# Equality!

A field, in abstract algebra, is a system where:

1. Addition works the way you expect.
2. Multiplication also works the way you expect.

We can make this more precise.

A field obeys the following axioms:

A1: Addition is commutative: for all x and y, x + y = y + x
A2: Addition is associative: for all x and y, (x + y) + z = x + (y + z)
A3: The field contains an additive identity: there exists a member 0 such that for all x, x + 0 = x
A4: Each member has an additive inverse: for each x there is a -x such that x + -x = 0

M1: Multiplication is commutative: for all x and y, x * y = y * x
M2: Multiplication is associative: for all x and y, (x * y) * z = x * (y * z)
M3: The field contains an multiplicative identity not equal to the additive identity: there exists a member 1 ≠ 0 such that for all x, x * 1 = x
M4: Each member other than the additive inverse has an multiplicative inverse: for each x ≠ 0 there is an x-1 such that x * x-1 = 1

D1: Multiplication distributes over addition: for all x, x, and z, x * (y + z) = (x * y) + (x * z)

That’s the whole thing. The most familiar fields are the rational numbers, real number, and complex numbers [1]. There is also a collection of finite fields. For each prime number P, the field consists of the numbers 0 through P-1. Addition and multiplication are normal, except that if the answer would be P or greater, you replace it with the remained you’d get dividing it by P. This is called modulo arithmetic. For instance, in the field with members 0, 1, 2, 3, and 4, 2 * 3 = 1, because 6 divided by 5 leaves the remainder 1. Here are the addition and multiplication tables for this field. You can verify all the axioms, if you have the patience.

 + 0 1 2 3 4 0 0 1 2 3 4 1 1 2 3 4 0 2 2 3 4 0 1 3 3 4 0 1 2 4 4 0 1 2 3
 * 0 1 2 3 4 0 0 0 0 0 0 1 0 1 2 3 4 2 0 2 4 1 3 3 0 3 1 4 2 4 0 4 3 2 1

Note one oddity of the above: in axiom M3, we have to specifically assume that 0 ≠ 1, because there is no way to prove it. In fact, if it were true, we’d get a system which is is one way interesting and in another trivial:

Lemma: For all x, x * 0 = 0
We know that 0 + 1 = 0, so x = x * 1 = x * (1 + 0)
Using D1, x = (x * 1) + (x * 0) = x + (x * 0)
Substituting equals for equals into 0 = -x + x,
we get 0 = -x + x + (x * 0) = 0 + (x * 0) = x * 0
QED

Using the lemma, if 0 = 1 then for all x, x * 1 = 0, or x = 0. So all numbers are equal to zero, and thus to each other. The result is a very complete system in which there is one answer to almost all questions. Is there a square root of two? Yes, it’s 0. Is there a square root of -1? Yup, 0. Can you divide by 0? Sure, and the result is 0.

Any resemblance to your least favorite ideology, which considers every issue a nail to be attacked with the same sledgehammer, is purely coincidental.

1. There’s a four-dimensional analog of the complex numbers called the quaternions, where in addition to 1 and i you have j and k, but it’s not quite a field because multiplication isn’t commutative. i * j = k, but j * i = -k

### Mike Schilling

Mike has been a software engineer far longer than he would like to admit. He has strong opinions on baseball, software, science fiction, comedy, contract bridge, and European history, any of which he's willing to share with almost no prompting whatsoever.

1. So if everything equals nothing, and nothingness is pure consciousness (consciousness without an object), then everything is pure consciousness? Duuude.

• Mike Schilling

Nothing is everything
Everything is nothing is
Break the fences, nothing is

• If you combine The Who and Sartre, you’re going to get some awesome Abbott and Costello skits.

• “Who’s on First?”
“We are alone.”
“What?”
“Second base.”

• Now I feel sick to my stomach.

• Mike Schilling

“Oh, that’s our shortstop!”

2. Kazzy

A post about equality? When did we start getting political on MD?

• Kazzy

And the way in which you draw a false equivalency between the left side of an equation and the right side of the equation is… troubling.

3. Oh, I meant to add, this was awesome and I want more like this.

4. Surely it’s possible, in theory, to prove that 1 and 0 aren’t equal.

Or is that something that only a n00b would say?

• Mike Schilling

This is pure mathematics; check your intuition at the door. If you can create a model where A1-An are true but An+1 is false, then you can’t prove An+1 from the others.

• Ken

Nonsense. I’m sure the Fifth Postulate can be proved from the others. Why, it even looks like a theorem.

Actually, hyperbolic geometry would be kind of fun. I like the equivalents to the Fifth Postulate, the ones that seem so obvious but turn out to be false in hyperbolic geometry. A rectangle exists. There is no upper limit to the area of a triangle. Pythagoras’ Theorem.

5. Kimmi

This does not count as mindless diversions.
I must be very mindful to understand what you’re talking about.

Nonetheless, I approve.

6. Alan Scott

I vote more math posts!

• Mike Schilling

Any requests? The real difficulty with these is coming up with topics.

• J@m3z Aitch

Explain and refute the incompleteness theorem in no more than 100 words.

• Heh… “Provide your own proofs of the Riemann hypothesis, using at least 3 different methods, with diagrams.”

• Mike Schilling

Heinlein used to do stuff like this. Anything he disapproved of (Cantor’s theory of transfinites, uncertainty, incompleteness) would have been proved to be complete nonsense sometime before a story was set, and some character would laugh at the stupid 20th-Centuryites for believing that crap.

• Kimmi