# The Colours of Numbers

August 11, 2003 5:28 PM Subscribe

The colour of numbers - For the math geeks out there (which I'm not - maybe his theories will be shot down in flames), Karl Palmen has discovered that numbers can be assigned one of eight "colours", related to their prime factors. He goes on to show the interesting mathematical properties of these colours. A novel way of playing with numbers. Software is on offer.

okay, sorry, brief chromo-lexical synasthesia brain spasm...

posted by jokeefe at 5:33 PM on August 11, 2003

posted by jokeefe at 5:33 PM on August 11, 2003

(Actually, I've always considered 4 to be more of a "Puce", jokeefe).

posted by Jimbob at 5:35 PM on August 11, 2003

posted by Jimbob at 5:35 PM on August 11, 2003

Whoa, I sense a tin foil hat in the author of this page's immediate future...

posted by jokeefe at 5:44 PM on August 11, 2003

posted by jokeefe at 5:44 PM on August 11, 2003

Neato. Tempting. Clearly nutzoid, but neato and tempting.

posted by gleuschk at 5:54 PM on August 11, 2003

posted by gleuschk at 5:54 PM on August 11, 2003

"2" is so red. 3 is yellow. 4 is dark... purple or brown. 5 is a sortof whitish blue. 1 is a sortof generic dark. I don't know what six is. 7 is clearly sortof an orange brown, eight being fiery straight out. 9 slightly darker color of five.

Maybe we should do a survey.

posted by weston at 6:07 PM on August 11, 2003

Maybe we should do a survey.

posted by weston at 6:07 PM on August 11, 2003

*Hence the square root of 2 is irrational.*

I've a horrible feeling that this actually makes sense to some mathematicians.

posted by dash_slot- at 6:09 PM on August 11, 2003

This looks like a generalization of Ulam's spiral, which is formed by writing the positive integers in a spiral and coloring primes & non-primes different colors. It's not uncompelling, like gleuschk said, but unfortunately there's probably not much to it.

Hmm, judging from that second URL, the guy I linked to seems to have an axe to grind... I think his point's still valid, though. You also have to remember that it's surprisingly easy to find meaning in seemingly random but infinite sequences: for example, my phone number happens to be in the digits of pi starting at position 100192, but there's nothing terribly cosmic about that.

posted by Johnny Assay at 6:09 PM on August 11, 2003

Hmm, judging from that second URL, the guy I linked to seems to have an axe to grind... I think his point's still valid, though. You also have to remember that it's surprisingly easy to find meaning in seemingly random but infinite sequences: for example, my phone number happens to be in the digits of pi starting at position 100192, but there's nothing terribly cosmic about that.

posted by Johnny Assay at 6:09 PM on August 11, 2003

And by the way, I only gave a cursory glance, but the guy may not be a nut case. I've actually used a scheme like this (not like the arbitrary synaesthesiac one I gave above) to illustrate abstract algebra/number theoretic concepts. Given a color (or other symbol) for a number and an addition table (modulo 10), you can suss out which color corresponds to our arbitrarily symbol'd and named number -- even if the columns aren't ordered correctly on the addition table. It's an excellent introduction to the concept of cycles. I think this guy is doing something like that.

posted by weston at 6:12 PM on August 11, 2003

posted by weston at 6:12 PM on August 11, 2003

If man is 5, then the devil is 6, and if the devil is 6,

posted by stavrosthewonderchicken at 7:27 PM on August 11, 2003

**then god is 7 then god is 7**!posted by stavrosthewonderchicken at 7:27 PM on August 11, 2003

This chicken's gone to heaven.

posted by stavrosthewonderchicken at 7:27 PM on August 11, 2003

posted by stavrosthewonderchicken at 7:27 PM on August 11, 2003

I also am not a mathematician, so I won't be surprised if somebody later reveals to us that this is some simple corrolary of a theorem from the middle ages. I do find it annoying that he doesn't even sketch proofs for any of his "conjectures."

A proof of the following should be provided, instead of just stating a belief of its truth:

Now, the

posted by dilettanti at 7:59 PM on August 11, 2003

A proof of the following should be provided, instead of just stating a belief of its truth:

*All numbers are uniquely colored.*That is, all natural numbers are consistently colored, and only one color is consistent with the rules for any number. Obviously, all numbers can be factored to primes, and their colors are assigned by the rules directly, so what must be shown is that exactly one "color multiplication table" will be consistent with the rules. But we know that black must be the "identity" color, since 1 is black. Then consider any product of primes. All prime factors will be black, red, blue or green. Any even number of factors of a particular color will cancel to black, so only the odd colors matter. All possible contingencies are covered in the rules, so the multiplication table he provides is consistent and unique up to permutations of the colors (or substitutions of other colors, naturally).Now, the

*real*question.*Who cares?*What interesting properties of the numbers are elucidated by coloring them? Do all cyan numbers share any interesting properties, aside from being cyan? Is there a more efficient algorithm for finding colors? Will it help in the search for Mersenne primes? Anything?posted by dilettanti at 7:59 PM on August 11, 2003

*"2" is so red. 3 is yellow. 4 is dark... purple or brown. 5 is a sortof whitish blue. 1 is a sort of generic dark. I don't know what six is. 7 is clearly sort of an orange brown, eight being fiery straight out. 9 slightly darker color of five.*

weston, I'm afraid contemplating other people's colour/number relations makes my brain feel like it's been dipped in fibreglass. Or like someone's raking fingernails over the blackboard of my frontal lobes. It's just

*weird*.

posted by jokeefe at 9:48 PM on August 11, 2003

dash_slot: The demonstration that root 2 is irrational actually makes sense. He gives a rule that assigns a color to every rational number. So if root 2 were rational, it would have a color assigned to it. The square of any number with a color is a number whose color is black. So if root 2 is rational, its square (a.k.a. 2) should be black. But according to the rules of coloring, 2 is red. This is a contradiction, so root 2 can't have a color (and thus can't be rational).

Of course, as dilettanti points out, what this guy is saying only works if every number has a color (which is pretty obvious, since you can get at least one color assignment by factoring the number into primes, finding the colors of the prime factors, then doing the color-multiplication on those colors) and if every number only has one possible color assignment. This would be a bit harder to prove if it weren't for the fact that the system of colors that the author defines is nothing more than a group of equivalence classes on natural (then positive rational) numbers. (If you have a multiplication table for the operation with an identity element and inverses, all you need in the table to know you have a (finite) group is that each row and each column contains each element exactly once.)

Since the color system is a group, you get the fact that the multiplication operation is associative for free. Also, since the multiplication table is symmetric across the (upper-right to lower-left) diagonal, the operation is commutative as well. Since the operation is commutative and associative, no matter what order and/or permutation you use to calculate a number's color from its prime factors, you're guaranteed to get the same thing.

Most of what's on the page I (personally) don't find very interesting. Almost any similar system is going to give some results that look interesting. The thing is, most of those results really just come from the complexity of the number system itself, not from the way the equivalence classes are set up. This sort of setup is just a less-expressive and less-precise way of conveying the same information as the number system and algebraic formulae already do. (As an aside, I'm sorry for inflicting that sentence on the English speaking world.)

The only interesting question on the page, to my mind, is Challenge 5. This comes to the same this as asking "Is the closure of the set of all black numbers equal to the set of all positive reals?" It's an interesting question, but the answer isn't too hard to get.

To prove that it is, all we need to prove is that in any positive open interval, there is a black number. I'm going to show that in any positive open interval, there is a number that is a square of a rational number (since such a number will always be a black number). (Take sqrt(x) to be the square root of x.)

Given the interval (a, b), with a > 0 and b > 0, consider the interval (sqrt(a), sqrt(b)). There is always a rational number in any open interval (really, just trust me; I'm tired of typing), call it x. Since sqrt(a) < x < sqrt(b), a < x^2 < b. So x^2 (the square of x) is in the interval (a, b). And since x is rational, so is x^2. QED

Wow. I'm a dork.

posted by samw at 11:46 PM on August 11, 2003

Of course, as dilettanti points out, what this guy is saying only works if every number has a color (which is pretty obvious, since you can get at least one color assignment by factoring the number into primes, finding the colors of the prime factors, then doing the color-multiplication on those colors) and if every number only has one possible color assignment. This would be a bit harder to prove if it weren't for the fact that the system of colors that the author defines is nothing more than a group of equivalence classes on natural (then positive rational) numbers. (If you have a multiplication table for the operation with an identity element and inverses, all you need in the table to know you have a (finite) group is that each row and each column contains each element exactly once.)

Since the color system is a group, you get the fact that the multiplication operation is associative for free. Also, since the multiplication table is symmetric across the (upper-right to lower-left) diagonal, the operation is commutative as well. Since the operation is commutative and associative, no matter what order and/or permutation you use to calculate a number's color from its prime factors, you're guaranteed to get the same thing.

Most of what's on the page I (personally) don't find very interesting. Almost any similar system is going to give some results that look interesting. The thing is, most of those results really just come from the complexity of the number system itself, not from the way the equivalence classes are set up. This sort of setup is just a less-expressive and less-precise way of conveying the same information as the number system and algebraic formulae already do. (As an aside, I'm sorry for inflicting that sentence on the English speaking world.)

The only interesting question on the page, to my mind, is Challenge 5. This comes to the same this as asking "Is the closure of the set of all black numbers equal to the set of all positive reals?" It's an interesting question, but the answer isn't too hard to get.

To prove that it is, all we need to prove is that in any positive open interval, there is a black number. I'm going to show that in any positive open interval, there is a number that is a square of a rational number (since such a number will always be a black number). (Take sqrt(x) to be the square root of x.)

Given the interval (a, b), with a > 0 and b > 0, consider the interval (sqrt(a), sqrt(b)). There is always a rational number in any open interval (really, just trust me; I'm tired of typing), call it x. Since sqrt(a) < x < sqrt(b), a < x^2 < b. So x^2 (the square of x) is in the interval (a, b). And since x is rational, so is x^2. QED

Wow. I'm a dork.

posted by samw at 11:46 PM on August 11, 2003

There seems to be a lot of "this is true in all cases, except..." which smacks of a flawed theory to me. However, its far too early for me to actually read the page properly.

posted by Orange Goblin at 1:32 AM on August 12, 2003

posted by Orange Goblin at 1:32 AM on August 12, 2003

*Wow. I'm a dork.*

Revel in your dorkness, samw. Your whole post is way over my head, but it's still one of my favorite comments I've ever seen on MeFi. Well-reasoned, explanatory. . . maybe even (how would I know?) exactly right.

posted by LeLiLo at 1:52 AM on August 12, 2003

QED indeed.

I for one, welcome

posted by garyh at 4:05 AM on August 12, 2003

I for one, welcome

**samw**as our mathematical overlord.posted by garyh at 4:05 AM on August 12, 2003

This is an abelian group of order 8 in which every member is its own inverse (black, which I will abbreviate 1, being the identity). This makes it (Z2)^3.

From how he defined it, the generators appear to be red (R), green (G), and blue (B).

The non-generating members of the group come from:

Y=RG

M=RB

C=GB

W=RGB

With this in mind, we can look at the multiplication table and verify that it is, indeed, (Z2)^3

It looks like his structure is well-defined, at least on the natural numbers. Every number can be uniquely factored into primes. Each prime factorization leads to a unique collection of R, G, and B, which are multiplied together in a unique way.

posted by CrunchyFrog at 6:47 AM on August 12, 2003

From how he defined it, the generators appear to be red (R), green (G), and blue (B).

The non-generating members of the group come from:

Y=RG

M=RB

C=GB

W=RGB

With this in mind, we can look at the multiplication table and verify that it is, indeed, (Z2)^3

It looks like his structure is well-defined, at least on the natural numbers. Every number can be uniquely factored into primes. Each prime factorization leads to a unique collection of R, G, and B, which are multiplied together in a unique way.

posted by CrunchyFrog at 6:47 AM on August 12, 2003

*really, just trust me; I'm tired of typing*

I wish more proofs included this phrase. I'd be a lot better at math if that were so.

posted by ook at 8:09 AM on August 12, 2003

Now that I've had a chance to think about it, the group he sets up is kind of interesting. It's abelian of order 8, but each of the elements (other than black) has order 2, so it's

(even,even,even) = black

(odd,even,even) = red

(even,odd,even) = green

(even,even,odd) = blue

(even,odd,odd) = cyan

(odd,even,odd) = magenta

(odd,odd,even) = yellow

(odd,odd,odd) = white

...and the dude even chose the visual color scheme corresponding to the way our eyes work: red when only the red cones are excited, magenta when red and blue cones are excited, etc.

So

On preview: What CrunchyFrog said.

posted by dilettanti at 8:11 AM on August 12, 2003

**Z**(2)⊕**Z**(2)⊕**Z**(2). In other words, it's isomorphic to the group of ordered triples where each element in the triple takes one of two values: say, even or odd. In fact, the coloring scheme is quite simple. Color the primes (red, green and blue), then assign colors to composites by determining whether the number of factors of a particular color is even or odd. Denote any integer*x*by its prime factors (r,g,b), where r,g,b ? {even,odd} and r=even if*x*has an even number of red prime factors, etc. Then all factorizations may be summarized:(even,even,even) = black

(odd,even,even) = red

(even,odd,even) = green

(even,even,odd) = blue

(even,odd,odd) = cyan

(odd,even,odd) = magenta

(odd,odd,even) = yellow

(odd,odd,odd) = white

...and the dude even chose the visual color scheme corresponding to the way our eyes work: red when only the red cones are excited, magenta when red and blue cones are excited, etc.

So

*any*method of coloring the primes in red, green and blue will result in a division of the natural numbers (rationals too) into eight equivalence classes that may be so colored. Question 1: How many different ways can you color the primes? Answer: infinite. Question 2: Does any one classification tell us something new about the natural numbers? Answer: I have no idea. No evidence yet.On preview: What CrunchyFrog said.

posted by dilettanti at 8:11 AM on August 12, 2003

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1 is white, 2 is blue, 3 is red, 4 is pink, 5 is yellow, 6 is brown, 7 is green, 8 is indigo, 9 is gold.

Any questions?

posted by jokeefe at 5:32 PM on August 11, 2003