A very interesting Mathematical Paradox

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xkcdfan
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Re: A very interesting Mathematical Paradox

Postby xkcdfan » Wed Dec 08, 2010 1:51 am UTC

Turtlewing wrote:1+1=2 is part of the defenition of addition (when applied to most common number systems). Asking to prove 1+1=2 without using the deffenition of 2 is asking for something like:

let there be a thing
let there be another thing
by the defenition of addition there are now more than 1 things, and less than 3 things. let's call this case "there are 2 things".
(prety stupid and useless, but to drive home the stupid and useless I made it seems even less formal and well thought out in the original post)

Asking for a proof of 1+1=2 without using the defenition of _addition_ on the other hand is actually "prove addition for the natural numbers". which is still pretty pointless since the proof of addition is basicly just a restatment of the defenition of counting, and counting is generally held to be a fundamental postulate that is well verified by observation. So the proof will likley involve a practical demonstration of counting, but at least this question is driving that something that is worthwile to do every now and then (questioning your assumptions).

Oh my god, why can't you spell? Get Firefox, ffs, it has a built-in spellchecker.

Also, there is a formal mathematical proof that 1+1=2. If you seriously think it's as inane as you've made it out to be, try Google. It's your friend.

Turtlewing
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Re: A very interesting Mathematical Paradox

Postby Turtlewing » Wed Dec 08, 2010 8:01 pm UTC

Sure you can prove 1+1=2.
here's one such example (emphasis mine):

P1. 1 is in N.
P2. If x is in N, then its "successor" x' is in N.
P3. There is no x such that x' = 1.
P4. If x isn't 1, then there is a y in N such that y' = x.
P5. If S is a subset of N, 1 is in S, and the implication
(x in S => x' in S) holds, then S = N.

Then you have to define addition recursively:
Def: Let a and b be in N. If b = 1, then define a + b = a'
(using P1 and P2). If b isn't 1, then let c' = b, with c in N
(using P4), and define a + b = (a + c)'.

Then you have to define 2:
Def: 2 = 1'


2 is in N by P1, P2, and the definition of 2.

Theorem: 1 + 1 = 2

Proof: Use the first part of the definition of + with a = b = 1.
Then 1 + 1 = 1' = 2 Q.E.D.

- Doctor Rob, The Math Forum


As you will notice, the proof requires a definition of "2", and a definition of "addition".

For clarity, here's the actuall proof with all the definitions removed for simplicity:
Theorem: 1 + 1 = 2

Proof: Use the first part of the definition of + with a = b = 1.
Then 1 + 1 = 1' = 2 Q.E.D.


Not a whole lot different from:
From the definition of addition:
1+1=2, QED

Is it?

[edit]
As to spelling, I can spell, with effort, if I so choose. I simply choose not to. I find generally speaking the benefit of correct spelling is not worth the effort required to fix spelling as bad as mine when there are no consequences for being ignored, also I find the reactions from those who attempt to correct my spelling and grammar amusing and revealing.

Besides, If people like me actually fixed all our spelling and grammar mistakes, what would Grammar/Spelling Nazis do for fun?
[/edit]

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xkcdfan
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Re: A very interesting Mathematical Paradox

Postby xkcdfan » Fri Dec 10, 2010 10:53 am UTC

You figured out how to use Google, which is a plus, but you took the first result you found, which is from a website that answers math questions from elementary schoolers. Try digging a little deeper. You can prove 1+1=2 without recursion.

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Re: A very interesting Mathematical Paradox

Postby lightvector » Fri Dec 10, 2010 11:26 pm UTC

xkcdfan wrote:You figured out how to use Google, which is a plus, but you took the first result you found, which is from a website that answers math questions from elementary schoolers. Try digging a little deeper. You can prove 1+1=2 without recursion.


I really don't see what the point of this is, since there's nothing "deeper". By this, I mean that any proof that 1+1=2 is not so much a logical deduction of some non-obvious fact, as much as it is a confirmation that whatever underlying axiomatic system we've chosen to model the natural numbers does indeed formalize our intuitive concept of the natural numbers. The way in which you prove 1+1=2, and whether the proof is completely trivial or whether it requires dozens of lines, is completely dependent on what axiomatic system, out of the many possible, that you decide to use.

It's like one standard definition of "3" in axiomatic set theory: 3 = {{},{{}},{{},{{}}}}. This definition doesn't reveal any deep properties of "3" or shed any light on what the number "3" is. It merely shows (when combined with additional definitions of basic arithmetic operations, etc.) that set theory is capable of encoding our intuitive idea of the counting numbers, and is otherwise a meaningless formality.

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Re: A very interesting Mathematical Paradox

Postby Arariel » Sat Dec 11, 2010 2:14 am UTC

It's only a "paradox" because it's in base 10.

Everything's better in base 9. We should totally switch to that.

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Re: A very interesting Mathematical Paradox

Postby phlip » Sat Dec 11, 2010 3:39 am UTC

Arariel wrote:It's only a "paradox" because it's in base 10.

Everything's better in base 9. We should totally switch to that.

0.888...9 = 1

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Re: A very interesting Mathematical Paradox

Postby Arariel » Sat Dec 11, 2010 4:11 am UTC

phlip wrote:
Arariel wrote:It's only a "paradox" because it's in base 10.

Everything's better in base 9. We should totally switch to that.

0.888...9 = 1

Damnit, is that eight eigths?

But I have a solution:
Base all-the-primes-multiplied-with-each-other.

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Re: A very interesting Mathematical Paradox

Postby t1mm01994 » Sat Dec 11, 2010 10:44 am UTC

So, your plan is basically to have a different single character for every number? Good luck finding enough symbols, and keeping them apart from eachother... Also decimals just fail as the first digit behind the integers is the amounts of 1/infinity, which is poorly defined.. Not gonna work bro ;)

Arariel
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Re: A very interesting Mathematical Paradox

Postby Arariel » Sat Dec 11, 2010 2:01 pm UTC

Decimal implies base 10.
Base all-primes would be "infinitesimal". :D

And yes, I was joking. :p

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Re: A very interesting Mathematical Paradox

Postby charonme » Sat Dec 11, 2010 4:06 pm UTC

what? this is still going on? what are the elementary school teachers doing?

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Re: A very interesting Mathematical Paradox

Postby Turtlewing » Mon Dec 13, 2010 7:38 pm UTC

xkcdfan wrote:Try digging a little deeper. You can prove 1+1=2 without recursion.


That pretty much misses the point. Why should I prove that 1+1=2 without using recusrion when the proof with recursion is so much easier, adiquately translatable into the silly non-formal language I wanted to use, and recursion was not disalowed by the challange?

Anyway, I've never seen a non-recursive (or inductive) proof of addition and I'm not sure ho to go about creating one, so I'm a bit skeptical to the existance of a valid one given the nature of the operation, but you may demonstrate your claim that one exists at your convinience (or not, since it's really not that important in the grand scheme of things and the (non)existence of such a proof really doesn't have much baring on the validity of the recursive deffenition and proofs based on it).

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xkcdfan
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Re: A very interesting Mathematical Paradox

Postby xkcdfan » Wed Dec 15, 2010 3:11 am UTC

Turtlewing wrote:
xkcdfan wrote:Try digging a little deeper. You can prove 1+1=2 without recursion.


That pretty much misses the point. Why should I prove that 1+1=2 without using recusrion when the proof with recursion is so much easier, adiquately translatable into the silly non-formal language I wanted to use, and recursion was not disalowed by the challange?

t1mm01994 wrote:
Macbi wrote:
xkcdfan wrote:
lightvector wrote:Then, we can easily have infinite sequences, followed by values that come after that, followed by more infinite sequences, and more values afterwards, and so on. For instance, if we want an infinite number of zeros, followed by a 1, this is just the sequence where all terms are 0 except the [imath]\omega[/imath]th term, which is 1. Of course, it might be hard to make the arithmetic have nice properties, but it *can* be done sensibly, if we are willing to step outside the real numbers.

No, that actually doesn't make sense.

Prove it.

Only if you prove, without using the definition of 2, that 1+1=2.

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Re: A very interesting Mathematical Paradox

Postby Loki-Pi » Thu Dec 23, 2010 10:04 pm UTC

Its fairly simple actually
0.9999(recurring) = 1 - 0.000(recurring)1
0.000(recurring)1 is infinite zeros followed by a 1, which is zero because they go on forever before hitting the 1
Hence 0.999(recurring) = 1
QED

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Re: A very interesting Mathematical Paradox

Postby JohnLeaf » Tue Dec 28, 2010 6:29 am UTC

Wait, this doesn't make sense. If you times x=0.9999... By ten (0.9999...X10), then you have, no matter how many nines, whether it's two, or infinity, 9.9999...9 minus 0.000...9 (9.9999...9-0.000...9).

Which means that if you minus it by x, then you get a number that equals 8.(?), not 9. That will leave you with a number, that when divided by 9, would equal the original, o.9999... .

This is the same as the fact that, no matter how far you go with 0.333..., when you times it by 3, it will NEVER equal one. It will equal a number infinitely close to 1, but it will not equal one.

0.999...*10=9.999...-0.000...9

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Re: A very interesting Mathematical Paradox

Postby douglasm » Wed Dec 29, 2010 6:33 am UTC

JohnLeaf wrote:This is the same as the fact that, no matter how far you go with 0.333..., when you times it by 3, it will NEVER equal one. It will equal a number infinitely close to 1, but it will not equal one.

So what, pray tell, is 1 / 3 * 3? Take 1 and divide by 3. Multiply the result of that by 3. What do you get, and does your choice of how to write down the intermediate result really change the final product? Do you seriously contend that one third time three does not actually equal one just because you choose to represent "one third" in decimal form?

Also, may I refer to you the various repetitions in this thread about putting a final digit after the non-existent end of an infinite sequence of digits and how much nonsense that idea is?

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Re: A very interesting Mathematical Paradox

Postby SANTARII » Wed Dec 29, 2010 2:51 pm UTC

JohnLeaf wrote:Wait, this doesn't make sense. If you times x=0.9999... By ten (0.9999...X10), then you have, no matter how many nines, whether it's two, or infinity, 9.9999...9 minus 0.000...9 (9.9999...9-0.000...9).

Which means that if you minus it by x, then you get a number that equals 8.(?), not 9. That will leave you with a number, that when divided by 9, would equal the original, o.9999... .

This is the same as the fact that, no matter how far you go with 0.333..., when you times it by 3, it will NEVER equal one. It will equal a number infinitely close to 1, but it will not equal one.

0.999...*10=9.999...-0.000...9

Anything infinitely close to something is equal to it. If it were not equal to it by any amount, it could be closer, and so not infinitely close to it.

Apart from the fact that you can't have something following an infinitely recurring pattern (otherwise it isn't infinitely recurring), let's say you theoretically have a 9 at the infinite place.
If you have a number, X, at N places past the decimal point, in denary, then that number is equal to X/10^N.
For instance, 0.005 = 5/10^3.
If you have something at the infinite place, for instance, 9, that will be equal to 9/10^infinity.
You are dividing 9 by an infinitely high number, this results in 0.
You may say it results in something infinitely close to 0, but anything infinitely close to 0 IS 0, there is no other number infinitely close to 0.
Insert witty comment.

Such a cliché.

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Re: A very interesting Mathematical Paradox

Postby JohnLeaf » Sat Jan 01, 2011 5:08 am UTC

SANTARII wrote:
JohnLeaf wrote:Wait, this doesn't make sense. If you times x=0.9999... By ten (0.9999...X10), then you have, no matter how many nines, whether it's two, or infinity, 9.9999...9 minus 0.000...9 (9.9999...9-0.000...9).

Which means that if you minus it by x, then you get a number that equals 8.(?), not 9. That will leave you with a number, that when divided by 9, would equal the original, o.9999... .

This is the same as the fact that, no matter how far you go with 0.333..., when you times it by 3, it will NEVER equal one. It will equal a number infinitely close to 1, but it will not equal one.

0.999...*10=9.999...-0.000...9

Anything infinitely close to something is equal to it. If it were not equal to it by any amount, it could be closer, and so not infinitely close to it.

Apart from the fact that you can't have something following an infinitely recurring pattern (otherwise it isn't infinitely recurring), let's say you theoretically have a 9 at the infinite place.
If you have a number, X, at N places past the decimal point, in denary, then that number is equal to X/10^N.
For instance, 0.005 = 5/10^3.
If you have something at the infinite place, for instance, 9, that will be equal to 9/10^infinity.
You are dividing 9 by an infinitely high number, this results in 0.
You may say it results in something infinitely close to 0, but anything infinitely close to 0 IS 0, there is no other number infinitely close to 0.


The flaw in this line of thinking, if you use that to support this, 'paradox', is that you must include the fact that if you look at, infinitely close to, as the same as, then .999... is automatically 1, and not .999... , which basically means that this is not a paradox, but a line of thinking that is in error. A simple estimation. It would be much more suitable to write,

0.999...=x
x*10=9.999...
9.999...-x~9
:. ~9x/9 ~ 1

And, .999 ~ 1

Where ~ is approximate, not proportional.


Hence, an obvious error in the line of thinking that .999...=1.

It is merely approximately 1.

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Re: A very interesting Mathematical Paradox

Postby xkcdfan » Sat Jan 01, 2011 5:22 am UTC

JohnLeaf wrote:It is merely approximately 1.

0.9 is approximately 1; it's 0.1 off from 1.
1.15 is approximately 1; it's 0.15 off from 1.

So how far off from 1 is 0.999...? Please give a real number as an answer.

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Re: A very interesting Mathematical Paradox

Postby JohnLeaf » Sat Jan 01, 2011 5:38 pm UTC

xkcdfan wrote:
JohnLeaf wrote:It is merely approximately 1.

0.9 is approximately 1; it's 0.1 off from 1.
1.15 is approximately 1; it's 0.15 off from 1.

So how far off from 1 is 0.999...? Please give a real number as an answer.

That wasn't the point I was trying to prove though. I was responding to the fact that was stated that,
Anything infinitely close to something is equal to it. If it were not equal to it by any amount, it could be closer, and so not infinitely close to it.


Which basically is saying that this isn't a paradox since 0.999... is REALLY close to one, it must be one. That's obviously not true, and while I might not be a genius. Far from it, I think that I made more sense than that, because that just says that this *paradox* is just a wrong way of thinking. And a friend of mine who is well read in the areas of math and Physics, told me that even if it IS infinite, 9.99... - 0.99... WILL equal something less than 9, but more than 8.

EDIT: If you want to go on to the point of minus, or adding, infinite numbers, that's a whole other problem that I can't answer.

Now on to something I discussed in depth last night with the same friend for an hour or so last night.

So what, pray tell, is 1 / 3 * 3? Take 1 and divide by 3. Multiply the result of that by 3. What do you get, and does your choice of how to write down the intermediate result really change the final product? Do you seriously contend that one third time three does not actually equal one just because you choose to represent "one third" in decimal form?


OK, to start, 1/3 * 3 = 1 This is true. It will always be true. On the other hand, in decimal form, 0.333... * 3 is NOT 1. it never will be. It's an approximation that we use to represent 1/3, but 0.333... does not equal 1/3. It's the same with 0.999... It doesn't equal one. Pretending it does is just fooling yourself in to thinking that you found some deep, awe inspiring, paradox, that's really just a foolish play on numbers.

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Re: A very interesting Mathematical Paradox

Postby t1mm01994 » Sat Jan 01, 2011 7:09 pm UTC

Basically, you are refusing to use the idea of infinite substraction and addition, or even the existance of infinite numbers, to then call it a "foolish play on numbers".
Dear friend, could you give me the value of pi?
Also, if 0.333... doesn't equal 1/3, then what is the decimal representation of 1/3?

And finally: Note that 0.333... isn't the sequence of 0,3 , 0,33 , 0,333 , but the limit of that sequence. That idea is the basis of the paradox as you call it.

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Re: A very interesting Mathematical Paradox

Postby phlip » Sun Jan 02, 2011 2:01 am UTC

JohnLeaf wrote:Which basically is saying that this isn't a paradox since 0.999... is REALLY close to one, it must be one. That's obviously not true,

However, what is "obviously true" and what is actually true are often two different things, especially when infinity is involved. That's all "paradox" means - the reality is counter to intuition, or expectation.

In this case, your intuition is that it's possible for two numbers to be infinitely close but unequal. But, in the real numbers as actually used by mathematicians, it's not.

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Re: A very interesting Mathematical Paradox

Postby Strange Quirk » Mon Jan 03, 2011 2:42 am UTC

Seriously, guys? Wasn't there a rule around here against these threads? (Oh, that was Mathematics. Still...) Or is the fact that it's a 5 year long thread excuse it from a lock?
If you disagree that 0.999... = 1, or are in an argument with someone who disagrees, go to the Wikipedia article on it. There are plenty of proofs and explanations there. If you or the person you're arguing with still doubts it after that, it's a good sign of some sort of brain damage or, at best, a complete lack of good mathematical understanding, and there is nothing any of us here will be able to do. So seriously, NO MORE POSTS ABOUT IT!


...
Ok fine, I can't resist. I'm sure this has been brought up already. For those that don't believe that 0.3333...=1/3:
0.3=3/10, right?
0.33=3/10+3/100, right?
0.333=3/10+3/100+3/1000, right?
0.333...=3/10+3/100+3/1000+..., right? They are EXACTLY equal.

Surely you know how to work with infinite arithmetic series, right? Go add it up.
x=0.333...=3/10+3/100+3/1000+... (Obvious, right?) (1)
x*10=3+3/10+3/100+3/1000+...(Multiply by 10) (2)
x*9=3 (Subtract (1) from (2) )
x=1/3 (Divide by 9)
1/3=0.3333 (Substitution)
Q.E.D.

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Re: A very interesting Mathematical Paradox

Postby jestingrabbit » Mon Jan 03, 2011 5:11 am UTC

Strange Quirk wrote:Seriously, guys? Wasn't there a rule around here against these threads? (Oh, that was Mathematics. Still...) Or is the fact that it's a 5 year long thread excuse it from a lock?


The way I see it, a good forum should be able to have a thread like this open somewhere. It should be possible to convince reasonable people of truths, and there should be a place, with as many good people hanging around, to have that happening. This is that place.

Plus, it does kinda get grandfathered in, as you suggested.

@JohnLeaf: should the set of real numbers depend upon their representation? If you use a base 3, you can definitely get a third, its simply written 0.1. It seems to me you're confusing the representation with the number. I also want to know whether pi has a decimal representation.
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Re: A very interesting Mathematical Paradox

Postby RonWessels » Mon Jan 03, 2011 4:21 pm UTC

Strange Quirk wrote:Surely you know how to work with infinite arithmetic series, right? Go add it up.
x=0.333...=3/10+3/100+3/1000+... (Obvious, right?) (1)
x*10=3+3/10+3/100+3/1000+...(Multiply by 10) (2)
x*9=3 (Subtract (1) from (2) )
x=1/3 (Divide by 9)
1/3=0.3333 (Substitution)
Q.E.D.

Arrrggghhh!! So close!

The last line (ok, second-last if you include the Q.E.D.) should be:
1/3 = 0.333... (Substitution)

It's fairly easy to prove (and fairly obvious) that 0.3333 [imath]\ne[/imath] 1/3 :oops:

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Re: A very interesting Mathematical Paradox

Postby SANTARII » Tue Jan 04, 2011 6:17 pm UTC

Spoiler:
JohnLeaf wrote:
SANTARII wrote:
JohnLeaf wrote:Wait, this doesn't make sense. If you times x=0.9999... By ten (0.9999...X10), then you have, no matter how many nines, whether it's two, or infinity, 9.9999...9 minus 0.000...9 (9.9999...9-0.000...9).

Which means that if you minus it by x, then you get a number that equals 8.(?), not 9. That will leave you with a number, that when divided by 9, would equal the original, o.9999... .

This is the same as the fact that, no matter how far you go with 0.333..., when you times it by 3, it will NEVER equal one. It will equal a number infinitely close to 1, but it will not equal one.

0.999...*10=9.999...-0.000...9

Anything infinitely close to something is equal to it. If it were not equal to it by any amount, it could be closer, and so not infinitely close to it.

Apart from the fact that you can't have something following an infinitely recurring pattern (otherwise it isn't infinitely recurring), let's say you theoretically have a 9 at the infinite place.
If you have a number, X, at N places past the decimal point, in denary, then that number is equal to X/10^N.
For instance, 0.005 = 5/10^3.
If you have something at the infinite place, for instance, 9, that will be equal to 9/10^infinity.
You are dividing 9 by an infinitely high number, this results in 0.
You may say it results in something infinitely close to 0, but anything infinitely close to 0 IS 0, there is no other number infinitely close to 0.


The flaw in this line of thinking, if you use that to support this, 'paradox', is that you must include the fact that if you look at, infinitely close to, as the same as, then .999... is automatically 1, and not .999... , which basically means that this is not a paradox, but a line of thinking that is in error. A simple estimation. It would be much more suitable to write,

0.999...=x
x*10=9.999...
9.999...-x~9
:. ~9x/9 ~ 1

And, .999 ~ 1

Where ~ is approximate, not proportional.


Hence, an obvious error in the line of thinking that .999...=1.

It is merely approximately 1.

"you must include the fact that if you look at, infinitely close to, as the same as, then .999... is automatically 1, and not .999..."
This is not true.
Yes, 0.999... IS automatically 1, but there is no reason to say that it is therefore NOT 0.999...

"which basically means that this is not a paradox"
Correct, it isn't a paradox.

"0.999...=x
x*10=9.999...
9.999...-x~9"
Incorrect.
It is not approximately 9, it is exactly 9.
x = 0.999...
9+x = 9.999...

"Hence, an obvious error in the line of thinking that .999...=1."
Where was the obvious error?
You merely stated that those values are approximations when they are not.
Insert witty comment.

Such a cliché.

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Re: A very interesting Mathematical Paradox

Postby RebeccaRGB » Wed Jan 05, 2011 8:46 am UTC

From earlier in this thread: ...999 = -1 because ...999 is the tens-complement representation of -1.

...997 = -3
...998 = -2
...999 = -1
...000 = 0
...001 = 1
...002 = 2

Just like

...1101 = -3
...1110 = -2
...1111 = -1
...0000 = 0
...0001 = 1
...0010 = 2

in binary.

So what would ...888 be? Or, if you want to think in binary because you think having only two digits renders this question moot, or twos-complement makes more sense to you than tens-complement, what about ...110110110?

EDIT: Due to the way our place-value system works, ...1111 (binary) would correspond to the infinite sum [math]\displaystyle\sum\limits_{k=0}^\infty 2^k[/math] which does not converge, of course. Is there any property of divergent sums that could help answer the above question? Say, a property that would be -1 for this particular sum?

EDIT 2: Aha!

http://en.wikipedia.org/wiki/Divergent_ ... ic_methods

For ...1111, c = 1 and r = 2, so G(c,r) = G(1,2) = 1 / (1 - 2) = 1 / -1 = -1. For ...999, c = 9 and r = 10, so G(c,r) = G(9,10) = 9 / (1 - 10) = 9 / -9 = -1. That works!

What about ...888? G(c,r) = G(8,10) = 8 / (1 - 10) = 8 / -9 = -0.888...

Hahahaha! ...888 = -0.888...! So, logically, ...999 = -0.999... = what else, -1! What goes around comes around!

EDIT 3: Apparently what I've been musing about here is the p-adic numbers. This was probably already mentioned earlier in this thread. Oh well.

http://en.wikipedia.org/wiki/P-adic_number

Also apparently there's no p-adic representation of e and the p-adic numbers can have zero divisors (when p is composite). What the hell? Math is weird.
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Re: A very interesting Mathematical Paradox

Postby phlip » Wed Jan 05, 2011 11:02 am UTC

SANTARII wrote:"which basically means that this is not a paradox"
Correct, it isn't a paradox.

Again, this is "paradox" meaning "counter to intuition"...

It's a paradox in the same way that, say, the Birthday Paradox is... a mathematical result that is counter to what you expect.

The idea that "paradox" means "wrong", or "self-inconsistent" comes from things like the liar paradox, or the grandfather paradox, where the actual mathematical result that's counter to intuition is that naive logic or naive time travel (respectively) are broken.

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Re: A very interesting Mathematical Paradox

Postby charonme » Wed Jan 05, 2011 2:51 pm UTC

jestingrabbit wrote:confusing the representation with the number
spot-on! The notation "0.999..." is truly different from the notation "1". However, they both denote the same exact number. But why there are no threads disputing "one"="uno"="eins" ?
Also, some other people seem to have a problem with the idea of a limit - they apparently don't comprehend that some limits can be exactly equal to a number (instead being merely "approximates").
The third problem is that some people invent their own (often nonsensical) notations, for example "0.000...9".

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Re: A very interesting Mathematical Paradox

Postby math » Sun Feb 20, 2011 5:26 am UTC

Let's put it this way.
0.99999... ends at infinity nines. So 9.999999...-0.999999...=8.999999...1. Because beyond infinity nines is a single digit that makes the whole problem be rounding, instead of definitive numbers.
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Re: A very interesting Mathematical Paradox

Postby xkcdfan » Sun Feb 20, 2011 7:09 am UTC

math wrote:Let's put it this way.
0.99999... ends at infinity nines. So 9.999999...-0.999999...=8.999999...1. Because beyond infinity nines is a single digit that makes the whole problem be rounding, instead of definitive numbers.

no

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Re: A very interesting Mathematical Paradox

Postby Ankit1010 » Sun Feb 20, 2011 11:54 pm UTC

The paradox is simply hard to reconcile without formal maths. It's very easily proven true using limits, infinite series or set/field theory and properties of the real numbers.

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Re: A very interesting Mathematical Paradox

Postby charonme » Mon Feb 21, 2011 3:29 pm UTC

math wrote:0.99999... ends at infinity nines. [...] beyond infinity nines is a single digit
So in other words, after the never-ending stream of nines ends, there is something else there? :wink:


the problem is that you expect a never-ending construct to end somewhere. It doesn't. It's neverending - which means it never ends, it has no end, it's endless, without an end. There can't be anything "after" its "end", because the supposed "end" does not exist. There's nowhere to put your "last 1" after the stream of 9. Anywhere you'd look to put your 1, you won't find a place for it, because there are never-ending nines. If (even "after infinite steps") you think you found a place for your 1, you actually just got so tired you can't see anymore and there are even more nines there.

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Re: A very interesting Mathematical Paradox

Postby Micali » Sat Feb 26, 2011 6:39 am UTC

SANTARII wrote:
JohnLeaf wrote:Wait, this doesn't make sense. If you times x=0.9999... By ten (0.9999...X10), then you have, no matter how many nines, whether it's two, or infinity, 9.9999...9 minus 0.000...9 (9.9999...9-0.000...9).

Which means that if you minus it by x, then you get a number that equals 8.(?), not 9. That will leave you with a number, that when divided by 9, would equal the original, o.9999... .

This is the same as the fact that, no matter how far you go with 0.333..., when you times it by 3, it will NEVER equal one. It will equal a number infinitely close to 1, but it will not equal one.

0.999...*10=9.999...-0.000...9

Anything infinitely close to something is equal to it. If it were not equal to it by any amount, it could be closer, and so not infinitely close to it.

Apart from the fact that you can't have something following an infinitely recurring pattern (otherwise it isn't infinitely recurring), let's say you theoretically have a 9 at the infinite place.
If you have a number, X, at N places past the decimal point, in denary, then that number is equal to X/10^N.
For instance, 0.005 = 5/10^3.
If you have something at the infinite place, for instance, 9, that will be equal to 9/10^infinity.
You are dividing 9 by an infinitely high number, this results in 0.
You may say it results in something infinitely close to 0, but anything infinitely close to 0 IS 0, there is no other number infinitely close to 0.


But then you're implying that you can have 0.000...01 of a number. 0.000...1 doesn't exactly exist... That sort of "number" is essentially the definition of a limit to 1. Something that is infinitely close to something else doesn't equal it.

Also another way I like to look at it is like a large glass filled with water. If you were to equally distribute that glass of water into three other glasses, each would be 33% of water from the original glass. But if you were to pour them all back into the original glass, you would have 100% of the water back in.

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Re: A very interesting Mathematical Paradox

Postby zachbarnett » Tue Mar 01, 2011 12:57 am UTC

Most resistance to [imath].999... = 1[/imath] stems from intuitions that are plausible but false for the real numbers (given how mathematicians have constructed the real numbers). Alternative number systems can be constructed that allow for the possibility of infitesimal non-zero numbers, and in such number systems, [imath].999 != 1[/imath].

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Re: A very interesting Mathematical Paradox

Postby Turtlewing » Tue Mar 01, 2011 9:51 pm UTC

zachbarnett wrote:Most resistance to [imath].999... = 1[/imath] stems from intuitions that are plausible but false for the real numbers (given how mathematicians have constructed the real numbers). Alternative number systems can be constructed that allow for the possibility of infitesimal non-zero numbers, and in such number systems, [imath].999 != 1[/imath].


At the same time I find this comment insightful and off target.

You could construct a number system in which .999... does not equal 1.0 However in many ways that would be like creating an English dialect where "bat" and "taco" are synonyms. It doesn't make the winged mammal a Mexican food item, it just means the symbols you're using have different meanings because you've defined them that way.

However where you are correct is in diagnosing the core problem with why people intuitively come to the wrong conclusion. That is most people don't use the real numbers in day to day life. They use a more practical minded number set were any number "too small to be worth worrying about right now" is an infinitesimal (this number set is useless for formal math (there's no consistent way to define those infinitesimals), but saves a lot of trouble in most people's day to day lives). For example a normal person mat treat .333... as equivalent to .333 Thus 0.000333... is an "infinitesimal". As such when they see .999... = 1.0 they inductively reason that: "this must be like all those other times I've seen two numbers separated by an infinitesimal where there were technically different numbers, but not in a meaningful way" and they try to explain this as round-off error, that .999... and 1.0 are separated by some hypothetical nonzero infinitesimal that isn't important enough to worry about but technically makes them different numbers. The trouble is this is the rare case where that's not true in the real numbers.

Unlike all the other numbers too small to care about which were infinitesimals in the "day to day numbers" but are finite numbers in the real numbers .999... and 1.0 are separated by the only day to day infinitesimal that is also an infinitesimal in the real numbers: 0; making them actually the same number by any reasonable definition of equality for the real numbers. Much like how the turkey doesn't get his dinner the night before Thanksgiving their induction has failed them because this once case is fundamentally different from all the others for which their reasoning would have worked just fine.

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Re: A very interesting Mathematical Paradox

Postby Ordinata » Thu Mar 03, 2011 1:28 pm UTC

Spoiler:
Turtlewing wrote:
zachbarnett wrote:Most resistance to [imath].999... = 1[/imath] stems from intuitions that are plausible but false for the real numbers (given how mathematicians have constructed the real numbers). Alternative number systems can be constructed that allow for the possibility of infitesimal non-zero numbers, and in such number systems, [imath].999 != 1[/imath].


At the same time I find this comment insightful and off target.

You could construct a number system in which .999... does not equal 1.0 However in many ways that would be like creating an English dialect where "bat" and "taco" are synonyms. It doesn't make the winged mammal a Mexican food item, it just means the symbols you're using have different meanings because you've defined them that way.

However where you are correct is in diagnosing the core problem with why people intuitively come to the wrong conclusion. That is most people don't use the real numbers in day to day life. They use a more practical minded number set were any number "too small to be worth worrying about right now" is an infinitesimal (this number set is useless for formal math (there's no consistent way to define those infinitesimals), but saves a lot of trouble in most people's day to day lives). For example a normal person mat treat .333... as equivalent to .333 Thus 0.000333... is an "infinitesimal". As such when they see .999... = 1.0 they inductively reason that: "this must be like all those other times I've seen two numbers separated by an infinitesimal where there were technically different numbers, but not in a meaningful way" and they try to explain this as round-off error, that .999... and 1.0 are separated by some hypothetical nonzero infinitesimal that isn't important enough to worry about but technically makes them different numbers. The trouble is this is the rare case where that's not true in the real numbers.

Unlike all the other numbers too small to care about which were infinitesimals in the "day to day numbers" but are finite numbers in the real numbers .999... and 1.0 are separated by the only day to day infinitesimal that is also an infinitesimal in the real numbers: 0; making them actually the same number by any reasonable definition of equality for the real numbers. Much like how the turkey doesn't get his dinner the night before Thanksgiving their induction has failed them because this once case is fundamentally different from all the others for which their reasoning would have worked just fine.

I think he's talking about nonstandard analysis, which is a formal model of the reals, different from the standard one. It extends the ordinary reals with an infite set of infinitesimals and infinites. The restriction to the ordinary reals is completely compatible with standard analysis, so it is not merely using the same names for something different. It allows you to do things like dividing differentials, which can make things in analysis a lot more straightforward (proving [imath]\frac{dy}{dx}\frac{dx}{dt}=\frac{dy}{dt}[/imath] is just a matter of canceling [imath]dx[/imath].) But it still doesn't let you divide by 0, so I say BAH!

Also, ofc .99... = 1. I think the misconception that it does not, stems from not quite grasping that the sequence of 9 is in fact never-ending, as in it NEVER ends. I think the intuitive idea is when one says "never-ending" is that one just means "really really long" but that it actually does end at some point.

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Re: A very interesting Mathematical Paradox

Postby skullturf » Thu Mar 03, 2011 11:22 pm UTC

Ordinata wrote:Also, ofc .99... = 1. I think the misconception that it does not, stems from not quite grasping that the sequence of 9 is in fact never-ending, as in it NEVER ends. I think the intuitive idea is when one says "never-ending" is that one just means "really really long" but that it actually does end at some point.


I partly agree, but I'd say something a bit different. My impression is that the people who say "0.999... can't equal 1" are thinking of 0.999... as a "process" or as something that's "changing", as opposed to thinking of it as one fixed entity. (One fixed entity that, counterintuitive though it may seem at first, has infinitely many digits after the decimal point, each one of which is 9. But we're talking about *a* number or *a* thing that has these infinitely many 9's.)

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Re: A very interesting Mathematical Paradox

Postby Jimmigee » Fri Mar 04, 2011 11:12 am UTC

I think skullturf has it nailed. People think "But it doesn't matter how far I go, 0.999... is always less than 1". They're mentally reading the number from left to right, and comparing it with 1.

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Re: A very interesting Mathematical Paradox

Postby krucifi » Fri Mar 04, 2011 12:28 pm UTC

skullturf HAS nailed it.

people are thinking 0.999.... is just a sequence where whenever you DO get to an end you just keep adding 9's as though those 9's will never reach the intended target of the so called "end". but there is you wont suddenly see where it stops only to see millions of 9's suddenly appear after it because somone finally decided to investigate.

the number is just another way of writing 1. we arent sayin 0.999.... is a seperate number sometime after 0.99999999999999 we are saying its as close to 1 as it can get making them equal. its not hard to see that 0.5 is the same as 1/2. this is the same thing essentially, in that the one number has two different decimal representations.
zenos paradox deals with something similar.
people just arent getting the idea of infinity. sure adding 9's to 0.99 will NEVER get to 0.999.... the normal way, but what about the infinite series? infinity is a strange thing the laws of numbers do not apply there lol. removing a 9 from this number changes NOTHING. adding a number likewise. so stop trying to add a 1 at the end of an infinite series subtraction.... because if you say the 9's in 0.999.... must end then so must the 0's in 1.000..... meaning there will STILL be nowhere for that 1 to go because there will always be one more 9 than there can be a zero, because zero is just an upgraded 9 from the previous decimal position.
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Re: A very interesting Mathematical Paradox

Postby charonme » Fri Mar 04, 2011 5:32 pm UTC

perhaps something like denying [imath]\displaystyle\sum\limits_{n=1}^\infty \frac{3}{4^n}=1[/imath] or more accurately [imath]\displaystyle\sum\limits_{n=1}^\infty \frac{9}{10^n}=1[/imath], but this was already mentioned here several times

But I'm not sure this was mentioned: could some 0.9...=1 deniers give this a thought if nothing here persuaded them? Real numbers don't have predecessors nor successors. If 0.9... != 1 then 0.9... would have to be < 1. Since there can be no conceivable number between 0.9... and 1 (can it?), 0.9... would have to be a predecessor of 1 (and 1 a successor of 0.9...). But since real numbers don't have predecessors nor successors, the original assumption (0.9... != 1) must be wrong.


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