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Does 1 Times 1 Equal 2?

May 22, 2018

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.


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.


\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


\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?


Lost in Complexity

May 19, 2018

Should we expect simplicity in a theory named for complexity?

Amer. Phy. Soc. interview source

Sabine Hossenfelder is a physicist at the Frankfurt Institute for Advanced Studies who works on quantum gravity. She is also noted for her BackRe(Action) blog. She has a forthcoming book Lost in Math: How Beauty Leads Physics Astray. Its thesis is that the quest for beauty and simplicity in physics has led to untestable theories and diverted attention from concrete engagements with reality.

Today we wonder whether her ideas can be tested, at least by analogy, in computational complexity.
Read more…

A New Proof Of An Ancient Result

May 5, 2018

Triangulating proofs to seek a shorter path

Cropped from 2016 Newsday source

Mehtaab Sawhney is an undergraduate student at MIT. His work caught my eye on finding his recent paper with David Stoner about permutations that map all three-term arithmetic progressions mod {n} to non-progressions. Here a progression is an ordered triple {(a,b,c)} where {a -2b + c = 0 \pmod{n}}. The paper addresses when such permutations can be found in certain small subgroups of {S_n} while I am interested in senses by which they are succinct. This made me curious about Sawhney’s other work.

Today Ken and I wish to report on Sawhney’s simple new proof of the famous triangle inequality in {\mathbb{R}^n}.
Read more…

Finding Coprime Pairs

April 29, 2018

Deferring or avoiding randomization

Great Discoveries in STEM source

Claude Bachet de Méziriac was a French mathematician of the early 1600s. He is the first person we know to have posed and solved the problem, given relatively prime (also called coprime) integers {x} and {y}, of finding integers {a} and {b} such that {ax + by = 1}.

Today we revisit some questions about generating coprime pairs deterministically.
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Tom Lehrer At 90

April 10, 2018

It makes a fellow proud to be a nerd

Tasmanian Archive source

Tom Lehrer is Emeritus Lecturer in Mathematics at Cowell College of the University of Santa Cruz. He is listed not in the Mathematics Department but in Humanities, for which he also lectured on musical theater. He was my first witness that effective input to the social conversation could start from conversancy in mathematics.

Yesterday was his 90th birthday and we hope he had a great one.
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The Entropy of Baseball

April 1, 2018

The most shocking existential fact about the universe?

Sports April Fools source

George Ruth Jr., the “Babe,” may have thought he had cosmic significance but no one knew it until now. He would have said it was all a joke anyway. He certainly loved pranks. As an April Fool’s joke during Florida spring training, he once let it be reported that he had slimmed down to 108 pounds and was beginning a new career as a jockey.

Today we report how the Babe—and every major-league player from David Aardsma and Henry Aaron to Edward Zwilling and Tony Zych—helped uncover a fact about the universe.
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Is There Momentum in Chess?

March 28, 2018

And if so, what units does it have?

Cropped from source

Fabiano Caruana has just won the 2018 chess Candidates Tournament. This earns him the right to challenge Magnus Carlsen for the world championship next November in London. A devastating loss on Saturday to the previous challenger, Sergey Karjakin, had seemed to presage a repeat of his last-round failure in the 2016 Candidates. But Caruana reversed the mojo completely with powerful victories in Monday’s and Tuesday’s last two rounds to win the tournament by a full point.

Today we congratulate Caruana on his triumph and hail the first time an American will challenge for the world championship since Bobby Fischer in 1972. And we ask, is there really such a thing as being “in” or “out of” form in chess and similar pursuits?
Read more…