# The Unapologetic Mathematician

## Integers

I’m back from Ohio at the College Perk again. The place looks a lot different in daylight. Anyhow, since the last few days have been a little short on the exposition, I thought I’d cover integers.

Okay, we’ve covered that the natural numbers are a commutative ordered monoid. We can add numbers, but we’re used to subtracting numbers too. The problem is that we can’t subtract with just the natural numbers — they aren’t a group. What could we do with $2-3$?

Well, let’s just throw it in. In fact, let’s just throw in a new element for every possible subtraction of natural numbers. And since we can get back any natural number by subtracting zero from it, let’s just throw out all the original numbers and just work with these differences. We’re looking at the set of all pairs $(a,b)$ of natural numbers.

Oops, now we’ve overdone it. Clearly some of these differences should be the same. In particular, $(S(a),S(b))$ should be the same as $(a,b)$. If we repeat this relation we can see that $(a+c,b+c)$ should be the same as $(a,b)$ where we’re using the definition of addition of natural numbers from last time. We can clean this up and write all of these in one fell swoop by defining the equivalence relation: $(a,b)\sim(a',b')\Leftrightarrow a+b'=b+a'$. After checking that this is indeed an equivalence relation, we can pass to the set of equivalence classes and call these the integers $\mathbb{Z}$.

Now we have to add structure to this set. We define an order on the integers by $(a,b)\leq(c,d)\Leftrightarrow a+d\leq b+c$. The caveat here is that we have to check that if we replace a pair with an equivalent pair we get the same answer. Let’s say $(a,b)\sim(a',b')$, $(c,d)\sim(c',d')$, and $(a,b)\leq(c,d)$. Then $a'+b+c+d'=a+b'+c'+d\leq b+b'+c'+c$
so $a'+d'\leq b'+c'$. The first equality uses the equivalences we assumed and the second uses the inequality. Throughout we’re using the associativity and commutativity. That the first inequality implies the second follows because addition of natural numbers preserves order.

We get an addition as well. We define $(a,b)+(c,d)=(a+c,b+d)$. It’s important to note here that the addition on the left is how we’re defining the sum of two pairs, and those on the right are additions of natural numbers we already know how to do. Now if $(a,b)\sim(a',b')$ and $(c,d)\sim(c',d')$ we see $(a+c)+(b'+d')=(a+b')+(c+d')=(b+a')+(d+c')=(a'+c')+(b+d)$
so $(a+c,b+d)\sim(a'+c',b'+d')$. Addition of integers doesn’t depend on which representative pairs we use. It’s easy now to check that this addition is associative and commutative, that $(0,0)$ is an additive identity, that $(b,a)+(a,b)\sim(0,0)$ (giving additive inverses), and that addition preserves the order structure. All this together makes $\mathbb{Z}$ into an ordered abelian group.

Now we can relate the integers back to the natural numbers. Since the integers are a group, they’re also a monoid. We can give a monoid homomorphism embedding $\mathbb{N}\rightarrow\mathbb{Z}$. Send the natural number $a$ to the integer represented by $(a,0)$. We call the nonzero integers of this form “positive”, and their inverses of the form $(0,a)$ “negative”. We can verify that $(a,0)\geq(0,0)$ and $(0,a)\leq(0,0)$. Check that every integer has a unique representative pair with ${}0$ on one side or the other, so each is either positive, negative, or zero. From now on we’ll just write $a$ for the integer represented by $(a,0)$ and $-a$ for the one represented by $(0,a)$, as we’re used to.

March 18, 2007 Posted by | Fundamentals, Numbers | Leave a comment