $m_1 / n_1$ is understood to be a real number ? Not just the integer part ?

How just one decide that $1/3 = 0.333333\ldots$ and not some other real number ?

mez
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Shuzheng
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    So far you're only constructing the rationals, not the reals. Later, when you construct the reals, you will identify some of them with the rationals you already know. But that comes only once you have real numbers to identify with them. – hmakholm left over Monica Jul 20 '14 at 07:21
  • You can indentify an integer $z$ with $(z,1)$. Where is the problem? – Peter Franek Jul 20 '14 at 07:49
  • One does not really say that $n$ and $(n,1)$ are strictly equal. However, there is a natural injective morphism from $\mathbb{Z}$ to $\mathbb{Q}$ as you just constructed it, where $n$ is sent to $(n,1)$. In this way we identify $\mathbb{Z}$ with this particular subset of $\mathbb{Q}$. – Joachim Jul 20 '14 at 08:16
  • Thank you very much. So it is first after we have constructed the reals that we identity $(1,3) \in \mathbb Q$ with $0.333333 \ldots$ ? (that is the real number corresponding to division of $1$ by $3$ viewed as reals). And we say that $\mathbb Z$ is contained within $\mathbb Q$ by means of a injective homomorphism ? And given elements $a \in \mathbb Z$ and $b \in \mathbb Q$ we have that their decimal expansion is equal viwed as reals and therefore we say they represent the same real value ? (Is this correct @HenningMakholm ?). I dont think Wikipedia explain this well. – Shuzheng Jul 22 '14 at 07:17
  • @NicolasLykkeIversen: Note that decimal expansions are not the defining characteristic of the reals (unless you're taking a rathet unusual path to them). It is true that every element of $\mathbb Z$ maps directly to the same real as the do if you map them via $\mathbb Q$, but being "the same real" is not a matter of its decimal expansions. – hmakholm left over Monica Jul 22 '14 at 07:55

1 Answers1

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Let $f: \Bbb{Z} \to \Bbb{Q}: \ z \to z/1$. Show that this is an injective ring homomorphism. Note that $z/1$ refers to an equivalence class of objects, for instance when you're working in arithmetic with: $$ \dfrac{3}{5} = \dfrac{21}{35} = \dfrac{3\cdot 7}{5\cdot 7} $$ In particular, notice the equals. So you've already worked with such a structure.

If $\frac{a}{b}$ is a non-integer rational and your base expansions (eg. $0.333333\dots$) are in base $c$ (eg. base $c = 10$, or "decimal"). Then $x = \frac{a}{b} = \lfloor \dfrac{a}{b} \rfloor + \dfrac{a \pmod b}{b} = I_1 + \dfrac{r_1}{b} = I_1 + \dfrac{1}{c}(\lfloor \dfrac{c \cdot r_1}{b} \rfloor+ \dfrac{c\cdot r_1 \pmod b}{b}) = \dots $. Thus $\frac{a}{b}$ produces an infinite base $c$ expansion if the sequence $$r_1 = a \pmod b, \\ r_2 = c\cdot r_1 \pmod b, \\ r_3 = c \cdot r_2 \pmod b, \\ \dots $$ reaches zero at some finite point in time. However if $c^k r_1 \equiv c^k a = 0 \pmod b$ and $c$ is a unit $\pmod b$, in other words $\gcd(c,b) = 1$, then the above can be multiplied by $c^{-k}$ and so $a = 0\pmod b$. But we already said $\frac{a}{b}$ was non-integer. Therefore whenever the bottom of the fraction $b$ is relatively prime to the base of the expansion representation $c$, you'll get an infinite base-$c$ expansion. Prove that for rationals this expansion always starts repeating a sequence after enough places passed the decimal. This should be true since there are a finite number of remainders $\pmod b$. Another important thing that's sort of related is that this sort of base expansion representation is not uniquely determined, for instance $1 = .9999999\dots$.

Daniel Donnelly
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  • Thank you very much. So it is first after we have constructed the reals that we identity $(1,3) \in \mathbb Q$ with $0.333333 \ldots$ ? (that is the real number corresponding to division of $1$ by $3$ viewed as reals). Is the method for expressing $\frac a b$ the standard way of doing it ? Why are $1 = 0.99999 \ldots$ ? Is this because viewed as a series $0.99999 \ldots$ converge to $1$ ? – Shuzheng Jul 22 '14 at 07:14
  • $1 - 0.\bar{9} = 0.00000\dots000\dots $ (you never get to $\dots0001.$). – Daniel Donnelly Jul 22 '14 at 12:04
  • @NicolasLykkeIversen Made mistake, please see bolded. – Daniel Donnelly Jul 22 '14 at 12:08