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How Did Anyone Do Math in Roman Numerals?

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Re: How Did Anyone Do Math in Roman Numerals?

#111
Off topic rant:

It bugs me when people mix our standard numerals with Roman numerals, such as 12MM to mean twelve million. They are different numerals and the meaning is not defined when they are used together.

And Roman numerals are not like SI suffixes, meaning they are not multiplicative; Roman numerals are additive, so MM is two thousand, not one million. Also, M is an SI suffix, so 12M means twelve million and 12MM just looks like a typo.

Obviously people do not use SI suffixes may not feel the same way, this is just my pet peeve because I use SI suffixes in science.

Re: How Did Anyone Do Math in Roman Numerals?

#112
post #96

Earlier quoted context omitted.

Yes, that's what I said. But you chose a simple case. Try subtracting more complex numbers, specifically something more complex for the subtrahend. E,g, 42 - 13, which is XLII - XIII. Not only to you have to convert, but you have to convert both sides, and to either a similar format that can be subtracted, or to a lowest common format as suggested above (which is not roman numerals, BTW, so you need to know two syste…

For many cases one can ignore division, but yes, as the article said that generally required an abacus. Multiplication doesn't in general require an abacus if you're trained in roman numerals. Let's take your "more complicated numbers", 42 * 13. Expand XLII * XIII = X * XLII + XLII + XLII + XLII, the first requires you to know a times table to see that it's CDXX, the rest you'd add mentally, LLL - XXX = CXX, so it's…

I think we can all agree addition is fairly easy. Multiplication, since it can be explained in terms of addition, it also not really that hard. Subtraction and division are harder, as I originally noted, and by extension division, are harder, as there is not one simple rule for conversion before subtraction that won't require additional conversions unless you go to the simplest form, which is unwieldy (along string of I's).

From my example, which was meant for subtraction, not addition or multiplications, we get the following:

XLII - XIII

so we convert to an optimized form for arithmetic (which is, as I've noted a second format you need to know and keep track of, since it's not valid Roman numeral format

XXXXII - XIII

then we note that we have to borrow from the X "place", so let's do that

XXXVIIIIIII - XIII

now we can actually subtract

(XXXX - X) + V + (IIIIIII - III) = XXX + V + IIII = XXXVIIII

Now we have to convert back to valid Roman numerals

XXXVIV = XXXIX

This is using common current arithmetic techniques. I'm not sure whether this is the method they would have employed, or whether it would have been a different method that could have been easier or harder depending on how it combines with the attributes of this number system. Important things to note about the method shown are a pre and post conversion step (or steps, if you don't get every sub-conversion right initially or don't go to a lowest common format), and then the arithmetic itself requires more borrows.

Re: How Did Anyone Do Math in Roman Numerals?

#113

Earlier quoted context omitted.

The equivalent to II + II = IIII isn't 2 + 2 = 4, it's 2 + 2 = 22. Simply jamming the symbols together gives you the right answer.

Right, much like I + V is IV. (There is more help; it's not as simple as people have been portraying, though...)

V + I = VI though.

You just have to put the larger number first for the addition.

Re: How Did Anyone Do Math in Roman Numerals?

#114

I always thought that roman numerals would be a simpler way to do basic arithmetic and might lend itself more to simple commerce. For example: III represents 3 things, so III + II = IIIII For simple commerce application that is simpler, I just have to then remember that IIIII = V, and VV = X and XXXXX = L, LL = C. Armed with just those simple rules I could probably get by in the market place in Rome. With Arabic numb…

>1 is one thing, 2 represents 2 things, 3 represents 3 things and so on

I follow you thus far.

>Then I have to remember that 2 + 2 = 4, and 3 + 2 = 5

No you don't.

You have to remember that 1 + 1 = 2; 2 + 1 = 3; ...; 9 + 1 = 10; and then the rules repeat themselves, respecting columns for addition. All mathematics between 1 and 10 like 4 + 5 are already known at this point.

Roman numerals, on the other hand, give you no easily repeatable pattern to follow as the order of magnitude increases. I + I = II, III + II = V, V + V = X, X + X + X + X + X = L, LL = C, and I only got this far because of what you said. What comes next? C * 5 = x_1, x_1 * 2 = x_2, x_2 * 5 = x_3, ..., x_n * 5 = x_n+1, x_n+1 * 2 = x_n+3

The complexity is unbounded. Sure, if you constrain yourself to "getting by in the market place in Rome", that's one thing, but even then I would imagine arguments around arithmetic could go either way.

Re: How Did Anyone Do Math in Roman Numerals?

#115
post #24

Since we have a lot of math experts here I thought I'd ask a question I was always wondering about: Is there an inherent advantage or disadvantage to using the decimal system as we do? Somehow I think octal or hexadecimal would be easier but I am not sure.

I was about to post that in my opinion base-12 is superior to base-10. But someone beat me to it. In a six-fi novel I'm writing, an advanced alien civilisation uses base-12. As to your question specifically regarding base-16 instead of base-12, it depends. Decimal itself is just a bizarre choice, most likely due to humans having literally ten digits. In decimal we can represent exact fractions of 1/2, 1/5, and 1/10 (…

> In a six-fi novel I'm writing, an advanced alien civilisation uses base-12

Six-fi? Typo or genre I'm not aware of?

Re: How Did Anyone Do Math in Roman Numerals?

#116
Some aspects of 'math' came along with arabic, but Roman Numerals work fine for their situation, at that time.

In conversation, it's just the I-M/0-1000 range for instance, learning how to notate that number is a different issue.

Notation is somewhat straight-forward: When you get to 4, the number undergoes a state change to 'subtract from the next number, dropping everything behind it, and we're in a new state so "the" number is changed to 5' so III becomes IV. Otherwise, just keep adding 1. That should work, adding to the previous rules as the state changes. Going to X leaves the V rules in place.

So it would be a XXXII-bit computer, for example.

Re: How Did Anyone Do Math in Roman Numerals?

#118
post #115

Earlier quoted context omitted.

I was about to post that in my opinion base-12 is superior to base-10. But someone beat me to it. In a six-fi novel I'm writing, an advanced alien civilisation uses base-12. As to your question specifically regarding base-16 instead of base-12, it depends. Decimal itself is just a bizarre choice, most likely due to humans having literally ten digits. In decimal we can represent exact fractions of 1/2, 1/5, and 1/10 (…

> In a six-fi novel I'm writing, an advanced alien civilisation uses base-12 Six-fi? Typo or genre I'm not aware of?

Haha, yeah, sci-fi.

Luckily I'm not yet at the copy-editing stage ;-)

Re: How Did Anyone Do Math in Roman Numerals?

#119
post #86

Earlier quoted context omitted.

Curious, why do you have to remember that 2 + 2 = 4 & 3 + 2 = 5? Once you know the values the symbols represent, at that point isn't it similar in simplicity to roman numerals? II + II = IIII 2 + 2 = 4 Don't see how the latter problem lends itself to any more memorization beyond symbols

Because if you don't memorize your 'primitive' algebraic rules you'll end up just doing a pullback with roman numerals in the middle. Integers don't have any values, they're mathematical objects with certain properties. Asking about 'the value of 5' doesn't make sense unless you're trying to convert to another, already known, number system. What is 2 + 5? Well 2 is II and 5 is V which is IIIII. So then we have IIIIII…

I just want to point out that it's humorously appropriate for a user named "Spivak" to mention pullbacks.

Re: How Did Anyone Do Math in Roman Numerals?

#120

Earlier quoted context omitted.

The equivalent to II + II = IIII isn't 2 + 2 = 4, it's 2 + 2 = 22. Simply jamming the symbols together gives you the right answer.

Right, much like I + V is IV. (There is more help; it's not as simple as people have been portraying, though...)

You just need to add a sort to to get VI from that -- something explained in the article regarding multiplication.
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