The unusual notation ...999 = -1

[xouper]

The following change of topic was posted near the bottom of page eleven of another thread and I feel deserves a thread of its own. For context, the relevant posts are repeated here. I posted a link to this thread in that other thread and I'm hoping we can continue the conversation here instead of there.

<div style="border: solid black 1px; margin: 0 32px 0 0; padding: 24px; background-color: #fff8f0;">Originally posted by 69dodge

Let x = . . .9999.
10x = . . .9990.
x - 10x = 9.
-9x = 9.
x = -1.

So, . . .9999 = -1.

</div>

Followed by:

<div style="border: solid black 1px; margin: 0 32px 0 0; padding: 24px; background-color: #fff8f0;">Originally posted by xouper

Interestingly, that very thing was published in The College Mathematics Journal, Volume 26, Number 1, January 1995, page 11-15, "The Repeating Integer Paradox," by Paul Fjelstad.

Although that notation does not represent traditional integers, there is something consistent about ...999 = -1. This can be shown using the notion of congruences.

Also, consider two's complement arithmetic as used by computers. In a 16 bit word, FFFF = -1. If the wordsize is infinite then ...FFFF is still equal to -1.

Two's complement can also be written in decimal. For example, in an 4 digit representation, 9999 = -1. It doesn't take much imagination to see that the limit condition is ...999 = -1.

The numbers represented by the notation ...PPP are called p-adics. I don't really know much about them, except that p-adics are not real numbers.

http://mathworld.wolfram.com/p-adicNumber.html</div>

Followed by:

<div style="border: solid black 1px; margin: 0 32px 0 0; padding: 24px; background-color: #fff8f0;">Originally posted by clk

S = 1 + 2 + 4 + 8 + 16 + 32 + 64 +...
so
2S = 2 + 4 + 8 + 16 + 32 + 64 +...
so
2S = S-1
solve for S to get:
S = -1
so...
1 + 2 + 4 + 8 + 16 + 32 + 64 +... = -1 !
How do you like them apples?

Obviously, this problem can be easily solved correctly by making S equal to infinity, which is the proper sum. Then, subtracting 1 from infinity still gives you infinity, so in the end, you get infinity equals infinity, which is a true statement.</div>

And now I can post my reply to clk:

I assume everyone recognizes that your notation is the decimal equivalent of the following binary notation

S = ...111
10S = ...1110
S-10S = 1
S = -1

So in binary ...111 = -1

Which is consistent with what I posted previously about p-adics. I assume the computer math gurus among us will immediately see the beauty of this equality.

For any integer base B greater than 1, and X=B-1, then ...XXX = -1. Keeping in mind that we are not talking about real numbers here, but rather an unusual extention to the integers.

 

[patoco12]

More fun...

 

[xouper]

patoco12: More fun...
http://www.csds.uidaho.edu/~poconnell/math2.JPG

Hey, don't just tease us with some simple "two's complement" notation, tell us more.

BTW, when are you gonna put up a photo?

 

[patoco12]

Hey, don't just tease us with some simple "two's complement" notation, tell us more.

It's just a good way to convince people that two's complement wasn't just invented out of thin air.

 

[Trollbane]

Let x = . . .9999.
10x = . . .9990.
x - 10x = 9.
-9x = 9.
x = -1.

So, . . .9999 = -1.

</div>

Too easy:

x-10x is not 9. Its something like -8999.....9991.

 

[Wile E. Coyote]

Bad math hurt. Stop bad math! Stop!

With this kind of algebra it would be simple to prove that 2 + 2 = 5, for sufficiently large values of 2.

 

[chulbert]

Re: Re: The unusual notation ...999 = -1

Originally posted by xouper x-10x is not 9. Its something like -8999.....9991.

I think the thing you're missing is that:

(10)(...999) < (...999)

edit - fixed me engrish

 

[Trollbane]

Re: Re: Re: The unusual notation ...999 = -1

I think the thing you're missing is that:

(10)(...999) < (...999)

edit - fixed me engrish

Ummm.. Nope... 10 times any number is ten times larger than the constant (...999) in this case, unless the number is zero that is. I think you are mixing up a standard multiplying and powers here...

10*(...999)=(...999)+(...999)+ etc. > (...999) while

(...999)^10=(...999)(...999) etc < (..999) If (...999)<1

/edited to add.. The 10*c can be smaller ofcourse, but only if c<0

 

[bjornart]

You're both missing that ...999 is actually the same as infinity.
...999 = ...888 = ...777 and so on.

The exampel ...999 = -1 is fun because it appears to have something in common with two's complement in binary, when in fact it doesn't. Well, it sort of does, by treating ...999 as something else than infinity.

 

[Trollbane]

You're both missing that ...999 is actually the same as infinity.
...999 = ...888 = ...777 and so on.

The exampel ...999 = -1 is fun because it appears to have something in common with two's complement in binary, when in fact it doesn't. Well, it sort of does, by treating ...999 as something else than infinity.

I disagree here.

...999 wouldnt be infinity since by definition infinity cannot end and thats what that string does . If it was infinity a better way to state it would be:

999...

If it is considered as infinity thought the problem becomes quite weird in the statement:

infinity-10*infinity=infinity

and

infinity/9=infinity

so in the end we get a useless result:

infinity=infinity

 

[Martin]

Hrmm...

...999.999...

=...999 + 0.999...

= -1 + 1

= 0

 

[xouper]

Trollbane: Too easy:
x-10x is not 9. Its something like -8999.....9991.

What rules of addition are you using here? Please show all your steps.

 

[xouper]

Martin: Hrmm...
...999.999...
=...999 + 0.999...
= -1 + 1
= 0

Excellent.

 

[Martin]

Excellent.

It seemed like a bit of a mindf*ck when I posted it, but now that I think on it some more it's intuitively obvious. Just try multiplying by 10. Or anything else, for that matter.

 

[xouper]

bjornart: You're both missing that ...999 is actually the same as infinity.
...999 = ...888 = ...777 and so on.

I think the mathematicians who play with these kinds of "numbers" are quite mindful of that. Nonetheless, the mathematics of p-adics is internally consistent. Check out the link I posted previously to Mathworld on this topic if you want to see some of the formal math behind these "numbers". Working with p-adics in a non-prime number base (such as decimal) may be a bit more problematic, however.

The exampel ...999 = -1 is fun because it appears to have something in common with two's complement in binary, when in fact it doesn't.

Why do you say that? The congruences modulo to the number base are the same.

Consider the following infinite sequence (ordered list) of integers:

S = { 9, 99, 999, 9999, ... }

We can create another sequence by taking the congruence mod 10ⁿ of each element in the sequence:

<nobr>T = { 9 (mod10¹), 99 (mod10²), 999 (mod10³), 9999 (mod10⁴), ... }</nobr>

We can rewrite this sequence T = {-1, -1, -1, -1, ... }

The limit of the sequence T is thus -1, since no matter how far down the ordered list you go, the result does not diverge from -1.

The limit of the sequence S is ...999 with an infinite number of 9s to the left, and since the limit of the congruences of the sequence is -1, it is safe to say that ...999 is congruent to -1.

For notational simplicity, this is sometimes shortened to the notation ...999 = -1 when talking about p-adics. Since it is reasonable to invent a new number system by extending the integers to include infinite "numbers" like ...999, it is also reasonable to extend the definition of the equal sign in that number system.

In two's complement arithmetic, we ignore the overflow. With p-adics, we do the same thing. For example, when we add 1 to FFFF the result would normally be 10000, but if the wordsize is only 16 bits, we ignore the overflow to the 17th bit and the result is really the congruence mod 2¹⁶.

1 + FFFF <font face="symbol">º</font> 0 (mod 16⁴)

1 + 9999 <font face="symbol">º</font> 0 (mod 10⁴)

Seems to me there is a lot in common here, so maybe I missed your point?

Well, it sort of does, by treating ...999 as something else than infinity.

Exactly. By not stopping at the obvious notion that ...999 is not a real number and has no finite value in the traditiona sense, we can still ask questions about the nature of those kinds of "numbers". The results may seem counter-intuitive, which is typical when dealing with infinity, but p-adics are a legitimate and consistent number system, just not what we're used to.

 

[xouper]

Martin: It seemed like a bit of a mindf*ck when I posted it, but now that I think on it some more it's intuitively obvious. Just try multiplying by 10. Or anything else, for that matter.

Exactly. "If it walks like a duck, smells like a duck, ... " then it might just be a duck.

 

[Trollbane]

What rules of addition are you using here? Please show all your steps.

If we have a terminating sequence of 9s in a row that has a lenght of n. Multiplying the chain would result in a chain thats lenght is n+1 and ends with 0. Thus looking at the last digits would yield:

(...9999) for the normal chain

(...99990) for the multiplied, not (...9990)

Substract and you get (-899...991)

 

[Martin]

(...99990) for the multiplied, not (...9990)

The two are identical. In each, the ellipsis represents an infinite string of nines. The fact that you choose to pull out four instead of three makes no difference. ...999 = ...9999 = ...99999 etc etc.

 

[patoco12]

If we have a terminating sequence of 9s in a row that has a lenght of n. Multiplying the chain would result in a chain thats lenght is n+1 and ends with 0. Thus looking at the last digits would yield:

We have a terminating sequence of 9s, but the length is infinity.

 

[xouper]

Trollbane:
(...99990) for the multiplied, not (...9990)

Substract and you get (-899...991)

You didn't explain what you did between those two steps. That is the missing piece I was asking about. Also, as Martin already mentioned, ...99990 is the same as ...990.

For example, when I subtract ...9990 from ...999 I get

   ...999
- ...9990
  _______
  ...0009

which is the same as 9.