Is 1/3 Really Equal to 0.333...? Find Out the Easier Way with This Trick!

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The discussion centers on the mathematical equivalence of 1/3 and 0.333..., leading to the conclusion that 1 equals 0.999... when multiplied by three. Participants debate the simplicity and clarity of using 1/3 = 0.333... as a starting point versus directly proving 1 = 0.999... through algebraic manipulation. Some argue that the former approach is more intuitive for learners, while others emphasize the need for rigorous proof from first principles. The conversation touches on the representation of rational numbers and the nature of decimal expansions, highlighting that both forms are equally valid yet can be perceived differently. Ultimately, the discussion illustrates the complexities and nuances in understanding infinite decimals and their equivalences.
  • #61
arildno,

No I do not have a problem with that. The fractional or ratio representation of those numbers in terms of a decimal number does not carry the same continued aspects as such numbers like 1/3 does.

My logic was more keen on trying to resolve the discrepencies between fractioanl or ratio representation and the decimal form. The hope was that the representations were more correct.

Ironically, this has a long history to it! The old greeks didn't really like those irrational numbers and they could not trisect an angle. Of course, neither can we!

russ,

I see what you are saying but

1-0.99999...=0 still does not hold true and hence why I introduced the infinitesimal to resolve the discrepency. It is an interesting curiosity, isn't it? This special case where a ratio is not exactly the same as it's continued decimal form.

Halls,

I have to disagree. Perhaps this is a best way of representing the equation:

1.00000...\neq0.99999... because

1.00000...-0.99999...\neq 0

This is the tautology I'm batting around.

Remember these are just some small thoughts of a random poster!:biggrin:

Please excuse any misuse of language, it's Saturns' day!
 
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  • #62
polyb said:
russ,

I see what you are saying but

1-0.99999...=0 still does not hold true and hence why I introduced the infinitesimal to resolve the discrepency. It is an interesting curiosity, isn't it? This special case where a ratio is not exactly the same as it's continued decimal form.
No, it isn't curious at all, because 1-0.9...=0 does hold! If it didn't, then there'd be a number other than 0 that 1-.9... equaled. Since there isn't, it must equal 0.

This thread will probably need to be closed: polyb, you are arguing against definitions here. There's just nowhere to go with an argument like that.
 
  • #63
Polyb,
I have seen and heard your same argument so many times that it is losing its humor.

Here you are, in a forum, arguing basic architecture of the Real number system with mathematicians. An analogy of this is if you were to argue the height of a building with the man holding the blue prints.

A correct approach would be to attempt to ask questions about and learn why what seems obvious to you is wrong.

My history of this is that it does no good to present you with mathematical proofs, because you do not have the mathematical sophistication to understand them. We can talk till we are blue in the face, nothing changes.

So the word is, change your approach or this thread will be locked.
 
  • #64
OK Integral,

Explain why

1.00000...=0.99999...

Please!

My reasoning tells me that for all pratical purposes it is true, in reality that is usually well within the tolerances of observation. But when I look at that equation I see the discrepency of an infinitesimal.

Integral said:
My history of this is that it does no good to present you with mathematical proofs, because you do not have the mathematical sophistication to understand them. We can talk till we are blue in the face, nothing changes.

Please enlighten me if you are so inclined, otherwise how am I to posit that you have an explanation?

Why would this line of reasoning kill the thread? In essence this boils down to the concept of an infinitesimal, which by no means a simple idea.

Does this have more to do with mixing the idea of real numbers with that of integers?

Physical consideration:

Theoretically you can approach the speed of light but only asymptotically, you never reach c. So in the physical concept c does not eaqual 0.999...c. Perhaps this is where I am drawing some of this reasoning and why I am posting the thought.
 
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  • #65
It has nothing to do with an infinitesimal. It has to do with limits; something you don't appear to understand.
 
  • #66
If only we used a base-9 system... there would be no arguments

1/3 = .3 :)
 
  • #67
krab said:
It has nothing to do with an infinitesimal. It has to do with limits; something you don't appear to understand.

elaborate and enlighten please!:biggrin:

If only we used a base-9 system... there would be no arguments

1/3 = .3 :)

Too funny!

Hey, what else is there to do on a staurday night??!
 
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  • #68
My friends and I argued this forever, and came to the conclusion that

1/3 = .333... mathematically simply based on definition, but that there is no such thing as .333... in anything except math, so really it must be right because the only context in which it is possible to have .333... contains the definition that .333... = 1/3
 
  • #69
polyb said:
Explain why

1.00000...=0.99999...
'Cause it's defined that way for the reals. There are no infinitesimally small numbers in the reals, so the two representations point to the same real number, the multiplicative identity.

If you won't take anyone's word here, in addition to the dozens of demonstrations, please read the following completely. It's wriiten by the 1998 Field's Medal winner, Tim Gowers.

http://www.dpmms.cam.ac.uk/~wtg10/decimals.html
 
  • #70
I can easily see hours being dwindled away on this topic.

Can be fun though!

I'm guessing that perhaps the thing here is the fact that the ratio of two whole numbers or integers(?) being defined in terms of a real number. Perhaps this is where I need to exercise some more discrimination.

As for the limiting process, zeno never reaches one, he is always an infinitesimal away. Correct?

I guess Einstein never reaches c either!


The strange thing to me is that certain ratios don't have the '...' and others do. For the ones that do, odd things like this topic arise.

How come I have the nagging suspicion that trisecting an angle has something to do with this?:rolleyes:
 
  • #71
polyb said:
OK Integral,

Explain why

1.00000...=0.99999...

Please!

The decimal representation of real numbers is indeed redundant.
A decimal representation is a mapping from Z into the finite set of symbols {"0","1",..."9"} (if we work in basis 10), which you can call n(z), with the property that there exists a number z0 such that n(z>z0) = "0".
You can define the "partial sum":
S(m) = Sum_{k = 0}^m 10^{z0 - k} n(z0-k)
The real number that is represented by n(z) is then given by the limit:
R[n] = lim_{m \rightarrow \infty} S(m)

We have now two different "decimal representations":

n1 has z0 = 1 and n1(1) = "0", n1(0) = "1", n1(-1) = "0", n1(-2) = "0"...

n2 has z0 = 0 and n2(0) = "0", n2(-1) = "9", n2(-2) = "9",...

It turns out that, according to the above procedure, both n1 and n2 have the real number associated to them equal to 1 = R[n1] = R[n2].

cheers,
Patrick.
 
  • #72
Gokul43201 said:
'Cause it's defined that way for the reals. There are no infinitesimally small numbers in the reals, so the two representations point to the same real number, the multiplicative identity.

If you won't take anyone's word here, in addition to the dozens of demonstrations, please read the following completely. It's wriiten by the 1998 Field's Medal winner, Tim Gowers.

http://www.dpmms.cam.ac.uk/~wtg10/decimals.html

Thanks Gokul, that will give me something to mull over. On brief glance I like his addressing the subtlety involved on this topic.

As for infinitesimals, I realize that there are no infinitesimally small real numbers but these objects are often useful for dealing with things. Please excuse my lack of proper etiquette with this subject but that is why I'm here, to gain some understanding in a chatroom through some random strangers that use pseudonyms!


As for taking others words for it: all I am asking is for is some enlightenment on the subject and hence the posting, so far I have had some pretty curt responses while others have been more helpful. My gratitude is with the later.

The decimal representation of real numbers is indeed redundant.
A decimal representation is a mapping from Z into the finite set of symbols {"0","1",..."9"} (if we work in basis 10), which you can call n(z), with the property that there exists a number z0 such that n(z>z0) = "0".
You can define the "partial sum":
LaTeX graphic is being generated. Reload this page in a moment.
The real number that is represented by n(z) is then given by the limit:
LaTeX graphic is being generated. Reload this page in a moment.

We have now two different "decimal representations":

n1 has z0 = 1 and n1(1) = "0", n1(0) = "1", n1(-1) = "0", n1(-2) = "0"...

n2 has z0 = 0 and n2(0) = "0", n2(-1) = "9", n2(-2) = "9",...

It turns out that, according to the above procedure, both n1 and n2 have the real number associated to them equal to 1 = R[n1] = R[n2].

cheers,
Patrick.

Thanks Patrick, so now I have to ask:

So can we say R[n1]-R[n2]=0? Perhaps only in the case where the partial sum's upper bound goes to infinity?
 
  • #73
I find it interesting that , "because it is obvious" is your only proof, while from us you demand formal proof. Perhaps you should try proving your claims.

Here is my version of the ]proof[/URL] if you consider my construction carefully you may get a glimpse of why this equality holds.
 
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  • #74
polyb said:
As for infinitesimals, I realize that there are no infinitesimally small real numbers but these objects are often useful for dealing with things.
Look into the hyperreals then. Within the reals though, the only infinitesimally small number is 0.
 
  • #75
Integral said:
Here is my version of the ]proof[/URL] if you consider my construction carefully you may get a glimpse of why this equality holds.

That's neat :approve:
 
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  • #76
vanesch said:
That's neat :approve:
Thank you, I think so too. A significant point of interest is that I never do arithmetic on digits "at infinity" unlike the usual demonstrations (3 * .333... etc. ) . The worst part is having to resort to the nested interval theorem, This is where the lack of mathematical sophistication that I mentioned in my earlier post comes in. I have really only proved this for those who understand that theorem, ie preaching to the choir.

The key to understanding this is that the symbol .999... means an infinite number of 9s. Not some finite number of 9's approaching infinity but an infinite number. Perhaps the lack of understanding comes from the concept of "you can never reach infinity" While this is true in a physical sense, mathematically we can represent the concept of having reached infinity, one way to express this is the ellipsis at the end of a pattern of digits.
 
  • #77
In skimming over these posts I have seen people argue with the evidence that: .999... + 9 = 9.999...

Is this correct math?

By the way, I understand the answer to the subject question of .999... = 1, I just was wondering the above question as a separate topic.
 
  • #78
Hi, forgive me if I'm being redundant...

Whatever happened to the uber-simple:

x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1

?? If that hasn't been discussed yet, is it "correct"?
 
  • #79
Mathematically Legit?

Human Being said:
Hi, forgive me if I'm being redundant...

Whatever happened to the uber-simple:

x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1

?? If that hasn't been discussed yet, is it "correct"?

This is basically what I am asking about... is this a legitimate "proof"? It seems like there is something wrong about it that I can't quite see... like it is right because it is not actually proving anything, but only manipulating its own given information? Does anyone know?
 
  • #80
x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1

Sure this is correct, but it is not a proof it is a demonstration. The Theorems of Real Analysis which show that to be meaningful are the same ones which are needed in the proof of 1 = .999... Mainly you need to show that .999... is indeed a fixed number. if you assume that is a legal operation you have essentially assumed the result, therefor you are using circular logic.

Again, the proof I have posted above is nearly fundamental. First I show that .999... is locked inside of a set of nested intervals. I have specified the theorem used to prove that there can only exist 1 real number in the intersection of a infinitely nested set of intervals. The fact that both 1 and .999... exist inside of the SAME set of nested intervals means that the occupy the same location on the Real Number line, this means that they are equal.
 
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  • #81
Human Being said:
Hi, forgive me if I'm being redundant...

Whatever happened to the uber-simple:

x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1

?? If that hasn't been discussed yet, is it "correct"?
It's correct because it works. It's not obvious though, how one multiplies numbers that are represented by infinitely long sequences of digits, even if you're just multiplying by 10.
 
  • #82
Most importantly, this example is a good demonstration of why mathematicians define the decimals the way they do.
 
  • #83
Thank you to everyone that responded to correct my thinking on this topic. I apologize for the delayed reply. I now understand why you say 1=0.999...! This short discourse has given me a opportunity for an overdue, albeit brief, review of limit and number theory. THANK YOU!:biggrin:
I would like to review my thinking in order to make sure my thought is correct.
Using vanesch's suggestion, it appears that I was constructing both numbers as follows:
For 1.0..., I was thinking
S(m)_{(1.0...)} = \sum^m_{k=0}10^{z_{0}-k}n(z_{0}-k)
where
n_{1} has z_{0} = 1 and n_{1}(1) = "0", n_{1}(0) = "1", n_{1}(-1) = "0", n_{1}(-2) = "0"...
and for 0.9..., I was thinking
S(m)_{(0.9...)} = \sum^m_{k=0}10^{z_{0}-k}n(z_{0}-k)
where
n_{2} has z_{0} = 0 andn_{2}(0) = "0", n_{2}(-1) = "9", n_{2}(-2) = "9",...
Of course the flaw in my thinking had more to do with the fact that it was finite and I was only doing a patial sum. Hence why I said 'it is obvious', at least so I thought, that:
S(m)_{(1.0...)} \neq S(m)_{(0.9...)}
because
S(m)_{(1.0...)} - S(m)_{(0.9...)}\neq 0
which is true if you are only considering a finite or partial sum. This led me to conclude:
S(m)_{(1.0...)} - S(m)_{(0.9...)} = p^m
Where we would could say for this case that p=0.1, which would be true as long as the sum was partial. But this is not the case, for as pointed out by vanesch. For in order to qualify these as real numbers, the limit of m had to go to infinity, or
R(n)=lim_{m \rightarrow \infty} S(m)
in which the difference that I was seeing would go to zero, or
p^m \rightarrow 0
I suspect that this number p^m is correlated to the nested intravel http://home.comcast.net/~integral50/Math/proof2a.pdf" that Integral so thoughtfully provided(BTW, thanks Integral, this really helped me wrap my mind around this nugget plus it was quite neat! :biggrin:).
So there you have it, hopefully I have corrected my thinking. If you think that further correction is needed please post your thoughts!
There is one thing I did come across that would finalize it for me, which is the following:
Does
lim_{m \rightarrow \infty} [1^m = (0.9...)^m]?
Other thoughts:
Regarding the trisection of angle, after some thought I came to realize what had been nagging me. Though one can bisect an angle it seems that the only way to trisect an angle is to use a limiting proceedure of bisections. If anyone has any thoghts please let me know. I am also curious as to wether any of the greeks had worked out limiting proceedures to any extent.
As for the physical case of the speed of light and Einstein's wish to go that fast, I am curious how fast he would have to go in order to be in that intravel such that nature agrees to say c=0.99...*c. Are these the same energies needed to create a black hole? Is this a feasible thought? I realize that this is another topic that belongs elsewhere but I had to finish the thought.
Once again THANK YOU everyone that helped in this exercise!
http://simpler-solutions.net/pmachinefree/thinkagain/comments.php?id=1589_0_3_0_C"
 
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  • #84
It takes an infinite amount of energy to accelerate something to c, hence it's not possible.
 
  • #85
I digressed!:rolleyes:
 
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  • #86
ok, the math works by definitions
but can it exist that A not= B, but A infinitely close to B?
I conjecture something like that does exist in reality.o:)

Selak
 
  • #87
Selak3 said:
ok, the math works by definitions
but can it exist that A not= B, but A infinitely close to B?
I conjecture something like that does exist in reality.o:)
Selak
Define "infinitely close" in a rigorous manner and we'll see. :smile:
 
  • #88
Selak3 said:
I conjecture something like that does exist in reality.o:)
Selak
What has "exist in reality" to do with this? :confused:
 
  • #89
I guess the nature of reality does have something to do with this.

Is there a part of the universe that is indefinitely divisible? Ie: you keeping cutting
a part of reality in half but it never complete disappears? (ie, you get infinitely close
to 0 but never quite reaching it).

Would the definition of 1/x as x approaches to infinity need to be modified?
Would one need to modify mathematics in these cases?
Would one want mathematics to reflect reality?o:)
 
  • #90
No, the "nature of reality" has nothing whatsoever to do with mathematical definitions.
 

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