## Thursday, December 22, 2011

### Base twelve math - Division Rules

[I should point out that only after writing this whole thing out did it occur to me to look online to see what others have said about base twelve.]

When I was writing up the division rules I was reminded of something someone once said about different bases. Specifically, they thought it would have been better if we used base twelve because twelve has more divisors. Because I feel like using random symbols that probably won't even show up for most people, I'm going to introduce two symbols that are new to me instead of using the traditional method of dealing with new digits (which is to use Latin letters.) So Ѧ means ten and ϟ means eleven.

Thus 1ϟѦ base twelve would be 1 gross (144 base ten) plus twelve times eleven, plus ten. Which is the same as 286 base ten.

Given that we're all already trained in base ten it will obviously be harder for us to use base twelve regardless. For example, when I wanted to work out what 1/7th would be in base 12 (.186Ѧ35 repeating, if I did it right) most of what I did was using base ten.

Also, the person who said base twelve would be better was in no way referring to the division rules. Fractions possibly*, but not division rules. All of which is to say, what I'm doing here is completely pointless.

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0

This is exactly the same as base ten, but there was an error in my thing for base ten. Zero isn't a divisor of anything, by definition. I honestly don't know what that is. The only number it could be a divisor of is zero, and apparently saying that zero divides zero would be problematic for something. Thus zero is the only integer that doesn't divide itself (for any other integer X, X is a divisor of X) and also the only integer that doesn't divide zero (for any other integer X, X is a divisor of zero.)

It's not as if saying zero was a divisor of zero would make zero over zero stop being indeterminate, 0=c*0 for an infinite number of integers c, so 0/0 would still be impossible to solve, and you'd still have to resort to limits if you wanted some kind of answer. Still, there must be a reason of some sort.

The division rule for zero will be the same in every base.

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1

Exactly the same as base ten: is it an integer?

The division rule for 1 will be the same in every base.

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2

Exactly the same as base ten: is the last digit divisible by two?

Two is a divisor of ϟѦ03495Ѧ349ϟ3Ѧ091ϟ3495159Ѧ because two is a divisor of Ѧ.

This still works because 2 divides twelve. So if the last digit is 0, 2, 4, 6, 8, or Ѧ, the number is divisible by 2.

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3

This is the first thing that is different. Is the last digit divisible by three?

Three is a divisor of 81Ѧϟ125ϟ7ϟ934589 because three is a divisor of 9.

This works because 3 divides twelve. So if the last digit is 0, 3, 6, or 9, the number is divisible by 3.

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4

Is the last digit divisible by four?

Four is a divisor of Ѧ0466208ϟ490Ѧ34Ѧ090 because four is a divisor of 0.

This works because 4 divides twelve. So if the last digit is 0, 4, or 8, the number is divisible by 4.
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5

This is the first hard one, it's sort of like it's somewhere between the rules for eleven and seven in base ten. You break up the digits into groups of two, add the lowest, subtract the next lowest, add the next, subtract the one after, and so on.
Of course, you can't do that when there are only two digits, but in that case there's another rule. Actually, there's more than one. Use which ever one is easier. You can either multiply the twelves digit by 2 and add the ones digit, or you can multiply the ones digit by -2 and add the twelves digit.

So take a number, say: ϟѦ0943530048Ѧ513, then break it up:
-ϟѦ+09-43+53-00+48-Ѧ5+13
then cry because you don't have a calculator that operates in base twelve which means that you have to work out the addition and subtraction yourself. Of course, if you did have a base twelve calculator handy, you could just use that to see if it's divisible by five without a division rule.

I may very well be getting all of this wrong and I'm not going to do it in one step, but I think that, combing terms pairwsie I get this:
-ϟ1+10+48-92= -Ѧ1-46 = -127

Then repeat:
+1-27= -26

Now we use one of the other rules:
We could say -2*2 + (-6) = -4 -6 = -Ѧ which is divisible by five.
Or we could say -2 + -6*(-2) = -2 +10 = Ѧ which is likewise divisible by five.

So, in theory at least, it is divisible by five. Is that actually true? Do I have to check? Ok, I'll check (says the idiot who typed in a random 15 digit number.)  My calculator says that that's equal to 15199987645464555 base ten, so yes, it is divisible by five. Lucky random number punching I guess, it has to happen one out of five times.

The first rule works because five divides 101 base twelve, meaning 100 base twelve is -1 mod five. So it's just like the sevens rule in base ten, but with one less digit.
The first of the other rules is because 10 = 2 (mod 5) base twelve. So we can replace every instance of 10 with 2, as in -26 = -2*10 -6= -2*2 -6 (mod 5).
The second is another case of this being like seven in base ten. You see 5 divides 21 base 12, so we can use the same trick we used with seven in base ten. -20 = 1 (mod 5) base twelve. So that means that we can take the -2*10 -6 from before and multiply it by -2 to get -2*(-20) – 6*(-2) = -2 – 6*(-2) (mod 5).
So basically it comes down to which digit you'd rather multiply by two. If things would be easier multiplying the twelves digit, multiply it by two, if it would be easier to multiply the ones digit, multiply it by negative two. Here, with the digits being 2 and 6, either way is easy, but maybe another time you don't remember what ϟ times two is, so if that shows up you'd rather use a different digit. Also, generally, the math will be simpler if you multiply the smaller number.

It's possible my brain may have exploded in the process of making this description.

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6

Is the last digit divisible by six?

6 is a divisor of Ѧ1ϟ2034931841ϟ04984203491040340 because six is a divisor of 0.

This works because 6 divides twelve. So if the last digit is 0 or 6, the number is divisible by 6.

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7

The first part of the rule is the same as in base ten, break things into groups of three, alternate adding and subtracting them.
The second part isn't all that far from what it was in base ten. Instead of subtracting twice the last digit from the rest, you add three times it to the rest.
In rare circumstances it might be useful to know that you can add the last digit to five times the rest as an alternate test.

ϟѦ0943530048Ѧ510
-ϟ+Ѧ09-435+300-48Ѧ+510 = 9ϟѦ-135+042 = 907
90+7*3 = 90+19 = Ѧ9  I have no idea if that is divisible by seven, so we repeat the second rule:
Ѧ+9*3 = Ѧ+23 = 103
10+3*3 = 10+9 = 19 we already discovered this is divisible by seven, but this feels like a good time to use the last rule.
1*5+9 = 12 That's just 7*2, but say we didn't know that. Either the second or the third rule would do.
For the second rule 1+2*3 = 1+6 = 7
For the third rule 1*5+2 = 5+2 = 7

I already checked that this was indeed divisible by seven (which is why it's so close to the number I used for 5.) So our rule has worked.

Ok, so why do these things work?
The first thing is because seven is a divisor of 1001 base twelve just like it's a divisor of 1001 base ten. So the same rule applies. The second thing is because 30 = 1 (mod 7) base twelve. We've made use of this sort of thing a lot by this point, so I'm thinking I don't need to explain it again. The third thing is because 10 = 5 (mod 7) base twelve. It's not going to be useful most of the time, but if you've got a two digit number and the twelves digit is a lot smaller than the ones digit, as in the case of the 19 we bumped into above, it might be useful.

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8

Are the last two digits divisible by 8?
Of course, we've got a question of how to tell. You take the ones digit and add it to four times the twelves digit.

If we're faced with 1Ѧ3963ϟѦ07Ѧ9071938451Ѧ49853403ϟ4, we only have to look at ϟ4.
How would one know whether ϟ4 is divisible by eight?
(Other than the fact I made sure it is, I mean.)
So we go to rule two, ϟ*4 + 4 = 38 + 4 = 40. If that's not obvious we repeat.
4*4+0 = 14. 1*4 + 4 = 8. So it's divisible by eight.

The first rule works because eight divides 100 (which is, I guess, pronounced, “one gross.”) It's the same reason that we only have to look at the last two digits when looking at 4 in base ten.
The second rule is because 10 = 4 (mod 8) base 12.

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9

Are the last two digits divisible by 9?
To deal with the two digit number we add the ones digit to 3 times the twelves.

So we look at 1Ѧ3963ϟѦ07Ѧ9071938451Ѧ49853403ϟ3 and ignore everything but the ϟ3.
ϟ*3 + 3 = 39 + 3 = 30, and now it's occurring to me that maybe things of the form ϟX are bad examples for this kind of rule.
Anyway, repeat to get 3*3 + 0 = 9. Done.

Lets do another example. Consider nine squared. That's 69 in base 12, but nothing about it jumps out at me as saying it's divisible by nine, so if we just randomly came across it we'd probably need to check instead of instantly knowing it was 9 squared and thus divisible by nine.
6*3 + 9 = 16 + 9 = 23.
2*3+3 = 9

This works for pretty much the exact same reasons as the eight rules. Nine divides a gross, so we only have to look at the last two digits. 10 = 3 (mod 9) base twelve, hence the multiplying the twelves digit by three.

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Ѧ (ten)

The lazy rule: check two and five.
The complex annoying rule?
This is where we flashback to the rules we used with twelve in base ten. Except worse.
We can add the last digit two twice what remains when we remove it. Or, if we don't want to multiply such a big number, we can do something really screwy with the number. Take the first digit, add to it twice the twelves, digit, four times the gross digit, negative two times the next digit, negative four times the next digit, two times the next digit, and so on.
The sequence is 2, 4, -2, -4, and repeat. Starting with the twelves digits and moving up the higher digits.

Do I have to do an example?

41ϟ4Ѧϟ459529784
4*4+1*2+ϟ*(-4)+4(-2)+Ѧ*4+ϟ*2+4*(-4)+5(-2)+9*4+5*2+2*(-4)+9*(-2)+7*4+8*2+4
Which is a massively intimidating problem. Hopefully it works out like this:
14+2–38–8+34+1Ѧ–14–Ѧ+30+Ѧ–8–16+24+14+4 = 5Ѧ
5*2+Ѧ = Ѧ+Ѧ = 18, though really we can see it's divisible by Ѧ at the Ѧ+Ѧ step.
1*2 + 8 = Ѧ and we are done.

That wasn't nearly as bad as I expected it to be (at least in part because my random numbers aren't as random as they could be, so there was some useful canceling.) It would have been better if I hadn't misread a sign and been forced to redo the addition and subtraction as a result.
Anyway, I'm thinking that it's not worse than the rule for 12 base ten.

The explanation is, entirely about twelve being 2 modulo ten. So twelve squared is 4, and twelve cubed is 8, but 8 = -2 (mod Ѧ) and I would much rather multiply something by two than eight. Also, having signs alternate can keep the sum from getting too high. In the above I added up 15 things, many of them two digit numbers, and the result was only a two digit number because six of those things were negative. Returning from the digression, twelve to the fourth power is 2*(-2) = -4 (mod Ѧ). To get twelve to the fifth power we multiply by two again, but remember what I said about not liking to multiply by eight when I can multiply by two? Well that goes for negative eight as well, and -8 = 2 (mod Ѧ).
So that's where the 2, 4, -2, -4, pattern comes from.

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ϟ (eleven)

Add up all the digits, is the result divisible by ϟ?

41ϟ4Ѧϟ459529789
4+1+ϟ+4+Ѧ+ϟ+4+5+9+5+2+9+7+8+9 = 83
8+3 = ϟ. So eleven does divide it.

This works for the exact same reason the rule for nine works in base ten. It will work for 10-1 regardless of what base you happen to be in. In any given base 10 = 1 (mod 10-1).

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10 (twelve)

Is the last digit zero?

41ϟ4ϟ459529780 is clearly divisible by 10.

Again, this is a rule that works for all bases. That's basically what being in base whatever means. 14514X = 14514*10 + X, and X is always less than 10, so the only way 10 divides the number is if X=0.

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11 (thirteen)

Take the ones digit, subtract the twelves digit, add the gross digit, subtract the next digit, and so on. Is the result divisible by 11?

41ϟ4Ѧϟ459529789
4-1+ϟ-4+Ѧ-ϟ+4-5+9-5+2-9+7-8+9 = 11
So it is divisible by 11 (thirteen.)

This is exactly the same as the rule for 11 base 10 and will be the same for 11 in any base. 10 = -1 (mod 11) in every base (that's what 11 means after all) so the rule will always work.

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So, to address the question of whether or not it's easier, here's what we've got:

In any base the rules for 0, 1, (10 – 1), 10, and (10 + 1) will be the same. (Though in base two 10-1=1 and so the rule for 10-1 is redundant.) That means that for any base from three to twelve more than a third of the fourteen numbers we looked at were going to follow those rules. They are neither easier nor more difficult.

In base ten, 2 and 5 have extremely easy rules, you just look at the last digit. (The same is true of 10 but we already covered that.)  In base twelve 2, 3, 4, and 6 fall into this extremely easy category. We've doubled the size of this category. This is an example of more divisors of the base making things easier since the reason these are easy is because they are divisors.

In base ten, 4 has you look at the last two digits. In base twelve the same rule applies to 8 and 9. The size of this category has likewise doubled. Again, more divisors makes things easier.  In this case it's because they're multiple of divisors.

All that's left are five, seven and ten.

Seven is a lot like how it was in base ten.  Five and ten are where things turn out harder, because five is no longer a factor of the base, and ten is no longer the base.  So the change isn't all making things easier.  Still, ten in base twelve seems to fall in the same category as things like six and twelve in base ten, so it doesn't seem to be making things harder so much as shifting where the difficulty lies.

The rule for five in base twelve doesn't have a good analog in base ten.  (Because 101 base ten is a prime number.)  It's easier than the seven rule in base ten and harder than 11 rule in any given base, but similar to both, and that's probably about as precise a comparison as we can make.

So overall it seems like things are quite a bit easier.  Several things have been made easier, other things haven't changed much, only two things have been made harder, and in the case of one of those it's arguable that the difficulty has been shifted onto it rather than created.

I have no idea what difference any of this would make in practice, though.

I do wonder about the importance of even and odd numbers. What if it weren't easy to figure them out? Consider, for example, any odd numbered base. To know that a number is even you'd have to add up all the digits and see if that was even. For a large number you'd probably have to repeat the process multiple times. If that were the case, would we still think in terms of even and odd?  Would we still have names for them? What if, instead of being able to tell whether something was even or odd at a glance, you could tell that it was divisible by three at a glance? Would that have changed our thinking?

In base 12, both would be things you could tell immediately upon seeing a number. Would that make a difference in the way one thinks?

Might we have a trichotomy of divisible by three, one more than divisible by three, and one less than divisible by three to go alongside the existing dichotomy of divisible by two and one more-or-less than divisible by two?  If we did, would it, in any way, matter?

Consider the fact that people actually say, “Two is the only even prime,” as if that's any more meaningful than saying, “X is the only prime divisible by X,” where X is any given prime.  Well yes, two is the only prime number divisible by two, and three is the only prime number divisible by three.

Anyway, there really was no point to this, and given that I kept slipping from base twelve to base ten and back again, there's a fair chance the math is wrong in places.

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* So, fractions. If I knew how to make a table this would be easier to communicate. Consider fractions of the form 1/X where X is a single digit number.

In base ten three such fractions terminate after a single digit. 1/2 = .5, 1/5 = .2, and 1/10 = .1
In base twelve five such fractions do. 1/2 = .6, 1/3 = .4, 1/4 = .3, 1/6 = .2, 1/10 = .1

I don't know about anyone else, but I find myself facing thirds more often than I'm facing fifths. (I have yet to meet a recipe that calls for a fifth of a cup, for example.) I definitely come across quarters more than fifths. So base twelve seems to be winning there.

In base ten there is one such fraction that terminates after two digits: 1/4 = .25
In base twelve two such fractions do: 1/8 = .16, 1/9 = .14

In base ten 1/3, 1/6, 1/7, 1/8, and 1/9 fall into neither of these categories.
In base twelve 1/5, 1/7, 1/Ѧ, and 1/ϟ fall into neither.

Which is to say, even though it has fewer numbers to deal with, base ten has more things that fail to be nice. Of course 1/8 base ten does terminate, after only three digits at that**.  So if we count that as being nice then they both have the same rate of failing to be nice, though base twelve still has more nice things.

Base ten has two things that repeat the same digit infinitely: 1/3 and 1/9.
Base twelve has but one, that being 1/ϟ.

Anyway, just looking at 1/2, 1/3, 1/4 you can see why someone might want to do math in base 12.  .4 is a much nicer thing to work with than .33333333333333and-so-on.  .3 is nicer than .25 as well.

** By now you've probably noticed a similarity between the number of digits we have to look at in a division rule for X and the number of digits in 1/X.  I'm not sure how obvious it is though.  It sort of makes me wonder if I had to list the one digit fractions and two digit fractions or if everyone already knew.