**Maintainer:**admin

(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.)

*1*Defining 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 something^{1} 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 group^{2}. 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.