Theorem B.46.
Let \(x,y,z,w\in\rats\text{.}\) If \(x=y\) and \(z=w\text{,}\) then \(x+z=y+w\) and \(xz=yw\text{.}\)
Section B.16 Obtaining the Rationals from the Integers
We consider the set \(\rats\) of all ordered pairs in \(\ints\times \ints\) of the form \((x,y)\) with \(y\neq 0\text{.}\) Elements of \(\rats\) are called rational numbers, or fractions. Define an equivalence relation, denoted \(=\text{,}\) on \(\rats\) by setting \((x,y)=(z,w)\) if and only if \(xw=yz\text{.}\) Here we should point out that the symbol \(=\) can be used (and often is) to denote an equivalence relation. It is not constrained to mean “identically the same.”
When \(q=(x,y)\) is a fraction, \(x\) is called the numerator and \(y\) is called the denominator of \(q\text{.}\) Remember that the denominator of a fraction is never zero.
Addition of fractions is defined by
\begin{equation*}
(a,b)+(c,d) = (ad+bc,bd),
\end{equation*}
while multiplication is defined by
\begin{equation*}
(a,b)(c,d) = (ac,bd).
\end{equation*}
As was the case with integers, it is important to pause and prove that both operations are well defined.
Addition and multiplication are both associative and commutative. Also, we have the distributive property.
Theorem B.47.
Let \(x,y,z\in Q\text{.}\) Then
- \(x+y=y+x\) and \(xy=yx\text{.}\)
- \(x+(y+z) = (x+y)+z\) and \(x(yz)=(xy)z\text{.}\)
- \(x(y+z)=xy+xz\text{.}\)
The additive inverse of a fraction \((a,b)\) is just \((-a,b)\text{.}\) Using this, we define subtraction for fractions: \((a,b)-(c,d)=(a,b)+(-c,d)\text{.}\)
When \((a,b)\) is a fraction, and \(a\neq 0\text{,}\) the fraction \((b,a)\) is the reciprocal of \((a,b)\text{.}\) The reciprocal is also called the multiplicative inverse, and the reciprocal of \(x\) is denoted \(x^{-1}\text{.}\) When \(y\neq0\text{,}\) we can then define division by setting \(x/y=xy^{-1}\text{,}\) i.e., \((a,b)/(c,d)=(ad, bc)\text{.}\) Of course, division by zero is not defined, a fact that you probably already knew!
As was the case for both \(\nonnegints\) and \(\ints\text{,}\) when \(n\) is a positive integer, and \(0\) is the zero in \(\rats\text{,}\) we define \(0^n=0\text{.}\) When \(x=(a,b)\) is a fraction with \(x\neq0\) and \(n\) is a non-negative integer, we define \(x^n\) inductively by (i) \(x^0=1\) and (ii) \(x^{n+1}=xx^n\text{.}\)
Theorem B.48.
If \(x\in\rats\text{,}\) \(x\neq0\text{,}\) and \(m,n\in\ints\text{,}\) then \(x^mx^n=x^{m+m}\) and \((x^m)^n=x^{mn}\text{.}\)
Many folks prefer an alternate notation for fractions in which the numerator is written directly over the denominator with a horizontal line between them, so \((2,5)\) can also be written as \(\frac{2}{5}\text{.}\)
Via the map \(g(x) = (x,1)=\frac{x}{1}\text{,}\) we again say that the integers are a “subset” of the rationals. As before, note that \(g(x+y) = g(x)+g(y)\text{,}\) \(g(x-y)=g(x)-g(y)\) and \(g(xy)=g(x)g(y)\text{.}\)
In the third grade, you were probably told that \(5 =\frac{5}{1}\text{,}\) but by now you are realizing that this is not exactly true. Similarly, if you had told your teacher that \(\frac{3}{4}\) and \(\frac{6}{8}\) weren’t really the same and were only “equal” in the broader sense of an equivalence relation defined on a subset of the cartesian product of the integers, you probably would have been sent to the Principal’s office.
Try to imagine the trouble you would have gotten into had you insisted that the real meaning of \(\frac{1}{2}\) was
\begin{equation*}
\frac{1}{2} =\langle(\langle(s(s(0)),s(0))\rangle,
\langle(s(s(0)),0)\rangle)\rangle
\end{equation*}
We can also define a total order on \(\rats\text{.}\) To do this, we assume that \((a,b),(c,d)\in\rats\) have \(b,d>0\text{.}\) (If \(b\lt 0\text{,}\) for example, we would replace it by \((a',b')=(-a,-b)\text{,}\) which is in the same equivalence class as \((a,b)\) and has \(b'>0\text{.}\)) Then we set \((a,b)\leq (c,d)\) in \(\rats\) if \(ad\leq bc\) in \(\ints\text{.}\)
Subsection B.16.1 Integer Exponents
When \(n\) is a positive integer and \(0\) is the zero in \(\rats\text{,}\) we define \(0^n=0\text{.}\) When \(x\in\rats\text{,}\) \(x\neq 0\) and \(n\in\nonnegints\text{,}\) we define \(x^n\) inductively by (i) \(x^0=1\) and \(x^{k+1}=xx^k\text{.}\) When \(n\in\ints\) and \(n\lt 0\text{,}\) we set \(x^n=1/x^{-n}\text{.}\)
Theorem B.49.
If \(x\in\rats\text{,}\) \(x\neq0\text{,}\) and \(m,n\in\ints\text{,}\) then \(x^mx^n=x^{m+m}\) and \((x^m)^n=x^{mn}\text{.}\)
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