Math problem

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MindDetective

MindDetective

In fact I will go 1 step further and offer my opinion. 0.9999... + 9 definitely preserves all of the infinite 9s in the decimal places. 0.9999... * 10 may not preserve them all, effectively losing the 9 in the infinity place. In Chad Sexington 's proof, the assumption is that 9.9999... has preserved all of infinity of the 9s in the decimal place, which to me implies it is more accurate to say that 9 has been added rather than 10 has been multiplied.

*edit* Argh, I should have said I would go 0.9999... steps further. Opportunity missed.
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stienman said:
I thought the fundamental question was: do we have free will, or are we merely meat robots?
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No, the answer to this question informs us to the answer on that one.
 
Krisken

Krisken

It really is as simple as 1/3 of 1 is .333(repeating), .333(repeating) times 3 is .999(repeating), 1=.999(repeating). Not that my brain accepts this.
 
MindDetective

MindDetective

Krisken said:
It really is as simple as 1/3 of 1 is .333(repeating), .333(repeating) times 3 is .999(repeating), 1=.999(repeating). Not that my brain accepts this.
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That is because there is some rounding going on here as well. It is again occurring in the infinity place, so is excused (or goes unnoticed).
 
tegid

tegid

MindDetective said:
In fact I will go 1 step further and offer my opinion. 0.9999... + 9 definitely preserves all of the infinite 9s in the decimal places. 0.9999... * 10 may not preserve them all, effectively losing the 9 in the infinity place. In Chad Sexington 's proof, the assumption is that 9.9999... has preserved all of infinity of the 9s in the decimal place, which to me implies it is more accurate to say that 9 has been added rather than 10 has been multiplied.

*edit* Argh, I should have said I would go 0.9999... steps further. Opportunity missed.
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No, the answer to this question informs us to the answer on that one.
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Wait, then where you have the problem is with the infinity of decimal places. 0.999... x 10 =9.999... because, since there's an infinity of 9's after the coma, you don't lose any. There isn't a last 9 that gets shifted out of place. You can accept that without assuming that 0.99...=1 (except it's an immediate consequence so it's normal to equate the two).

Let's try it with a different periodicity in the decimals: Would you accept that 0.898989... x 10 = 8.989898... and 0.898989... x 100 = 89.8989... preserving an infinity of 89's in the decimal places? (although in the case where we multiply by 10 they get shifted in position, if you identify them by that)?
 
Chad Sexington

Chad Sexington

If you have a pattern of 110110110... that repeats infinitely, are there enough zeroes to pair up one 0 to each 1? (yes.) Infinity, man. You can't lose it.
 
MindDetective

MindDetective

tegid said:
Wait, then where you have the problem is with the infinity of decimal places. 0.999... x 10 =9.999... because, since there's an infinity of 9's after the coma, you don't lose any. There isn't a last 9 that gets shifted out of place. You can accept that without assuming that 0.99...=1 (except it's an immediate consequence so it's normal to equate the two).

Let's try it with a different periodicity in the decimals: Would you accept that 0.898989... x 10 = 8.989898... and 0.898989... x 100 = 89.8989... preserving an infinity of 89's in the decimal places? (although in the case where we multiply by 10 they get shifted in position, if you identify them by that)?
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Sure you do. You lose it in the infinity place! That's like saying infinity minus 1 is infinity. It's true but we can't do much with that information. You've effectively shifted infinity over by 1 decimal place. But we don't have a notation for that, so we round it off and say good enough.
 
strawman

strawman

MindDetective said:
Well, that's an oversimplification. You don't have to accept it as true, only as close enough.
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MindDetective said:
Sure you do. You lose it in the infinity place! That's like saying infinity minus 1 is infinity. It's true but we can't do much with that information. You've effectively shifted infinity over by 1 decimal place. But we don't have a notation for that, so we round it off and say good enough.
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You keep using that word. I do not think it means what you think it means.
 
MindDetective

MindDetective

stienman said:
The word infinity. The following provides several proofs, but more interestingly a few posts point out that this depends on what 0.999... actually means, what real numbers are (and aren't), and the fact that you can go both ways depending on your assumptions. This touches on one of the foundations of math:

http://math.stackexchange.com/questions/11/does-99999-1
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I honestly am not sure what you're getting at. I thought you were arguing that they were equal but now it seems like you are saying "it depends on how concepts like infinity are defined", which really doesn't put you in much of a place to tease me about how I'm using the word myself.
 
strawman

strawman

I'm arguing that we've reached the point in the discussion where we should realize we're arguing semantics.

This all depends on whether 0.999... is actually unending. If it never ends, then there is no "infinity place" and your solution doesn't work. It must equal 1 simply because there is no number between 0.999... and 1, therefore they are the same number.

If, however, your version of infinity allows for an "infinity place" and is thus countably infinite then there is a number between 0.999... and 1, thus they are separate and distinct numbers.[DOUBLEPOST=1355349944][/DOUBLEPOST]
Krisken said:
How does math define infinity?
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Which version of math?

:awesome:
 
Chad Sexington

Chad Sexington

Krisken said:
So do mathematicians who focus on different fields punch each over over this kind of thing?
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Well, no, not really. It's just that sometimes you do things with complex numbers, sometimes with real numbers, natural numbers... Someone doing calculus would know that .999...=1 in the real number system, while also knowing that there are systems wherein you can have an infinite infinitesimal number .999... that is not 1.
 
tegid

tegid

So, I(we)'m using infinity as neverending and Mind Detective is using it as very, very large, so large you can't wrap your head around it?[DOUBLEPOST=1355354121][/DOUBLEPOST]
stienman said:
If, however, your version of infinity allows for an "infinity place" and is thus countably infinite then there is a number between 0.999... and 1, thus they are separate and distinct numbers.
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Is that what countably infinite means? I'd have said that the 'regular' 0.999... has a countable infinite amount of 9's. I don't even know if it applies, since it's not a set.
 
strawman

strawman

Well, I'm not sure infinity actually applies here. You can count the number of nines after the decimal place, but it's actually not useful. Sure, there are an infinite number of them, but who cares?

Along the same lines is wondering whether 0.999... is rational. Can it be represented by an integer fraction?
 
Chad Sexington

Chad Sexington

MindDetective I've thought about this and I think I've come to the conclusion I lack the expertise to give a simple enough explanation. I was originally caught off guard that you didn't agree with .999...=1, and thought you just didn't like my example (which I continue to defend as not flawed! :D). I suppose the only argument I have left to make is that if you agree 1/3 + 1/3 + 1/3 = 3/3, do you then think 3/3 is only approximately 1? Or do you think .333... is only approximately 1/3? If you accept that 1/3 = .333... then surely you must accept that 3/3 = .999... and therefore .999... = 1, not approximately but absolutely.

But I think that is essentially covered in this thread and I suspect you feel 1/3 is only aproximately .333... that somewhere in the infinite digits there is a change that accounts for the missing ....1 at the end. I feel like it's something I understand clearly and yet I'm reminded of Einstein's "If you can't explain something simply, you don't understand it well enough." so I maybe just have to admit I'm not the man to explain this to you.

But I'm right, dammit! I'M RIIIIIIIIIIIGHT. :D
 
MindDetective

MindDetective

stienman said:
It must equal 1 simply because there is no number between 0.999... and 1, therefore they are the same number.
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See, this part bugs me. It doesn't seem necessary to me that a number has to be in between them for them to be different. A question that popped into my head as I was driving around doing errands this evening was spurred by seeing this argument on the link you gave. What is the closest number to 1 smaller than 1 that is not 1?[DOUBLEPOST=1355365630][/DOUBLEPOST]
Chad Sexington said:
MindDetective I've thought about this and I think I've come to the conclusion I lack the expertise to give a simple enough explanation. I was originally caught off guard that you didn't agree with .999...=1, and thought you just didn't like my example (which I continue to defend as not flawed! :D). I suppose the only argument I have left to make is that if you agree 1/3 + 1/3 + 1/3 = 3/3, do you then think 3/3 is only approximately 1? Or do you think .333... is only approximately 1/3? If you accept that 1/3 = .333... then surely you must accept that 3/3 = .999... and therefore .999... = 1, not approximately but absolutely.

But I think that is essentially covered in this thread and I suspect you feel 1/3 is only aproximately .333... that somewhere in the infinite digits there is a change that accounts for the missing ....1 at the end. I feel like it's something I understand clearly and yet I'm reminded of Einstein's "If you can't explain something simply, you don't understand it well enough." so I maybe just have to admit I'm not the man to explain this to you.

But I'm right, dammit! I'M RIIIIIIIIIIIGHT. :D
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It only approximates 1/3 because it only ever CAN approximate it. When you take 1 and divide it by 3, you get a 3 with a remainder of 1, which is then divided by 3, etc. After each 3 is a remainder of 1 that still needs dividing. The remainder is essentially dropped when we say "it is 3s for infinity", partly because the way we use infinity is a short hand for rounding. In my mind it is 3 for infinity with a remainder of 1, even at the "end" of infinity (I know, I know, there is no "end"...but that is why the remainder should still be considered as hanging around, I think!)
 
papachronos

papachronos

Chad Sexington said:
Your setup makes no sense. In step two you don't multiply both sides by 10. You add 9 to the left and multiply the right by 10. That's not equivalent.
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MindDetective said:
This is it exactly. It is the exact same problem with your set up.
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MindDetective , the method you used to "break" the proof is incorrect. The initial setup doesn't multiply one side by ten and add nine to the other side and call it equivalent:


if x = a then 10x = 10a is true for every value of a. More generally:

if x = a then (base)*x = (base)*a.

For any rational number (i.e. one that can be expressed as a fraction, be that in numerator/denominator format or by using a radix point), multiplying by that number by its representational number base is equivalent to shifting the radix point one space to the right.

Therefore, x = 0.999... <=> 10x = 9.999... is a valid statement. The only assumption implicit in the second statement is that everything is happening in base 10.
 
MindDetective

MindDetective

papachronos said:
MindDetective , the method you used to "break" the proof is incorrect. The initial setup doesn't multiply one side by ten and add nine to the other side and call it equivalent:


if x = a then 10x = 10a is true for every value of a. More generally:

if x = a then (base)*x = (base)*a.

For any rational number (i.e. one that can be expressed as a fraction, be that in numerator/denominator format or by using a radix point), multiplying by that number by its representational number base is equivalent to shifting the radix point one space to the right.

Therefore, x = 0.999... <=> 10x = 9.999... is a valid statement. The only assumption implicit in the second statement is that everything is happening in base 10.
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You should read the rest of the thread because I've been arguing that the "working" proof does exactly the same thing I did. It only adds 9 to one side and multiples the other side by 10, not multiple both sides by 10.[DOUBLEPOST=1355366735][/DOUBLEPOST]My argument is thus:
.99999...=x
9.99999...=9+x
9.99999... - .99999...=9 and 9+x - x=9
9 = 9
x is still = .99999...
 
papachronos

papachronos

MindDetective said:
You should read the rest of the thread because I've been arguing that the "working" proof does exactly the same thing I did. It only adds 9 to one side and multiples the other side by 10, not multiple both sides by 10.
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That's just it, though - the working proof doesn't do this. Well, okay, it does, but only as a side effect. It takes the more general proof and applies it to a specific value of a.

Working proof: x = a <=> 10x = radixpointshift(a)
Your "broken" proof: x = a <=> 10x = 9 + a

The two methods are incompatible. The second is only valid for a = 1 (which is what you're saying), otherwise algebraic law is broken. The first is applicable to any value of a, and algebraic law is never broken.
 
papachronos

papachronos

MindDetective said:
My argument is thus:
.99999...=x
9.99999...=9+x
9.99999... - .99999...=9 and 9+x - x=9
9 = 9
x is still = .99999...
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Problem with that method is that you remove x from the proof entirely in the third step, so you end up with an identity statement which doesn't prove anything.

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