The unusual notation ...999 = -1

[xouper]

The following change of topic was posted near the bottom of page eleven of another thread (http://www.randi.org/vbulletin/showthread.php?s=&threadid=29713&perpage=40&pagenumber=11) 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.

:D </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? :D

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...


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

[xouper]

patoco12: More fun...
http://www.csds.uidaho.edu/~poconnell/math2.JPGHey, don't just tease us with some simple "two's complement" notation, tell us more. :D

BTW, when are you gonna put up a photo (http://www.mrc.uidaho.edu/mrc/team/printPeople.php?ID=219)?

[patoco12]

Originally posted by xouper
Hey, don't just tease us with some simple "two's complement" notation, tell us more. :D


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

[Trollbane]

[QUOTE]Originally posted by xouper [B]
<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.

:D </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]

Originally posted by Trollbane
[QUOTE]Originally posted by xouper [B]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]

Originally posted by chulbert

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]

Originally posted by 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.

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
= 0Excellent. :D

[Martin]

Originally posted by xouper
Excellent. :D 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]

Originally posted by xouper
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]

Originally posted by Trollbane
(...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]

Originally posted by Trollbane


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
_______
...0009which is the same as 9.

[Martin]

Originally posted by xouper
Exactly. "If it walks like a duck, smells like a duck, ... " then it might just be a duck. No doubt someone will be along presently to insist that it's really a turkey.

[Trollbane]

Originally posted by patoco12


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

A terminating sequence of 9s that has infinite lenght is equally logical to a square circle. If it terminates at any point it is not infinite. If the number is infinite it is pointless to say it has any value and any calculation made with that number can have any value you want it to have.

If you quantisize infinity it no longer is infinity, since infinity is the largest possible number and for a quantisized number there always is a n+1.

/edited a bit for spelling

[bjornart]

Originally posted by Trollbane


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...


What's better with that? Both of them are infinity. The good thing about the first one is that numbers are best 'read' from the desimal point outwards, so the nines there can be read as 9, 90, and 900. While the nines in 999... are infinity, infinity and infinity.

[patoco12]

Originally posted by Trollbane


A terminating sequence of 9s that has infinite lenght is equally logical to a square circle. If it terminates at any point it is not infinite.


When I said it terminates, I meant it terminats at the decimal point but has NO starting point. I could have easily said that it starts at the decimal point and has no termination point. Sorry about my confusing terminology.



If the number is infinite it is pointless to say it has any value and any calculation made with that number can have any value you want it to have.

If you quantisize infinity it no longer is infinity, since infinity is the largest possible number and for a quantisized number there always is a n+1.



You are confusing the value of the number with the length of the number. This caused 400+ posts in the .9~=1 forum.

I REALLY don't want to start another argument over infinties and such; that would be missing the point of this forum topic.

[Trollbane]

You cant use a logical system, like math, to evaluate things from outside of its scope. By the way this is going we might as well start discussing what came first the chicken or the egg.

Alternatively the same proof could be used to prove that:

x=..1111
10x=..1110
x-10x=1
...1111=-1/9

etc..

But it is a limitation of math, but rather a false dilemma, since you cant take the start nor the end of an infinity, since it has neither. This is going a bit on the philosophy side though.

But the key point is that if any chain ends or starts at any point it cannot be infinite.

/edited to clarify a bit and to add

Now the bloody Matrix tagline- Everything that has a beginning has and end- is in my head... I really need a drink.

[bjornart]

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

Maybe because I can hardly remember what 'congruences' mean. :)

Or maybe because, like Trollbane pointed out, you can use that same 'logic' to prove that:

x = ...777

10x = ...7770

x-10x = 7

-9x = 7

x = -7/9

Which made sense in the .999... debate, but which, at least to me, doesn't make sense when it comes to ...999.



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.

Just out of curiousity, and to avoid reading a lot about p-adics. How does ...777 = -7/9 fit into that system?

[LW]

Originally posted by Trollbane
You cant use a logical system, like math, to evaluate things from outside of its scope.

But, as xouper wrote, you can extend the system to handle things that you can't handle in the original.

Let's consider the set of natural numbers for while, that is, the numbers 0, 1, 2, ...

When you consider natural numbers, the expression 2 - 3 doesn't make any sense. If I have two apples I can't give you three of them.

But the expression starts to make sense when we extend the natural numbers to integers and we get 2 - 3 = -1.

This p-adic system is a way to extend rational numbers. It is not the conventional way and they are not real numbers that are used in the mathematical or everyday sense. So you don't have to expect them to behave like real numbers do.

But it is a limitation of math, but rather a false dilemma, since you cant take the start nor the end of an infinity, since it has neither. This is going a bit on the philosophy side though.

What we have here is a sequence of digits that is of infinite lenght. Such a sequence can have a start point. For example, the ordered sequence of all integers is infinite, but it has start, namely zero.

As an example of an infinite sequence that has an end but not beginning, consider the set of negative integers: ..., -3, -2, -1. This sequence ends in -1 but doesn't have a beginning.

[slimshady2357]

Posted by Trollbane...since you cant take the start nor the end of an infinity, since it has neither. This is going a bit on the philosophy side though.

I hope you can see how wrong that is now.

Think of a rope of infinite length, but you have one end in your hand. :D

Or with the Reals, the interval from 0 to 1 has a start AND an end point and yet contains an infinity of numbers.

I'm interested in what xouper thinks of Bjorn's ...777 = -7/9 result. Maybe I should just go read the link xouper provided! :p :D

Adam

[xouper]

Trollbane: You cant use a logical system, like math, to evaluate things from outside of its scope.Sometimes it's amusing when someone who doesn't seem to have a firm grasp of mathematics tries to tell mathematicians what they can or cannot do. And sometimes it's simply boring.

[Trollbane]

Originally posted by xouper
Sometimes it's amusing when someone who doesn't seem to have a firm grasp of mathematics tries to tell mathematicians what they can or cannot do. And sometimes it's simply boring.

Thank you for telling me how firm of a grasp I have at maths.. I just suck at the stupid theoretics/mental masturbation part. Come to think of it I might just use this proof on the pipeline project.. Come to think of it in thermodynamics calculations ...999=-1 is almost an acceptable margin of error :)

[xouper]

Trollbane: Thank you for telling me how firm of a grasp I have at maths.. I just suck at the stupid theoretics/mental masturbation part.You're welcome. And thank you for admitting you have no business criticizing those who are good at the stupid theoretics/mental masturbation part.

Edited to add - If it wasn't for the people who are good at the stupid theoretics/mental masturbation part, mathematics would be nothing more than an alchemy of numbers.