Friday, November 25, 2011

Defining the integers within set theory

(The first half of the class was spent going over the fall 2010 final, question 4. The second half was spent going over a question from the fall 2011 midterm, and using that as a starting point to define the integers using set theory.)

1Defining the integers¶

Start with $\omega \times \omega$. Define the equivalence relation^? $E$ on $\omega \times \omega$ by $(a, b) E (c, d) \iff a + d = b + c$. The set of equivalence classes - $(\omega \times \omega) / E$ - is then defined. (Each equivalence class has infinitely many ordered pairs.) We can call this set $\mathbb{Z}$, and define addition on $\mathbb{Z}$ by:

$$[(a, b)] + [(c, d)] = [(a + c, b+d)]$$

Check that this addition preserves something¹ about the equivalence classes:

$$[(a, b)] = [(a', b')],\, [(c, d)] = [(c', d')] \vdash [(a +c, b+d)] = [(a'+c', b'+d')]$$

This is an abelian group². Now we need something to act like zero, for which we can use $[(0, 0)]$ (or, really $[(\alpha, \alpha)]$ for any $\alpha$ because apparently $[(a, b)]$ stands for $a - b$?). For an additive inverse for $[(a, b)]$, we just use $[(b, a])$, because that results in the zero.

We can now check associativity and commutativity, if we so desire. Or, we could define multiplication:

$$[(a, b)] \times [(c, d)] = [(ac + bd, ad+bc)]$$

The identity for multiplication would have to be $[(1, 0)]$ or any $[(\alpha + 1, \alpha)]$.

Check that $<$ is a total order of $\mathbb{Z}$:

If $n \neq 0$, $m \neq 0$, $[(n \cdot m, 0)] > [(0, 0)]$. $[(n, 0)] > [(0, 0)]$ etc what?

N.B. Apparently there is a natural embedding of the natural numbers into the integers, whatever that means. Not sure why this was brought up.