Edit made on January 20, 2014 by ColinWright at 20:32:28
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HEADERS_END
This page is in the process of being _
re-written, so you may want to come _
back and visit again later.
Previously in this series:
* 1st: Constructing the Counting Numbers
* 2nd: Constructing the Integers
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(The rational numbers, that is.)
Suppose we know about integers but not about rational numbers.
That means we can solve equations like x+5=3, but not necessarily ones like 3x=7.
(Of course we can solve some such equations; 3x=6, for instance.)
As when constructing the integers, we would like to build on our
knowledge of the integers to make a more general sort of number
for which we can solve these equations.
We use exactly the same idea as when constructing the integers:
consider pairs /(a,b)/ where EQN:b\neq{}0. (The idea is that this
means a/b.) Say that /(a,b)=(c,d)/ exactly when /ad=bc./ (That is:
say that they are "equivalent" when this holds, and then work with
equivalence classes of pairs instead of just with pairs.)
The rules for doing arithmetic with these are exactly the ones
you learned in school for doing arithmetic with fractions. For
instance, /(a,b)+(c,d)=(ad+bc,bd)./
As before, it turns out that the arithmetic operations are well
defined; if you apply them to different elements of the same
equivalence classes, the result is in the same equivalence class.
When constructing the integers we found that /(a,a)/ "is zero";
here we find that /(a,a)/ "is 1".
When constructing the integers we found that /(b,a)/ "is the negative of"
/(a,b);/ here we find that /(b,a)/ "is the reciprocal of" /(a,b)./
When constructing the integers we found that /(a,0)/ "is /a";/
here we find that /(a,1)/ "is /a"./
Our collection of equivalence classes of pairs of integers acts
just as we would like the rational numbers to act.
So, now we can solve any linear equation /ax+b=c./ Want to solve
more complicated equations? Well, then it's time to think about
constructing the reals...
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This page uses:
* Equivalence class
* Equivalence relation
* Embedding