A very interesting Mathematical Paradox

[sabbathxxl]

Some of you may have seen this before. I would like to hear your views/responses to this.

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x = .9999...
10x = 9.999...
10x - x = 9.00
9x = 9
x = 1

Basically, we have shown that .999... = 1.

---

Here's another look at the paradox using fractioned examples.

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 _
.1 = 1/9 and so on.
 ___
.123 = 123/999
    _
So .9 = ?/9

It must be 9/9, or 1.

Thoughts?

[wisnij]

It's no paradox. 0.999... does equal 1.

There are many real values which can be written in different ways; we say that their representation is not unique. The most familiar case is with rational numbers. We say that 1/2 and 3/6 represent the same value, even though they superficially appear different. The former is simply in lowest terms, with all common factors removed. (For that matter, 0.5 is also equivalent.) Similarly, 0.999... and 1 name the same real number, the same point on the number line, even if they are written differently.

(Edit: wikilink, with various proofs.)

[Marrow]

its just like infinity divided by anything is 0

Also you should need the same number of digits used. if the 9's continue infinitely you need to use the same number of infinite 9's for x and 10x
So even if you cut x off at a predetermined number of decimals you are actually going to end up with 8.9991. If you used went to the ten thousanths place with x = .9999 then 10x still = 9.999 but if you say that 10x is 9.9999 and x is .9999 you can't get 10x.9999 to = 9.9999 because you didn't have that last digit when you did the math. All you are doing is rounding that last 9 up by one which has a chain effect all the way up to the first significant digit. All you are doing is proving the tendancy to round and use the closest real number to a string that will continue indeffinitely.

[Shoofle]

its just like infinity divided by anything is 0

I don't understand, and more significantly I doubt what you say. Perhaps you mean anything divided by infinity is zero? In any case, infinity is not a number and these statements are meaningless. You should use limit notation.

[Marrow]

its just like infinity divided by anything is 0

I don't understand, and more significantly I doubt what you say. Perhaps you mean anything divided by infinity is zero? In any case, infinity is not a number and these statements are meaningless. You should use limit notation.

Yes, I always confuse which number is divided by which when I say it rather that writing it down and knowing if I should expect a number larger or smaller than 1

[Teaspoon]

1 - 0.999... = 0.000...1

That's infinite 0s followed by a 1. But if you have infinite 0s, the entire set of decimal places are filled by 0s and the 1 hangs out there as the digit following the infinitieth digit, giving it a value that's one tenth of infinitesimal. Anything smaller than infinitesimal is also infinitesimal (the opposite end of that 10*infinity=infinity concept), which means it is too small to have a defined value and can thus be added or subtracted to another value without changing it.

[wisnij]

Also you should need the same number of digits used. if the 9's continue infinitely you need to use the same number of infinite 9's for x and 10x
So even if you cut x off at a predetermined number of decimals you are actually going to end up with 8.9991. If you used went to the ten thousanths place with x = .9999 then 10x still = 9.999 but if you say that 10x is 9.9999 and x is .9999 you can't get 10x.9999 to = 9.9999 because you didn't have that last digit when you did the math. All you are doing is rounding that last 9 up by one which has a chain effect all the way up to the first significant digit. All you are doing is proving the tendancy to round and use the closest real number to a string that will continue indeffinitely.

It doesn't work that way. The string of 9s is infinitely long; there is no last digit. Every digit 9 in x gets matched up with a digit 9 in 10x, and they all cancel just as described initially.

1 - 0.999... = 0.000...1

That's infinite 0s followed by a 1.

It doesn't work that way, either. You can't follow an infinite sequence of zeroes with a one... the zeroes never end, so there's no place for the one to go. More generally, there are no infinitesimal numbers other than 0 in the standard real numbers.

[Teaspoon]

"The string of 9s is infinitely long; there is no last digit. Every digit 9 in x gets matched up with a digit 9 in 10x, and they all cancel just as described initially."

We have a hotel with infinitely many rooms and they're all full. Then another 9 buses arrive, each carrying infinitely many passengers... Infinity really does kick up odd behaviour from the human brain, doesn't it?

And that bit about not being able to follow an infinite sequence of zeroes with a one? That's exactly the point I was making. There's no place for that trailing one, so it isn't actually there; hence the difference between 1 and 0.<infinite sequence of nines> is indistinguishable from zero and the two numbers must be equal.

[phlip]

1 - 0.999... < ε

What more is there to say?

[wisnij]

"The string of 9s is infinitely long; there is no last digit. Every digit 9 in x gets matched up with a digit 9 in 10x, and they all cancel just as described initially."

We have a hotel with infinitely many rooms and they're all full. Then another 9 buses arrive, each carrying infinitely many passengers... Infinity really does kick up odd behaviour from the human brain, doesn't it?

And that bit about not being able to follow an infinite sequence of zeroes with a one? That's exactly the point I was making. There's no place for that trailing one, so it isn't actually there; hence the difference between 1 and 0.<infinite sequence of nines> is indistinguishable from zero and the two numbers must be equal.

Granted, but your description of how infinitesimals work still wasn't right. There's an algebra called the dual numbers which take the form z = a + bε, where ε is an infinitesimal. Clearly, adding an infinitesimal (or a multiple thereof) is not meaningless in this system; a + bε ≠ a for nonzero b, even though |bε| < 1 no matter how large b is. However, this ε value doesn't exist in the standard real numbers. There aren't any infinitesimals in R, other than zero.

In other words, the reason that the "1 - 0.999... = ε" is wrong isn't because infinitesimals wouldn't work that way, it's just that R happens to not contain any suitable value of ε.

[Teaspoon]

My brain's doing odd things today, but just bear with me a little more. I've never done any mathematical study involving infinitesimals before. My higher education consists of dropping out of second-year software engineering because too many of the mandatory subjects were electronics-based rather than cool stuff like mathematics or programming concepts and I just couldn't maintain interest.

Anyway...

We're talking about real numbers here, so we have no infinitesimal numbers other than zero, which can be expressed as 0.000...

0.000...1=0.000...*0.1=0*0.1=0. Or, in human-speak, that number is a tenth of the only real infinitesimal number, which makes it equal to the only real infinitesimal number because infinites are crazy, hence it's equal to zero.

I sort of understand about dual numbers. I've never encountered them before, but I'm guessing they're treated roughly the same as complex numbers, with ε instead of i as the non-real part. I suppose the dual numbers are a bit nicer to work with because one could at least say "1+2ε > 1+ε", whereas the value of i is fucked up enough that the only legal comparison is equality.

Does anybody ever have reason to do algebra involving real, imaginary and infinitesimal numbers? What do they call it when they do?

Also, what's the popular opinion on the value of infinity*ε? Infinity? ε? 0? 1?
Hell, why not i?

[Factitious]

There are only countably many real numbers with terminating decimal expansions. Each of those also has a nonterminating decimal expansion.

So if we wanted to get rid of the confusing situation in which two expansions are used to represent the same number, the sensible thing to do might be to restrict ourselves to the nonterminating ones. That way, every real number has exactly one way it can be written, and they're all in the same nonterminating format. And this wouldn't have any effect on the vast majority of real numbers, so it's a trivial reform, right?

[Factitious]

I just realized something odd. Most of the nonterminating representations for numbers with terminating representations end in an infinite string of nines:

2 = 1.999...
7.5 = 7.4999...
-2.718 = -2.717999...

But zero doesn't fit this pattern. Is "0.000..." acceptable? It looks kind of silly to me.

[wisnij]

I sort of understand about dual numbers. I've never encountered them before, but I'm guessing they're treated roughly the same as complex numbers, with ε instead of i as the non-real part. I suppose the dual numbers are a bit nicer to work with because one could at least say "1+2ε > 1+ε", whereas the value of i is fucked up enough that the only legal comparison is equality.

Yes! Dual numbers can be ordered, which is really cool.

Also, what's the popular opinion on the value of infinity*ε? Infinity? ε? 0? 1?
Hell, why not i?

Infinity is not a number. That operation is not defined.

I just realized something odd. Most of the nonterminating representations for numbers with terminating representations end in an infinite string of nines:

2 = 1.999...
7.5 = 7.4999...
-2.718 = -2.717999...

But zero doesn't fit this pattern. Is "0.000..." acceptable? It looks kind of silly to me.

You could equally well write 2 = 2.000..., 7.5 = 7.5000..., etc. If you like, you can consider every finite-length real expansion to be terminated with an infinite string of zeroes, which are left off simply for notational convenience. Zero itself doesn't fit because it is the origin, and infinite sequences of 9s always "round" away from the origin (e.g. 0.999... = 1, -0.999... = -1, etc).

[Andy]

I'm not sure if it stays consistent, but you can use a similar approach to prove that:

9 = -1 (that is ...99999.....999.0 = -1)

I can't see why it shouldn't be consistent... but I think it is pretty cool... like maths' version of two's complement

[Torn Apart By Dingos]

I just realized something odd. Most of the nonterminating representations for numbers with terminating representations end in an infinite string of nines:

That's actually the only way it can be. If you disallow infinite tails of 9s, the decimal expansion for all real numbers is unique.

I'm not sure if it stays consistent, but you can use a similar approach to prove that:

9 = -1 (that is ...99999.....999.0 = -1)

That's not a real number. 0.999... is defined as the sum of 9*10^-k for all k=1,2,3,..., and that sum converges to 1. Your "number" does not converge.

[wisnij]

I'm not sure if it stays consistent, but you can use a similar approach to prove that:

9 = -1 (that is ...99999.....999.0 = -1)

I can't see why it shouldn't be consistent... but I think it is pretty cool... like maths' version of two's complement

http://www.inwap.com/pdp10/hbaker/hakmem/hacks.html#item154

[Marrow]

Also you should need the same number of digits used. if the 9's continue infinitely you need to use the same number of infinite 9's for x and 10x
So even if you cut x off at a predetermined number of decimals you are actually going to end up with 8.9991. If you used went to the ten thousanths place with x = .9999 then 10x still = 9.999 but if you say that 10x is 9.9999 and x is .9999 you can't get 10x.9999 to = 9.9999 because you didn't have that last digit when you did the math. All you are doing is rounding that last 9 up by one which has a chain effect all the way up to the first significant digit. All you are doing is proving the tendancy to round and use the closest real number to a string that will continue indeffinitely.

It doesn't work that way. The string of 9s is infinitely long; there is no last digit. Every digit 9 in x gets matched up with a digit 9 in 10x, and they all cancel just as described initially.

Well if you say it doesn't work that way then you never actually can finish calculating 10x because there is always another 9 to multiply by 10 so you can never end up with a number to subtract the .9 from to get an answer. You can't say that one continues forever but another one can terminate at the same number of significant digits +1. By that it means the answer to 10x-.9999=Hold on I am not done with the last one yet.

[wisnij]

Also you should need the same number of digits used. if the 9's continue infinitely you need to use the same number of infinite 9's for x and 10x
So even if you cut x off at a predetermined number of decimals you are actually going to end up with 8.9991. If you used went to the ten thousanths place with x = .9999 then 10x still = 9.999 but if you say that 10x is 9.9999 and x is .9999 you can't get 10x.9999 to = 9.9999 because you didn't have that last digit when you did the math. All you are doing is rounding that last 9 up by one which has a chain effect all the way up to the first significant digit. All you are doing is proving the tendancy to round and use the closest real number to a string that will continue indeffinitely.

It doesn't work that way. The string of 9s is infinitely long; there is no last digit. Every digit 9 in x gets matched up with a digit 9 in 10x, and they all cancel just as described initially.

Well if you say it doesn't work that way then you never actually can finish calculating 10x because there is always another 9 to multiply by 10 so you can never end up with a number to subtract the .9 from to get an answer. You can't say that one continues forever but another one can terminate at the same number of significant digits +1. By that it means the answer to 10x-.9999=Hold on I am not done with the last one yet.

There's no need to enumerate all the infinite digits in 10x, so that's not a problem. Simply knowing that there are a (countable) infinity of digits is enough to be able to use certain properties of infinities.

Remember the Grand Hotel puzzle where everyone had to shift down one room to make room for a new guest? Now picture the opposite operation: the guy in room 1 leaving and everyone else shifting one room in the other direction. Multiplying by 10 — i.e. moving the decimal place one point to the right — is almost exactly analogous to the latter case. For every digit right of the decimal place in 0.999... there is a corresponding digit to the right of the decimal place in 9.999..., and it all matches up. You can subtract the one from the other to come out with the integer 9, nice and even.

[Factitious]

I just realized something odd. Most of the nonterminating representations for numbers with terminating representations end in an infinite string of nines:

That's actually the only way it can be. If you disallow infinite tails of 9s, the decimal expansion for all real numbers is unique.

Yeah. The part I thought was odd was zero being an apparent counterexample. As wisinj pointed out, this makes sense because the "rounding" goes away from the origin.

If we wanted to force uniqueness, I think disallowing terminating expansions is more elegant than disallowing infinite tails of 9s, just because it seems a bit ad hoc to allow .888... but not .999....

I'm not sure if it stays consistent, but you can use a similar approach to prove that:

9 = -1 (that is ...99999.....999.0 = -1)

That's not a real number. 0.999... is defined as the sum of 9*10^-k for all k=1,2,3,..., and that sum converges to 1. Your "number" does not converge.

Interestingly, it can be made to converge if you use a certain bizarre metric. The result is the p-adic numbers (http://en.wikipedia.org/wiki/P-adic_number).
"Whereas two decimal expansions are close to one another if they differ by a large negative power of 10, two 10-adic expansions are close if they differ by a large positive power of 10. Thus 3333 and 4333 are close in the 10-adic metric, and 33333333 and 43333333 are even closer."

[Factitious]

I sort of understand about dual numbers. I've never encountered them before, but I'm guessing they're treated roughly the same as complex numbers, with ? instead of i as the non-real part. I suppose the dual numbers are a bit nicer to work with because one could at least say "1+2? > 1+?", whereas the value of i is fucked up enough that the only legal comparison is equality.

Yes! Dual numbers can be ordered, which is really cool.

It works out to be a straightforward dictionary order, right? Very cool -- much neater than that mess with the complex numbers.

[Marrow]

Also you should need the same number of digits used. if the 9's continue infinitely you need to use the same number of infinite 9's for x and 10x
So even if you cut x off at a predetermined number of decimals you are actually going to end up with 8.9991. If you used went to the ten thousanths place with x = .9999 then 10x still = 9.999 but if you say that 10x is 9.9999 and x is .9999 you can't get 10x.9999 to = 9.9999 because you didn't have that last digit when you did the math. All you are doing is rounding that last 9 up by one which has a chain effect all the way up to the first significant digit. All you are doing is proving the tendancy to round and use the closest real number to a string that will continue indeffinitely.

It doesn't work that way. The string of 9s is infinitely long; there is no last digit. Every digit 9 in x gets matched up with a digit 9 in 10x, and they all cancel just as described initially.

Well if you say it doesn't work that way then you never actually can finish calculating 10x because there is always another 9 to multiply by 10 so you can never end up with a number to subtract the .9 from to get an answer. You can't say that one continues forever but another one can terminate at the same number of significant digits +1. By that it means the answer to 10x-.9999=Hold on I am not done with the last one yet.

There's no need to enumerate all the infinite digits in 10x, so that's not a problem. Simply knowing that there are a (countable) infinity of digits is enough to be able to use certain properties of infinities.

Remember the Grand Hotel puzzle where everyone had to shift down one room to make room for a new guest? Now picture the opposite operation: the guy in room 1 leaving and everyone else shifting one room in the other direction. Multiplying by 10 — i.e. moving the decimal place one point to the right — is almost exactly analogous to the latter case. For every digit right of the decimal place in 0.999... there is a corresponding digit to the right of the decimal place in 9.999..., and it all matches up. You can subtract the one from the other to come out with the integer 9, nice and even.

Arguments abound

I just don't understand why you believe that you don't have to do one thing to both parts and still come up with an equation(why does that word look like its spelled wrong, well its better than equasion like I put first also thinking it to be wrong?). You can't do one thing to one of the x's and not do it to the other, even if they are on the same side of the equasion x must still equal =. Very SIMPLE rule from grade school is that ANYTHING multiplied by 10 ends in a 0. This applies to .999 repeating as well.

The hotel is pointless to bring up but since you did I will argue it anyway. With the hotel you were trying to make x (the number of rooms) = y (the number of guests). Personally I would have just stuck the new guy at the end and left everyone else in the hotel alone because its his fault for getting there so late. If x=y in the hotel, why does x not have to be the same as x in this problem.

Tell my why you don't have to enumerate all the infinite digits in 10, not just that you don't and the last nine just drops off the unmultiplied x.

You can't have a countable infinity, that defies the concept of infinity, and anything will work out if you only use certain properties of something while ignoring others. I can turn lead into gold if I ignore certain properties of it having the wrong number of molecules and such if I just paint it yellow. Unfortunately this only works with psychology.

[Heirtopendragon]

There are so many easier ways to explain this.

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1/3 = .333...
x 3    x 3
3/3 = .999...
1 = .999...

[Marrow]

There are so many easier ways to explain this.

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1/3 = .333...
x 3    x 3
3/3 = .999...
1 = .999...

Its still rounding anway you look at it

.333+.666=.999
.333+.667=1.000

[Heirtopendragon]

how is it rounding?

1/3 x 3 is 3/3 or 1
.333... x 3 is .999...
but 1/3 = .333...

1 = .999...

[Factitious]

Very SIMPLE rule from grade school is that ANYTHING multiplied by 10 ends in a 0.

5.6*10 = 56, which ends in a 6. 1/3*10 = 3.333..., which doesn't end. The grade school rule works for integers, though.

Personally I would have just stuck the new guy at the end and left everyone else in the hotel alone because its his fault for getting there so late.

The hotel has infinitely many rooms, one for each natural number (1,2,3,4...). Each of those numbers is finite. What's the number of the room you want to put the new guy in?

You can't have a countable infinity, that defies the concept of infinity, and anything will work out if you only use certain properties of something while ignoring others.

http://en.wikipedia.org/wiki/Countable. Really, read that. I think you're misunderstanding the meaning of "countably infinite."

[wisnij]

Well if you say it doesn't work that way then you never actually can finish calculating 10x because there is always another 9 to multiply by 10 so you can never end up with a number to subtract the .9 from to get an answer. You can't say that one continues forever but another one can terminate at the same number of significant digits +1. By that it means the answer to 10x-.9999=Hold on I am not done with the last one yet.

There's no need to enumerate all the infinite digits in 10x, so that's not a problem. Simply knowing that there are a (countable) infinity of digits is enough to be able to use certain properties of infinities.

Remember the Grand Hotel puzzle where everyone had to shift down one room to make room for a new guest? Now picture the opposite operation: the guy in room 1 leaving and everyone else shifting one room in the other direction. Multiplying by 10 ・i.e. moving the decimal place one point to the right ・is almost exactly analogous to the latter case. For every digit right of the decimal place in 0.999... there is a corresponding digit to the right of the decimal place in 9.999..., and it all matches up. You can subtract the one from the other to come out with the integer 9, nice and even.

Arguments abound

I just don't understand why you believe that you don't have to do one thing to both parts and still come up with an equation(why does that word look like its spelled wrong, well its better than equasion like I put first also thinking it to be wrong?). You can't do one thing to one of the x's and not do it to the other, even if they are on the same side of the equasion x must still equal =. Very SIMPLE rule from grade school is that ANYTHING multiplied by 10 ends in a 0. This applies to .999 repeating as well.

Grade school math also teaches us that pi = 22/7. Later on in life we learn which things we learned were only approximations used to gloss over a more complicated truth.

Let's phrase it this way instead:

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(1)        x = 0.999...
(2)      10x = 9.999...
(3)    x + 9 = 0.999... + 9   [from eq. 1]
             = 9.999...
(4)      10x = x + 9          [from eqs. 2 & 3]
(5)  10x - x = 9
(6)       9x = 9
(7)        x = 1
(8) 0.999... = 1              [from eqs. 1 & 7]


Look, if you don't like the algebraic proof, take your pick of the others. They all come to the same conclusion: within the standard real numbers, 0.999... = 1. The proof using Dedekind cuts is particularly nice.

The hotel is pointless to bring up but since you did I will argue it anyway. With the hotel you were trying to make x (the number of rooms) = y (the number of guests). Personally I would have just stuck the new guy at the end and left everyone else in the hotel alone because its his fault for getting there so late. If x=y in the hotel, why does x not have to be the same as x in this problem.

Tell my why you don't have to enumerate all the infinite digits in 10, not just that you don't and the last nine just drops off the unmultiplied x.

There is no end! What part of "infinite" don't you understand? The rooms in the hotel continue forever, just as do the digits in 0.999.... There's no "last room" to stick the new guy in, and there's no last digit that can be dropped or appended to.

You can't have a countable infinity, that defies the concept of infinity, and anything will work out if you only use certain properties of something while ignoring others. I can turn lead into gold if I ignore certain properties of it having the wrong number of molecules and such if I just paint it yellow. Unfortunately this only works with psychology.

You have a lot of learning to do. An infinity is countable if it is equal to the cardinality of the set of integers. That is, every element of an infinite set of this type could be uniquely numbered ("counted") with an integer. The set of digits in a real number, for example, is countably infinite set. On the other hand, the set of all real numbers is uncountable, as Cantor proved.

[Marrow]

because of the "..." rounding, cutting off, either way you are not fully representing t... you know what fine, I could argue this forever, look 1/3, or any repeating decimal continues forever unless you trucate it, round it, or otherwise denote its infinite length. Its just like those commercials that say things are 99.9% effective because if they say 100% and for some reason it doesn't work just that one damn time people are going to bitch and whine much like I am doing right now. Its a good thing for me that you people are tolerant of my not knowing when to give up, haha. Maybe when I actually know what I am talking about I'll just be happy and agree. Right now I just have something inside fo me that just wants the reality that I have learned and how I want it to be shaped to be right and work for everyone else so it will work for me too. Witnesses just have to go and make you prove everything instead of just letting it work.

(This post has not been made by two clearly insane people.){If anyone actually know what I am basing that statement off of I will keep posting here or never post again based on their whim, first come first serve.}

[posiduck]

this isn't a case of rounding. It is a mathematical fact without rounding.

[Torn Apart By Dingos]

I'm not sure if it stays consistent, but you can use a similar approach to prove that:

9 = -1 (that is ...99999.....999.0 = -1)

That's not a real number. 0.999... is defined as the sum of 9*10^-k for all k=1,2,3,..., and that sum converges to 1. Your "number" does not converge.

Interestingly, it can be made to converge if you use a certain bizarre metric. The result is the p-adic numbers (http://en.wikipedia.org/wiki/P-adic_number).
"Whereas two decimal expansions are close to one another if they differ by a large negative power of 10, two 10-adic expansions are close if they differ by a large positive power of 10. Thus 3333 and 4333 are close in the 10-adic metric, and 33333333 and 43333333 are even closer."

That's kind of cool. However, it's not a real number, so Andy can't say "using a similar approach". He might as well have said that you can show that 6=-1, and afterwards explain that he was actually talking about Z mod 7.

On second glance, ...999 isn't a p-adic number at all. The article states it's a "10-adic number", but p-adic numbers must have p prime.

[phlip]

1/3 is approximately equal to 0.33 - this is rounding.
1/3 is equal to 0.33333.... with an infinite number of 3s after the decimal point. There is no rounding involved, and the equality can be proven with limits and such.

[Teaspoon]

Okay, time to run through it backwards to avoid this "rounding off" thing that Marrow's complaining about and still prove that 0.999... = 1 without rounding anything off!

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x = 0.999...
x+9 = 9.999...
(x+9)/10 = 0.999...


Let me remind the audience that I divided by 10, which means that there's definitely no 0 appended to the end of the sequence. Now I'm just going to substitute x for that 0.999... (anybody want to claim that the 0.999... in that last line is different in value to the 0.999... that we started with?) and run through a few more easy steps that I don't believe anybody here will find reason to argue against.

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(x+9)/10 = x
x+9 = 10x
9 = 9x
1 = x


The other option is to do the ol' limiting sum thing.

Every digit in 0.999... is one tenth the digit before it.
This means we have a geometric series with rate (r) 0.1 and first term (a) 0.9.
This means Tx=0.9*0.1^(x-1)
How do we find the limiting sum of a geometric series?
For 0<r<1, SL=a/(1-r) = 0.9/(1-0.1) = 0.9/0.9 = 1

Creepy, huh?

[Penguin]

There's also this "proof" which, while not constructed as a formal logical proof, certainly can help to drive the point home to some people:

Code: Select all

1/9 = .1111...
2/9 = .2222...
3/9 = 1/3 = .3333...
4/9 = .4444...
5/9 = .5555...
6/9 = 2/3 = .6666...
7/9 = .7777...
8/9 = .8888...
9/9 = 3/3 = 1 = .9999...


That help anyone who's having difficulty with the idea of doing math to infinity? All you have to do is follow the pattern, and it's inherently clear that .9999... and 1 are the same thing.

The first time I saw this in sixth grade I had the worst time wrapping my mind around the proofs involving arithmetic with an infinite sequence, but when I saw this version it instantly made sense.

[Marrow]

Sorry, there is no covincing by saying something the same way repeatedly. Would this also mean that 0.666... and 0.667 are equal by the same reasoning.

[wisnij]

Sorry, there is no covincing by saying something the same way repeatedly. Would this also mean that 0.666... and 0.667 are equal by the same reasoning.

No.

[Marrow]

Sorry, there is no covincing by saying something the same way repeatedly. Would this also mean that 0.666... and 0.667 are equal by the same reasoning.

No.

Yep, I quit.

[Binary.Tobis]

Sorry, there is no covincing by saying something the same way repeatedly. Would this also mean that 0.666... and 0.667 are equal by the same reasoning.

No.

Very to the point.

[Marrow]

Sorry, there is no covincing by saying something the same way repeatedly. Would this also mean that 0.666... and 0.667 are equal by the same reasoning.

No.

Very to the point.

Yes, if its that difinitive then there is no point in saying anything about 1.000-.333 being .667 and .999-.333 being .666. For my next trick, I will fail to trisect an angle.

[Shoofle]

.999 != 1
.999-.333 != 1-.333
.333 != .667
Sorry I couldn't use proper not equal to signs, I didn't have them.

[posiduck]

Sorry, there is no covincing by saying something the same way repeatedly. Would this also mean that 0.666... and 0.667 are equal by the same reasoning.

Marrow, the reasoning works like this:

x=.999...
10x=9.999...
10x-x=9.999...-.999...
9x=9
x=1

So, let's look at it with .666...
x=.666...
10x=6.666...
10x-x=6.666...-.666...
9x=6
3x=2
x=2/3, or .666...

So, no.