Does 1 Times 1 Equal 2?

Gödel’s Lost Letter and P=NP 2018-08-02

A diversion in mathematical consistency

Cropped from source

Terrence Howard is an actor and singer who has been in a number of films and TV series. He was nominated for an Academy Award for his role in the movie Hustle & Flow. He currently stars in the TV series Empire.

Today Ken and I want to talk about his claim that ${1 \times 1 = 2}$ .

Apparently he thinks ${1 \times 1 = 2}$ , and has said:

This is the last century that our children will have to be taught that one times one is one.

We all know that ${1 \times 1 = 1}$ of course. Or does it? Of course it does. Howard, because of his visibility as an actor, has a platform to explain his ideas about arithmetic and mathematics. He claims to have a new theory of arithmetic that will change the world. About his discovery, he said in a Rolling Stone interview:

If Pythagoras was here to see it, he would lose his mind. Einstein too! Tesla!

A pretty far-out claim.

Reaction

The overwhelming respond to his claim has been predictable. Quoting Howard’s justification from the same interview—

How can it equal one? If one times one equals one that means that two is of no value because one times itself has no effect. One times one equals two because the square root of four is two, so what’s the square root of two? Should be one, but we’re told its two, and that cannot be.

—one commenter in a Reddit thread retorted:

No. No we are not. It’s like he hasn’t heard of decimals or something.

And it gets even nastier. He was on The View the other day and I must admit that I saw him there. The View, for those not into daytime TV, is a talk show that was created by Barbara Walters and Bill Geddie. It is on each day and has guests such as Howard on fairly often. Howard explained on the show how ${1 \times 1 = 2}$ . The hosts of The View are, of course, not math experts, so they listened politely to Howard and later thanked him for his comments.

I have no idea if they believed him or not. But I was shocked to hear that someone had seriously suggested that ${1}$ times anything is not the same thing. Indeed.

The Laws

So I started to write this post above his claim. My plan was to show that his claim led to a contradiction. This would of course show that he was wrong in his claim. But a funny thing happen—I did not get a contradiction. Let me explain.

My plan was to use the usual rules of arithmetic to show that ${1 \times 1 = 2}$ is wrong. The rules I planned to use were the standard ones:

Commutative Law: ${ A \times B = B \times A}$ .

Associative Law: ${ A \times (B \times C) = (A \times B) \times C}$ .

Distributive Law: ${ A \times (B + C) = A \times B + A \times C}$ .

On this page there is a nice explanation why the distributive law is useful:

You probably use this property without knowing that you are using it. When a group (let’s say 5 of you) order food, and order the same thing (let’s say you each order a hamburger for $3 each and a coke for $1 each), you can compute the bill (without tax) in two ways. You can figure out how much each of you needs to pay and multiply the sum times the number of you. So, you each pay (3 + 1) and then multiply times 5. That’s 5(3 + 1) = 5(4) = 20. Or, you can figure out how much the 5 hamburgers will cost and the 5 cokes and then find the total. That’s 5(3) + 5(1) = 15 + 5 = 20. Either way, the answer is the same, $20.

My only comment is that what world is a burger $3 and a coke $1? Perhaps at some fast food places, but I live in New York City and this off by a large factor.

Is Howard Wrong?

This famous passage involving Humpty Dumpty in Lewis Carroll’s Through the Looking-Glass (1872) applies just as much to mathematics as to words:

“I don’t know what you mean by ‘glory’,” Alice said. Humpty Dumpty smiled contemptuously. “Of course you don’t—till I tell you. I meant ‘there’s a nice knock-down argument for you!’ ” “But ‘glory’ doesn’t mean ‘a nice knock-down argument’,” Alice objected. “When I use a word,” Humpty Dumpty said, in rather a scornful tone, “it means just what I choose it to mean—neither more nor less.” “The question is,” said Alice, “whether you can make words mean so many different things.” “The question is,” said Humpty Dumpty, “which is to be master—that’s all.”

What I realized is that Howard could be right that ${1 \times 1}$ is ${2}$ . Really. Let’s agree to use ${\otimes}$ for what Howard means by multiplication. So call

$\displaystyle A \otimes B$

the Howard product of ${A}$ and ${B}$ . Now we could define ${1 \otimes 1}$ any way we want—like Humpty Dumpty in Lewis Caroll’s world.

Let’s see what happens if ${1\otimes1=2}$ . Now

$\displaystyle 1 \otimes 2 = 1 \otimes (1+1) = 1\otimes1 + 1\otimes1$

by the distributive law. Thus ${1 \otimes 2= 2 + 2 = 4}$ . But now

$\displaystyle 1\otimes4 = 1\otimes2 + 1\otimes2 = 4 + 4 = 8.$

Also

$\displaystyle 1 \otimes (3+1) = 1\otimes 3 + 1\otimes 1 = A+2$

So ${A+2 = 8}$ and ${A=6}$ . In general we get that

$\displaystyle 1 \otimes n = 2n.$

Now ${ 2 \otimes n = (1+1)\otimes n = 1\otimes n + 1\otimes n = 4n}$ . Continue in this manner and get

$\displaystyle A \otimes B = 2AB.$

In general the operation ${A \otimes B}$ is commutative, associative, and distributive.

$\displaystyle A \otimes B = 2AB$

and

$\displaystyle A \otimes (B+C) = 2A(B+C)$

But this is same as

$\displaystyle A \otimes B + A \otimes C = 2AB + 2AC = 2A(B+C).$

And likewise for ${(B+C) \otimes A}$ . Let’s check the associative law:

$\displaystyle A \otimes (B \otimes C) = A \otimes (2BC) = 4ABC.$

$\displaystyle (A \otimes B) \otimes C = 2AB \otimes C = 4ABC.$

The Point

The point is that Howard can define product in a new way. His system that has ${A \otimes B = 2AB}$ is just fine. It has all the usual basic properties of arithmetic but is really nothing new. Rather than a radical new system of arithmetic his system is just a kind of renormalization of the standard one. It is like changing feet to yards.

Ken adds that Howard’s ${\otimes}$ behaves like a parallel complexity measure for multiplication gates. It maps any ring onto the ideal of even elements in the ring—which of course can be the whole ring. Whether it has comparable utility to the idea of the field with one element is anyone’s guess.

Beyond the basic operations being the “same” there are some differences in Howard’s ${\otimes}$ . Note that in his system the notion of primes is different from the usual. But I thought I would stop here.

Open Problems

Do you think Howard will agree? Is it more useful to point out that Howard’s product is really just our old friend renormalized than to try and argue that he is wrong? What do you think?