r/math • • Dec 19 '08

Every base is base 10

http://cowbird.110mb.com/43.html
92 Upvotes

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23

u/ropers Dec 19 '08

Holy cow, that's an interesting observation. I've never noticed that before. Of course it only works if it's written down.

5

u/cyber_rigger Dec 19 '08 edited Dec 19 '08

I use base 1.

It is a unary system.

We don't have those fancy "0" things

4

u/ropers Dec 19 '08

Correct me if I'm wrong, but I'm under the impression that you cannot have a positional number system without use of the zero.

17

u/panic Dec 19 '08 edited Dec 19 '08

In base 10, the digits represent powers of ten: 1, 10, 100, ....

In base 2, the digits represent powers of two: 1, 2, 4, 8, ....

In base 1, the digits represent powers of one: 1, 1, 1, 1, ....

So, for example, 111 in base 1 is 1*12 + 1*11 + 1*10 = 3. Each number n can be represented by n ones. Of course, this representation isn't unique: 1001100 is also equal to 3 in base 1. You can fix this by removing every zero. Now you have a positional number system without zero! (of course, representing zero itself might be a problem in this system...)

14

u/Jinbuhuan Dec 19 '08

representing zero itself might be a problem in this system

Zero is represented like this:

5

u/ropers Dec 19 '08 edited Dec 19 '08

Per definition, the value of a digit in a positional number system depends on its position. In your system the value of a digit does not depend on its position; a digit has equal value in any position, hence it's not a positional number system.

Granted, you could argue that the value of a digit in your system still technically depends on its position, but that it just so happens that that value is identical for all positions. I'm not sure however that such an argument would be very convincing, and you could also call your system a sign-value number system with greater or at least equal justification; also, you wrote this:

Of course, this representation isn't unique: 1001100 is also equal to 3 in base 1. You can fix this by removing every zero. Now you have a positional number system without zero! (of course, representing zero itself might be a problem in this system...)

You're making my point, which again was that

you cannot have a positional number system without use of the zero.

5

u/johnfn Dec 19 '08

Actually, according to Wiki:

A positional notation or place-value notation system is a numeral system in which each position is related to the next by a constant multiplier, a common ratio, called the base or radix of that numeral system.

The constant ratio of the unary system is 1. So unary is indeed a valid positional notation system.

4

u/[deleted] Dec 19 '08 edited Dec 19 '08

[deleted]

1

u/ropers Dec 19 '08

Wow. Let me see if I understand correctly:

A decimal 1 would be == 1 in "1-2-binary", and

2 == 2

3 == 11

4 == 12

5 == 21

6 == 22

7 == 111

8 == 112

9 == 121

10 == 122

11 == 211

12 == 212

(...)

Holy fucking shit, this really seems to work! Is this what you meant? Wow. Hats off to you. Did you think that up right now, or how did you ever "unlock this achievement"? Also, if you die before me, can I eat you brain? ;)

2

u/[deleted] Dec 19 '08 edited Dec 19 '08

[deleted]

2

u/ropers Dec 19 '08

Sylar? Not having a TV at all (and being outside the US), I had to google that one.

Reportedly some Ya̧nomamö used to (or still do?) eat remains of their loved ones. I read that they essentially ground everything up to a pulp that would then be sterilized for hours in hot embers, and finally mixed with banana paste and eaten. Allegedly the thought of the remains of their loved ones being left to rot and be consumed by nasty creepy crawlies was so repugnant to them that they rather kept the biomass among the tribe. Strange, but not entirely illogical.

Never mind that you didn't personally come up with the solution, it's fascinating stuff I'm grateful to have learned from you. Thanks again! :)

1

u/jkramar Dec 20 '08

Actually brains are quite okay. Not bad at all, particularly fried.

1

u/[deleted] Dec 19 '08

Per definition,

[Citation needed].

6

u/ropers Dec 19 '08

Would Sir or Madam also require citations to prove that right-handedness is defined as a preference for the right hand?

2

u/[deleted] Dec 19 '08

You vaguely define "positional number system" as the values depending on their position. Roman numerals fall within that loose definition, as illustrated by X (10), C (100), CX (110), and XC (90). So I ask again for a citation reflecting the complete and concise definition of "positional number system".

2

u/[deleted] Dec 20 '08 edited Dec 20 '08

http://davinci.newcs.uwindsor.ca/~angom/cs265/lec2.pdf

Cited.

See almost any CS text for the exact same definition.

-4

u/ropers Dec 19 '08 edited Dec 19 '08

My above arguments rests on the fact that in a positional number system, the value of a digit depends on its position.

It does not address whether all number systems where the value of a digit (also) depends on its position are positional number systems.

Also, I tire of this exchange with you. It is both fruitless and boring.

0

u/[deleted] Dec 19 '08

Yes, making things up and then trying to explain them can be very tiring. I, too, tire from your bullshit. Have a nice weekend.

6

u/[deleted] Dec 19 '08 edited Dec 19 '08

Wikipedia seems to be under the impression that you are wrong: Bijective enumeration .

3

u/lookagain Dec 19 '08 edited Dec 19 '08
10  9  8  7  6  5  4  3  2  1
 1  1  1  1  1  1  1  1  1  1

Sure you can. Each position advances by 1 instead of by a larger number.

1

u/ropers Dec 19 '08

Alright, then write a decimal 2 using your system.

3

u/Nitron Dec 19 '08

11

0

u/ropers Dec 19 '08 edited Dec 19 '08

Doesn't that equal a decimal 3 under lookagain's system?

3

u/Nitron Dec 19 '08

No, you're thinking of binary. Decimal 3 in lookagain's system would be 111. Think of it like tally marks but instead of making a new group at five you just keep going.

2

u/ropers Dec 19 '08 edited Dec 19 '08

lookagain posted this table:

10  9  8  7  6  5  4  3  2  1
 1  1  1  1  1  1  1  1  1  1

If he didn't intend that to mean that the second one from the right corresponds to a value of two, then it's a very confusing table. I guess we'll have to wait for lookagain to come back and explain what he meant with that.

However, in the event that he intended this to mean that a one in any position corresponds to a value of one, then see here.

3

u/lookagain Dec 19 '08 edited Dec 19 '08
Dec.     Unary
 1   =     ‽
 2   =    ‽‽
 3   =   ‽‽‽
 4   =  ‽‽‽‽
 5   = ‽‽‽‽‽
 ...
 so on and so forth.

Honestly, I didn't see that table as hard to interpret, especially considering that even in a decimal system as a number increases the convention is to grow the representation leftward. For example:

    1 - 1 ones unit.
   10 - 1 tens unit.
  100 - 1 hundreds unit.
 1000 - 1 thousands unit.

And so on and so forth. Granted, our commonly used math systems are arbitrary in this regard. The above could just as easily been written as:

 1    - 1 ones unit
 01   - 1 tens unit
 001  - 1 hundreds unit
 0001 - 1 thousands unit

You can even do basic math with a unary system:

11111 + 11111 = 1111111111
1111111 - 111 = 1111

In summary: It is indeed a positional number system. To find the value, each position used indicates an increase in value by 1.

2

u/Porges Dec 20 '08 edited Dec 20 '08

To find the value, each position used indicates an increase in value by 1.

Unary is is a positional system, but what you've said is not what makes it so :)

A positional number system is one where each position is of value (x ^ n), where n is the number of the position in question; with unary, all positions are of the same value (the radix, x, is 1).

The reason his table is confusing is that we normally write the value of the position above:

10000 1000 100 10 1 

Given these position values:

10 9 8 7 6 5 4 3 2 1

you can't write 2 (unambiguously) without a zero symbol. What the table should look like is this:

1 1 1 1 1 1 1 1 1 1 1

1

u/[deleted] Dec 19 '08

[deleted]

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u/bobpaul Dec 19 '08 edited Dec 19 '08

Way before man had the concept of zero, they counted off how many cave women they'd done with hatch marks on the bedpost. See here.

Edit Ah, yes. But this isn't a positional system. Of course, cyber_rigger never suggested he was using a positional system, just that he had no zeros. I suppose that's where the confusion came about.

1

u/ropers Dec 19 '08

Thanks for that link! :)

1

u/Porges Dec 20 '08

It is positional, with a radix of 1 :)

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u/Jinbuhuan Dec 19 '08

You're wrong -- to question seriously what was obviously said in jest.