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3.2 Integers and Rationals.

3.34   Definition (Integers in $F$.). Let $F$ be a field. We define an element $z$ in $F$ to be an integer in $F$ if and only if $z$ can be written as the difference of two natural numbers; i.e., if and only if

\begin{displaymath}z=q-p \mbox{ for some } p,q\in\mbox{{\bf N}}_F.\end{displaymath}

We denote the set of integers in $F$ by $\mbox{{\bf Z}}_F$.

3.35   Exercise. What are the integers in $\mbox{{\bf Z}}_5$?

3.36   Exercise. Let $F$ be a field. Show that for all $x,y\in F$,

\begin{displaymath}x\in\mbox{{\bf Z}}_F \mbox{ and } y\in\mbox{{\bf Z}}_F\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}x+y\in\mbox{{\bf Z}}_F\end{displaymath}

and that

\begin{displaymath}x\in\mbox{{\bf Z}}_F \mbox{ and } y\in\mbox{{\bf Z}}_F\mbox{$...
...e{1ex}\Longrightarrow\hspace{1ex}$}x\cdot y\in\mbox{{\bf Z}}_F.\end{displaymath}

Also show that $x\in \mbox{{\bf Z}}_F\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}-x\in\mbox{{\bf Z}}_F$.

3.37   Theorem. Let $F$ be an ordered field and let $-\mbox{{\bf N}}_F=\{-x\colon x\in\mbox{{\bf N}}_F\}$. Then

\begin{displaymath}\mbox{{\bf Z}}_F=\mbox{{\bf N}}_F\cup(-\mbox{{\bf N}}_F) \mbox{ and } \mbox{{\bf N}}_F\cap(-\mbox{{\bf N}}_F)=\{0\}.\end{displaymath}


\begin{displaymath}n\in\mbox{{\bf N}}_F\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}n=n-0\in\mbox{{\bf Z}}_F\end{displaymath}


\begin{displaymath}n\in-\mbox{{\bf N}}_F\mbox{$\hspace{1ex}\Longrightarrow\hspac...
...$\hspace{1ex}\Longrightarrow\hspace{1ex}$}n\in\mbox{{\bf Z}}_F.\end{displaymath}

Hence, $\mbox{{\bf N}}_F\subset\mbox{{\bf Z}}_F$ and $-\mbox{{\bf N}}_F\subset\mbox{{\bf Z}}_F$, so
\mbox{{\bf N}}_F\cup(-\mbox{{\bf N}}_F)\subset\mbox{{\bf Z}}_F.
\end{displaymath} (3.38)

Now suppose $n\in\mbox{{\bf Z}}_F$. Then $n=p-q$ where $p,q\in\mbox{{\bf N}}_F$. If $p-q\geq 0$, then $p-q\in\mbox{{\bf N}}_F$. If $p-q\leq 0$, then $q-p\geq 0$, so $q-p\in\mbox{{\bf N}}_F$, so $-(p-q)\in\mbox{{\bf N}}_F$, so $-n\in\mbox{{\bf N}}_F$, so $n\in-\mbox{{\bf N}}_F$. Therefore, $n\in\mbox{{\bf N}}_F$ or $n\in-\mbox{{\bf N}}_F$; i.e., $n\in\mbox{{\bf N}}_F\cup-\mbox{{\bf N}}_F$, so

\begin{displaymath}\mbox{{\bf Z}}_F\subset\mbox{{\bf N}}_F\cup(-\mbox{{\bf N}}_F).\end{displaymath}

This combined with (3.38) shows that $\mbox{{\bf Z}}_F=\mbox{{\bf N}}_F\cup(-\mbox{{\bf N}}_F)$. Since all elements of $\mbox{{\bf N}}_F$ are $\geq 0$, and all elements of $-\mbox{{\bf N}}_F$ are $\leq 0$, it follows that $\mbox{{\bf N}}_F\cap(-\mbox{{\bf N}}_F)\subset\{0\}$, and clearly $0\in\mbox{{\bf N}}_F\cap -\mbox{{\bf N}}_F$, so $\mbox{{\bf N}}_F\cap(-\mbox{{\bf N}}_F)=\{0\}$. $\mid\!\mid\!\mid$

3.39   Definition (Rational numbers in $F$.). Let $F$ be a field. Let
\begin{displaymath}\mbox{{\bf Q}}_F=\left\{{n\over m}\colon n,m\in\mbox{{\bf Z}}_F \mbox{ and } m\neq 0\right\}.\end{displaymath}

The elements of $\mbox{{\bf Q}}_F$ will be called rational numbers in $F$. We note $\displaystyle {0={0\over 1}\in\mbox{{\bf Q}}_F}$ and $\displaystyle {1={1\over 1}\in\mbox{{\bf Q}}_F}$.

3.40   Theorem. Let $F$ be a field. Then the set $\mbox{{\bf Q}}_F$ of rational numbers in $F$ form a field (with the operations of $F$).

Proof: The various commutative, associative and distributive laws hold in $\mbox{{\bf Q}}_F$, because they hold in $F$, and we've noted that the additive and multiplicative identities of $F$ are in $\mbox{{\bf Q}}_F$, and they act as identities in $\mbox{{\bf Q}}_F$ because they are identities in $F$. We note that $+$ and $\cdot$ define binary operations on $\mbox{{\bf Q}}_F$; i.e., the sum and product of elements in $\mbox{{\bf Q}}_F$ is in $\mbox{{\bf Q}}_F$. Let $a,b\in\mbox{{\bf Q}}_F$ write $\displaystyle {a={p\over q},b={r\over s}}$ where $p,q,r,s\in\mbox{{\bf Z}}_F$ and $q\neq 0,\; s\neq 0$. Then

a+b &=& {p\over q}+{r\over s}={{ps+qr}\over {qs}} \\
a\cdot b &=& {p\over q}\cdot {r\over s}={{pr}\over {qs}}

and $ps+qr, qs, pr$ are all in $\mbox{{\bf Z}}_F$ and $q\cdot s\neq 0$. Hence $a+b$ and $a\cdot b$ are in $\mbox{{\bf Q}}_F$. Also, $\displaystyle {-a=-\left({p\over q}\right)={{-p}\over q}}$ where $-p,q\in \mbox{{\bf Z}}_F$, so $-a\in\mbox{{\bf Q}}_F$ and

b\neq 0&\mbox{$\Longrightarrow$}& b={r\over s} \mbox{ where } ...
...ver r} \\
&\mbox{$\Longrightarrow$}& b^{-1}\in\mbox{{\bf Q}}_F.

Hence $\mbox{{\bf Q}}_F$ is a field. $\mid\!\mid\!\mid$

3.41   Definition (Even and odd.). . . . In exercise 3.31 we defined even and odd natural numbers. We now extend this definition to integers. Let $F$ be a field and let $x\in\mbox{{\bf Z}}_F$. We say $x$ is even if and only if $x=2y$ for some $y\in\mbox{{\bf Z}}_F$, and we say $x$ is odd if and only if $x=2z+1$ for some $z\in\mbox{{\bf Z}}_F$.

3.42   Remark. In exercise 3.32A you showed that in an ordered field, every element of $\mbox{{\bf N}}_F$ is even or odd, and no element of $\mbox{{\bf N}}_F$ is both even and odd. Since $\mbox{{\bf Z}}_F=\mbox{{\bf N}}_F\cup-\mbox{{\bf N}}_F$, it follows fairly easily that if $F$ is an ordered field, then every element of $\mbox{{\bf Z}}_F$ is even or odd, and no element of $\mbox{{\bf Z}}_F$ is both even and odd.

3.43   Exercise. A
Let $F$ be a field, and let $n\in\mbox{{\bf Z}}_F$. Show that

\begin{displaymath}n \mbox{ is even }\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}n^2 \mbox{ is even,} \end{displaymath}


\begin{displaymath}n\mbox{ is odd }\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}n^2\mbox{ is odd. }\end{displaymath}

Let $F$ be an ordered field and let $n\in\mbox{{\bf Z}}_F$. Show that

n^2\mbox{ is even } &\mbox{$\Longrightarrow$}& n\mbox{ is even...
n^2\mbox{ is odd } &\mbox{$\Longrightarrow$}& n\mbox{ is odd. }

I want to show that in any ordered field $F$, $2$ is not a square in $\mbox{{\bf Q}}_F$. To show this I will use the following lemma.

3.44   Lemma. Let $F$ be an ordered field. Then every element in $\mbox{{\bf Q}}_F$ can be written as $\displaystyle {{m\over n}}$, where $m,n\in\mbox{{\bf Z}}_F$ and $m,n$ are not both even.

Proof: Let $F$ be an ordered field, and let $r\in\mbox{{\bf Q}}_F$. Then $\displaystyle {r={m\over n}}$ where $m,n\in\mbox{{\bf Z}}_F$ and $n\neq 0$. Since $\displaystyle {r={{-m}\over {-n}}}$, we may assume without loss of generality that $n>0$. Then $n\in\mbox{{\bf N}}_F$ so we can write any element of $\mbox{{\bf Q}}_F$ in the form $\displaystyle {r={m\over n}}$ where $m\in\mbox{{\bf Z}}_F$, $n\in\mbox{{\bf N}}_F$ and $n\geq 1$. Let

\begin{displaymath}S = \left\{ q \in \mbox{{\bf N}}_F: \mbox{ for some }p \in \mbox{{\bf Z}}_F \left( r = {p\over q}\right) \right\}.\end{displaymath}

Then $n \in S$, since $r = {m\over n}$. By the least element principle, $S$ has a least element $k$. We have

\begin{displaymath}r = {p\over k} \mbox{ for some }p \in \mbox{{\bf Z}}_F.\end{displaymath}

Then $p$ and $k$ are not both even, since if $p = 2P$ and $k = 2K$ where $P$ and $K$ are in $\mbox{{\bf Z}}_F$, then

\begin{displaymath}r = {p \over k} = {2P\over 2K} = {P\over K},\end{displaymath}

and hence $K \in S$. But this is impossible because $K = {1\over 2}k < k$, i.e. $K$ is less than the least element for $S$. $\mid\!\mid\!\mid$

3.45   Theorem. Let $F$ be an ordered field. Then $2$ is not a square in $\mbox{{\bf Q}}_F$.

Proof: Suppose there were an element $r\in\mbox{{\bf Q}}_F$ such that $r^2=2$. By our lemma, we can write $\displaystyle {r={m\over n}}$ where $m,n\in\mbox{{\bf Z}}_F$, $m,n$ not both even. Now

r^2=2&\mbox{$\Longrightarrow$}&{{m^2}\over {n^2}}=2 \\
...grightarrow$}&m \mbox{ is even (since } n^2\in\mbox{{\bf Z}}_F).


\begin{displaymath}m \mbox{ is even } \mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}m=2k \mbox{ for some } k\in\mbox{{\bf Z}}_F,\end{displaymath}


r^2=2&\mbox{$\Longrightarrow$}& m^2=2\cdot n^2 \\
&\mbox{$\Lo...{ is even } \\
&\mbox{$\Longrightarrow$}& n\mbox{ is even. }

Thus the statement $r^2=2$ implies $(m$ is even and $n$ is even and $m,n$ are not both even), which is false. The theorem follows. $\mid\!\mid\!\mid$

3.46   Note.

When Plato (427?-347B.C.) wrote The Laws, he lamented that most Greeks at the time believed that all numbers were rational (i.e. that all lines are commensurable):

ATHENIAN: My dear Cleinias, even I took a very long time to discover mankind's plight in this business; but when I did, I was amazed, and could scarcely believe that human beings could suffer from such swinish stupidity. I blushed not only for myself, but for Greeks in general.

CLEINIAS: Why so? Go on, sir, tell us what you're getting at.


ATHENIAN: The real relationship between commensurables and incommensurables. We must be very poor specimens if on inspection we can't tell them apart. These are the problems we ought to keep on putting up to each other, in a competitive spirit, when we've sufficient time to do them justice; and it's a much more civilized pastime for old men then draughts.

CLEINIAS: Perhaps so. Come to think of it, draughts is not radically different from such studies.

ATHENIAN: Well, Cleinias, I maintain that these subjects are what the younger generation should go in for. They do no harm, and are not very difficult: they can be learnt in play, and so far from harming the state, they'll do it some good[39, book vii,820].

However, when Aristotle (384-322 BC) wrote the Priora Analytica, he assumed that his reader was familiar with the proof of theorem 3.45 just given. The following quotation would not be understood by anyone who did not know that proof.

For all who effect an argument per impossible infer syllogistically what is false, and prove the initial conclusion hypothetically when something impossible results from the assumption of its contradictory; e.g., that the diagonal of the square is incommensurate with the side, because odd numbers are equal to evens if it is supposed to be commensurate. One infers syllogistically that odd numbers come out equal to evens, and one proves hypothetically the incommensurability of the diagonal since a falsehood results through contradicting this.[4, 1-23, 41a, 23-31]

The meaning of the word ``rational'' has changed since the time of Euclid. He would have said that a line of length $\sqrt{2}$ was rational, but a rectangle of area $\sqrt{2}$ was irrational. The following quotation is from book X of The Elements[19, vol 3, p10, definitions 3 and 4].

Let then the assigned straight line be called rational, and those straight lines which are commensurable with it, whether in length and in square or in square only, rational, but those which are incommensurable with it irrational.

4. And let the square on the assigned straight line be called rational, and those areas which are commensurable with it rational, but those which are incommensurable with it irrational.

3.47   Warning. An early commentator on Euclid (quoted in [19, vol III page1]) suggested that perhaps

$\cdots$ everything irrational and formless is properly concealed, and, if any soul should rashly invade this region of life and lay it open, it would be carried away into the sea of becoming and be overwhelmed by its unresting currents.

3.48   Notation ( $\mbox{{\bf N}},\mbox{{\bf Z}},\mbox{{\bf Q}}$.) We have defined natural numbers $\mbox{{\bf N}}_F$ in any field $F$, and we've seen that the natural numbers in $\mbox{{\bf Z}}_5$ and the natural numbers in $\mbox{{\bf Q}}$ are quite different. However, if $F$ is an ordered field, then

\begin{displaymath}\mbox{{\bf N}}_F=\{0,1,2,3,4,\cdots\}\end{displaymath}

where the list contains no repetitions, since when we add a new term to the list we get something greater than every element already in the list. Hence if $F,G$ are two ordered fields then $\mbox{{\bf N}}_F$ and $\mbox{{\bf N}}_G$ are `` essentially the same". We will denote the natural numbers in an ordered field by $\mbox{{\bf N}}$, and call $\mbox{{\bf N}}$ `` the natural numbers". Since we defined $\mbox{{\bf Z}}_F$ in terms of $\mbox{{\bf N}}_F$, and we defined $\mbox{{\bf Q}}_F$ in terms of $\mbox{{\bf Z}}_F$, the integers in any two ordered fields are `` essentially the same" and the rationals in any two ordered fields are `` essentially the same". We will denote the integers in any ordered field by $\mbox{{\bf Z}}$, and call $\mbox{{\bf Z}}$ `` the integers".

\begin{displaymath}\mbox{{\bf Z}}=\mbox{{\bf N}}\cup -\mbox{{\bf N}}=\{0,1,-1,2,-2,3,-3,\cdots\}.\end{displaymath}

Similarly we will call the rational numbers in an ordered field $\mbox{{\bf Q}}$, and call $\mbox{{\bf Q}}$ `` the rational numbers"

\begin{displaymath}\mbox{{\bf Q}}=\left\{ {n\over m}\colon n,m\in\mbox{{\bf Z}},\; ,m\neq 0\right\}.\end{displaymath}

3.49   Remark. One can define formally what it means to say $\mbox{{\bf N}}_F$ and $\mbox{{\bf N}}_G$ are `` essentially the same," and one can prove that if $F,G$ are ordered fields, then $\mbox{{\bf N}}_F$ and $\mbox{{\bf N}}_G$ are `` essentially the same" (e.g., see [35, page 35]).

However, one can also construct ordered fields $F$ and $G$ such that $\mbox{{\bf N}}_F$ and $\mbox{{\bf N}}_G$ are radically different! (see [41]) The reason that both of these apparently contradictory things can happen is that our definition of $\mbox{{\bf N}}_F$ involves looking at the set of all inductive subsets of $F$, and our vague notions of set and function are just too imprecise to deal with this delicate question. The two quoted contradictory results are proved using different set theories, which are not consistent with each other, but both of which are more or less consistent with everything we've used about sets.

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Next: 3.3 Recursive Definitions. Up: 3. Induction and Integers Previous: 3.1 Natural Numbers and   Index