Building the Rationals

The integers allow addition and subtration – however, division is still tricky. For example, there is no \(x\) such that, \(2x=1\). If \(2x=1\), then \(x+x=1\), and so \(x<=1\). Since \(1+1=2\ne 1\), \(x<1\). However, \(0+0=1\), so \(0<x<1\).

Much like the integers, first consider how \(X/Y\) should behave and then formalize. Much like we did with the rationals, we want to think how $latex X/Y$ behave, and then formalize it.

  • \(X/Y=Z/W\) if and only if \(XW=ZY\)
  • \(X/Y+Z/W=(XW+YZ)/YW\)
  • \((X/Y)(Z/W)=(XZ)/(YW)\)

Formally define: for two pairs of integers, \((X,Y)~(Z,W)\) if \(XW=ZY\).

Exercise:

  • Prove this relationship is reflective, symmetric and transitive.

TODO: Add Hints (Issue #6)

Let \(\mathbb{Q}\) be the set of equivalence classes under this relationship.

Define multiplication and addition based on the earlier inspiration as:

\[[(X,Y)]+[(Z,W)]=[(XW+YZ,YW)]\]
\[[(X,Y)][(Z,W)]=[(XZ,YW)]\]
Exercise:
  • Prove these definitions depend only on the equivalence classes.

TODO: Add Hints (Issue #6)

Note that \([(X,1)]+[(Y,1)]=[(X+Y,1)]\) and \([(X,1)][(Y,1)]=[(XY,1)]\) and so \(e:\mathbb{Z}\to\mathbb{Q}\) defined by \(e(X)=[(X,1)]\) is an embedding. Much like earlier, it is sometimes useful to confuse \(e(X)\) and \(X\).

Note that if \([(X,Y)]\ne 0\) (or, really, \(e(0)\)), than \(X\ne 0\). In this case, \([(Y,X)]\in\mathbb{Q}\) and \([(X,Y)][(Y,X)]=[(XY,YX)]=[(1,1)]=e(1)\) So every element different from \(0\) has a multiplicative inverse.

In conclusion, the field axioms are satisfied:

  • \(a+0=a\)
  • \(a+(b+c)=(a+b)+c\)
  • \(a+b=b+a\)
  • For every \(a\) there is a \(-a\) such that \(a+(-a)=0\)
  • \(a1=a\)
  • \(a(bc)=(ab)c\)
  • \(ab=ba\)
  • For every \(a\ne 0\) that is a \(a^{-1}\) such that \(aa^{-1}=1\)
  • \(a(b+c)=ab+ac\)

The field of rationals is called \(\mathbb{Q}\).

The embedding of the natural numbers into the integers, combined with the embedding of the integers into the rationals, gives an embedding of the natural numbers into the rationals. Define a rational number, \(r\) to be “positive” if there are natural numbers that are not \(0\), \(n\) and \(m\), such that \(n=rm\).

Theorem: for every rational, \(r\), exactly one of those is true:
  • \(r=0\)
  • \(r\) is positive
  • \(-r\) is positive

Proof: Assume \(r=[(X,Y)]\) for \(X\) and \(Y\) integers. Since \(Y\ne 0\), by the structure of integers, either \(Y\) is natural or \(-Y\) is natural. If \(-Y\) is natural, by the definition of equivalence, \(r=[(-X,-Y)]\), so we can assume \(Y\) is natural.

If \(X=0\), then \(r=0\), and \(rm=0\) for every natural number \(m\).

If \(X\) is natural, then \(rY=X\).

If \(-X\) is natural, then \(-rY=-X\).

QED

Theorem: If \(r\) and \(s\) are positive, then so are \(r+s\) and rs.

Proof: If \(rm=n\) and sl=k, then \((rs)ml=nk\) and

\[(r+s)ml=rml+sml=rml+slm=nl+km\]

And \(ml\) and \(nl+km\) are natural numbers, since the natural numbers are closed under addition and multiplication.

QED

Define \(r\leq s\) if \(r=s\) or \(s-r\) is positive.

It is straightforward to verify that

  • \(x\leq y\) or \(y\leq x\) and if both are true, \(x=y\)
  • If \(x\leq y\) then \((x+z)\leq (x+z)\)
  • If \(x\leq y: and :math:`0\leq z\) then \(xz\leq yz\)

Togther with the field axioms, those are the axioms of an “ordered field”.

Exercise: Assume \(F\) is an ordered field, there is a function \(e:\mathbb{Q}\to F\), which is one-to-one (i.e., \(e(x)=e(y)\) implies \(x=y\)), preserves addition, multiplication, zero, one and order. (Such a function is called an “embedding”).

Hint-a-long: Define a sequence by recursion:

  • \(e(0)=0_F\)
  • \(e(n+1)=e(n)+1_F\)

Prove that \(n<m\) implies \(e(n)<e(m)\)

Define \(e(-n)=-e(n)\), and \(e(n/m)=e(n)/e(m)\).

The rationals, therefore, are the smallest ordered field.