If you want a headache at 12am

This is the same as arguing whether 1 divided by infinity is equal to zero.

I feel there is no right or wrong answer and it just depends on how you look at things and it depends on whether or not you think things like infinity and .999... are actual numbers or just mathamatical ideas that allow us to put a word to something that we cannot otherwise grasp.

'Infinity' was invented by us as was infinitly repeating numbers. The purpose of it is to help us understand things like calculus and other mathamatical ideas and models that help us in turn understand the real world.

I find it best to allow .9999... to equal 1 when it is needed to equal one and not get too hung up thinking about a number that doesnt even exist.

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Originally Posted by yasunobu13

What is the last piece? What does that mean? The question is does 0.999... = 1?

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no, it doesn't.

the last "piece" is the last number in the infinite line of numbers. only adding this last piece you can have 1.

read my example again, it very much proves that you can have 0.999... but unless you'll add the last "piece", it will never be 1 (and that is what he says, that you don't need the last "piece", that you're good without it.) Do you understand now?

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Originally Posted by yasunobu13

Do you agree that 0.999... is a specific number on the real line? If so, then all the proofs in those links I provided are all true.

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if 0.999... is a "specific number on the real line" (real line?) then I'm right and 0.999... != 1 since 1 is another number on that line.

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Originally Posted by yasunobu13

What specific part of the proofs do you disagree with or do not understand?

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if you'll not agree with me after this post, I'll start with his explanations.

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Originally Posted by MechaPiano

I'll start with his explanations.

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Forget everything from before. Start with this and tell me where the fallacy lies.

A: 1/3 = 0.333... [by definition of division]

B: 3 * (1/3) = 1 [by definition of multiplication and division]

C: 3 * (0.333...) = 0.999... [by definition of multiplication]

D: From A we have 1/3 = 0.333...
Multiply both sides by 3

E: 3 * (1/3) = 3 * (0.333...)

F: From B we have 3 * (1/3) = 1
Giving us 1 = 3 * (0.333...)

G: From C we have 3 * (0.333...) = 0.999...
Giving us 1 = 0.999...

how can any of you argue with the mathmatical PROOF??

Thanks for the headache. :D

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Originally Posted by yasunobu13

Forget everything from before. Start with this and tell me where the fallacy lies.

A: 1/3 = 0.333... [by definition of division]

B: 3 * (1/3) = 1 [by definition of multiplication and division]

C: 3 * (0.333...) = 0.999... [by definition of multiplication]

D: From A we have 1/3 = 0.333...
Multiply both sides by 3

E: 3 * (1/3) = 3 * (0.333...)

F: From B we have 3 * (1/3) = 1
Giving us 1 = 3 * (0.333...)

G: From C we have 3 * (0.333...) = 0.999...
Giving us 1 = 0.999...

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are you insane? how on Earth did you get 0.999... by multiplying 0.333... with 3? (and please don't tell me you used a regular calculator to do it)

edit:
just take a scientific calculator and first, divide 1 with 3 (you'll get 0.333...) and then multiply what you get with 3, you'll see that you'll have 1 (not 0.999...)

OMG do you know what rounding is? you must have a MASSIVE calculator if it can display an infinate number of decimal places!!!!!

what seperates one number from another?

all the numbers in-between.

so if you have 0.9999 to an infinate number of decimal places then there is NO number between it and 1, this is because you cannot put the "last piece", as you call it, on the end, because there is still and infinate number of 9's to put before it!

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Originally Posted by dudeqwerty

OMG do you know what rounding is? you must have a MASSIVE calculator if it can display an infinate number of decimal places!!!!!

what seperates one number from another?

all the numbers in-between.

so if you have 0.9999 to an infinate number of decimal places then there is NO number between it and 1, this is because you cannot put the "last piece", as you call it, on the end, because there is still and infinate number of 9's to put before it!

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I hate to be mean, but we've been trough this - read from the start. you're off the point.

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Originally Posted by MechaPiano

I hate to be mean, but we've been trough this - read from the start. you're off the point.

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maybe, but im still right.

what you said about "real line" (very basic maths) and what gerbic said about asymptotes/limits, is thinking about this in the wrong way.

you are thinking, ok i have 0.9 thats quite close to 1, so if i add a "9" to the end i get 0.99, even closer to 1, but the number we are using here is 0.9999.. to infinate decimal places they are already there, you do not add them one-by-one, it is its own number.

also you seem, from your last post, you are basing your argument on your use of a scientific calculator, computers round numbers of to avoid infinatly calculating answers.

think of this using a proper mathmatics, like all the proofs that have been show, and use your brain, dont rely on graphical interpertations or calculators which have pretty big rounding errors,

zlatan

Quote:

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Originally Posted by MechaPiano

are you insane? how on Earth did you get 0.999... by multiplying 0.333... with 3? (and please don't tell me you used a regular calculator to do it)

edit:
just take a scientific calculator and first, divide 1 with 3 (you'll get 0.333...) and then multiply what you get with 3, you'll see that you'll have 1 (not 0.999...)

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Thats because a scientific calculator recognises the equivilancy between 0.999... and 1, which is exactly the point he is trying to make.

Would you accept that 3*(a+b)=3a + 3b?
In which case you can treat o.333... as 0.3 + 0.03 + 0.003...
Multiply each part by 3 and get 0.9 + 0.09 + 0.009...
Therefore 0.333... multiplied by 3 = 0.999...
The rest of his proof follows as he showed.

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Originally Posted by dudeqwerty

maybe, but im still right.

what you said about "real line" (very basic maths) and what gerbic said about asymptotes/limits, is thinking about this in the wrong way.

you are thinking, ok i have 0.9 thats quite close to 1, so if i add a "9" to the end i get 0.99, even closer to 1, but the number we are using here is 0.9999.. to infinate decimal places they are already there, you do not add them one-by-one, it is its own number.

also you seem, from your last post, you are basing your argument on your use of a scientific calculator, computers round numbers of to avoid infinatly calculating answers.

think of this using a proper mathmatics, like all the proofs that have been show, and use your brain, dont rely on graphical interpertations or calculators which have pretty big rounding errors,

zlatan

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you're off the point again and you're wrong about what I do, how I do it and the whole thing. sorry.

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Originally Posted by MechaPiano

you're off the point again and you're wrong about what I do, how I do it and the whole thing. sorry.

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so why are you wrong then?

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Originally Posted by chris-sharpe999

Thats because a scientific calculator recognises the equivilancy between 0.999... and 1, which is exactly the point he is trying to make.

Would you accept that 3*(a+b)=3a + 3b?
In which case you can treat o.333... as 0.3 + 0.03 + 0.003...
Multiply each part by 3 and get 0.9 + 0.09 + 0.009...
Therefore 0.333... multiplied by 3 = 0.999...
The rest of his proof follows as he showed.

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you would be right if 0.333... had an end, but since it doesn't you can't do that. you need to "translate" 0.333... into 1/3 form and only then make your calculations.

no you dont, they are the same number!!!!!!!!!!!!!!!!!!!

You can do that, you just have to realise that the 9s you get from the multiplication extend to infinity, which is why I included the ...s, you can still multiply in seperate parts, you just need to make sure you add all those parts back together afterwards, giving 0.999...

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Originally Posted by chris-sharpe999

You can do that, you just have to realise that the 9s you get from the multiplication extend to infinity, which is why I included the ...s, you can still multiply in seperate parts, you just need to make sure you add all those parts back together afterwards, giving 0.999...

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I understand, but it's not that - it's just that you can't perform that kind of operation on a number in that form. You can do it if you have 0.25 or 0.9678 or something, but when you have an infinite number you need to "reform" it.

edit:
it's a rule, just like 0! is 1 ;)

No you don't. I'll try and show it in a different way, using a geometric series.
In a general geometric series, you have a "first term" a, a common ratio r, and the term number n. The nth term is a*r^(n-1), and the "sum to n" is (a[1-r^n])/(1-r). So the geometric series here is a=0.3 r=0.1, eg:
S1= 0.3
S2= 0.3 + (0.3*0.1) = 0.3 + 0.03 = 0.33
S3= 0.3 + (0.3*0.1) + (0.3*0.1*0.1) = 0.3 + 0.03 + 0.003 = 0.333

Take this series to infinity (which in this case is obvious but usually you use the forumula a/(1-r))
s∞ = 0.333... with 3s off to infinity

Now multiply this by 3. To do this we only need to multiply each term by 3, so we multiply a by 3. 0.3*3 = 0.9
Constructing the same series using a=0.9 would clearly give us 0.999...
BUT, using the formula a/(1-r) to find the sum as n->∞, we get
0.9 / (1-0.1) = 0.9 / 0.9
And any number divided by itself MUST =1
So therefore from the same infinite series we get 0.999... and 1, therefore they must be equal.

Quote:

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Originally Posted by MechaPiano

are you insane? how on Earth did you get 0.999... by multiplying 0.333... with 3? (and please don't tell me you used a regular calculator to do it)

edit:
just take a scientific calculator and first, divide 1 with 3 (you'll get 0.333...) and then multiply what you get with 3, you'll see that you'll have 1 (not 0.999...)

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You can't bring a calculator into a mathematical proof, at least not of this form. Once you do that, we're talking about discrete math. On a calculator 0.999... actually has an end point (on the order of 1/10^8 or 1/10^32, depends on the calculator).

But we're not dealing with discrete math here, we're dealing with the real number line. On that line, numbers exist smaller than the floating point machine epsilon that calculators and computers have.

Try the multiplication long hand. Start off with a series of 3's and multiply by 3.

Code:

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  0.33333333...
x          3
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  0.99999999...

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If I end it there, then yes, I would have to round up to one or truncate the number at some point. But, the symbol (...) means that the series of 3's (and 9's) continue on forever.

That is how I can multiply 3 * 0.333...

At some point entering it onto a calculator, I would have to stop entering 3's. That makes it discrete and unapplicable.

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Originally Posted by chris-sharpe999

No you don't. I'll try and show it in a different way, using a geometric series.
In a general geometric series, you have a "first term" a, a common ratio r, and the term number n. The nth term is a*r^(n-1), and the "sum to n" is (a[1-r^n])/(1-r). So the geometric series here is a=0.3 r=0.1, eg:
S1= 0.3
S2= 0.3 + (0.3*0.1) = 0.3 + 0.03 = 0.33
S3= 0.3 + (0.3*0.1) + (0.3*0.1*0.1) = 0.3 + 0.03 + 0.003 = 0.333

Take this series to infinity (which in this case is obvious but usually you use the forumula a/(1-r))
s∞ = 0.333... with 3s off to infinity

Now multiply this by 3. To do this we only need to multiply each term by 3, so we multiply a by 3. 0.3*3 = 0.9
Constructing the same series using a=0.9 would clearly give us 0.999...
BUT, using the formula a/(1-r) to find the sum as n->∞, we get
0.9 / (1-0.1) = 0.9 / 0.9
And any number divided by itself MUST =1
So therefore from the same infinite series we get 0.999... and 1, therefore they must be equal.

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:) you're doing the same thing only more complex, but this doesn't change anything. you can't do what you do with an infinite number, you just can't. you can't add it, you can't multiply it, you can't do anything with it as long as it in it's 0.whatever... form. I perfectly understand what you are pushing and it would be correct with any number that is not infinite.

I've proved 1 != 0.999... at least once already,
I'm tired of this and I'm not going to continue unless I'll have some backup (or at least some moral support).

You most definately CAN add/multiply infinite expansions, and there's nothing in that post to prove 1!=0.999...
For example, 0.111... + 0.7 = 0.811111...
or if that doesn't count, 0.111... + 0.111... = 0.222...
Which is exactly what we're doing above (3X = X + X + X) and there is no mathematical reason you can't do that.

ok, my last attempt would be to ask you to draw y=1/1-x function and see for yourself that x can never be 1, but it can be 0.999...

if that doesn't help, then I don't know what will :)

A perfectly valid point, but doesn't prove a thing. A vertical asymptote is different from a horizontal one (well, depending on which axis you define as domain and which as range, usually x:domain y:range). In the case of y=1/1-x you place your own restriction on the domain so you don't end up dividing by zero. It is still fine to say that as x->1, y->∞. In the case of y = 0.9(1- 0.1^[x-1])/0.9 = 1- 0.1^[x-1] (adapting the geometric series formula to work on a graph as y in terms of x) you don't have to place any limit on the domain, but as x->∞, 0.1^[x-1] ->0, y->1. So the limit is 0.999... AND 1. And your definately right that we should stop now, we're just going in circles (probably won't happen though).

graphical methods rely on rounding of infinatly long numbers, what is gonna help you understand that 0.99.. = 1?

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Originally Posted by MechaPiano

are you insane? how on Earth did you get 0.999... by multiplying 0.333... with 3? (and please don't tell me you used a regular calculator to do it)

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I would have to agree with Mech here.

0.333... * 3 does not equal 0.999...

If you multiple 0.333... * 3 there will ALWAYS still be an infinite amount of 3's that have yet to be multiplied by the 3. 0.999... is an infinte number of 9/(10^x) while 0.333... is an infintie number of 3/(10^x) in the process of being multiplied by 3. They're not the same and never will be.

Saying 0.333... * 3 = 0.999... means you've put a limit on 0.333... & 0.999...

0.333... * 3 = 0.999... will never be true unless at some point you stop multiplying 0.333... * 3 and state that they are the same at THAT POINT.

You can multiply an infinite representation. You don't multiply each part as a seperate action, in which case you wouldn't ever have multiplied every term, but the multiplication is one action applied to every term. It is only one action, so you don't have an infinite amount left to multiply because all terms go at once. This is a case where using a fractional representation IS a good way of showing it. Using a different example, so we don't bring the whole 0.999...=1 thing into it:
1/9 = 0.111...
3 * 1/9 = 3 * 0.111...
3/9 = 1/3 = 0.333...
Therefore 3 * 0.111... = 0.333... and that is a perfectly acceptable conclusion.

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Originally Posted by phires

I would have to agree with Mech here.

0.333... * 3 does not equal 0.999...

If you multiple 0.333... * 3 there will ALWAYS still be an infinite amount of 3's that have yet to be multiplied by the 3. 0.999... is an infinte number of 9/(10^x) while 0.333... is an infintie number of 3/(10^x) in the process of being multiplied by 3. They're not the same and never will be.

Saying 0.333... * 3 = 0.999... means you've put a limit on 0.333... & 0.999...

0.333... * 3 = 0.999... will never be true unless at some point you stop multiplying 0.333... * 3 and state that they are the same at THAT POINT.

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How do you people multiply?

Are you saying that, in order to multiply 3 * 0.333... I have to have intermediate steps?

At some point it becomes 0.9333...
Then it becomes 0.99333... and so on?
As though time must pass for any multiplication to occur.

This is ridiculous, what's the highest level math that you and Mecha have had?

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Originally Posted by MechaPiano

I've proved 1 != 0.999... at least once already,

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Actually, you failed to prove (or disprove) anything. What you described was a sum of an infinite geometric sequence. A well known example in fact. An example who's sum is, you guessed it, 1.

http://mathworld.wolfram.com/GeometricSeries.html
1/2 + 1/4 + 1/8 + ...

So r = 1/2. The sum of r^1 + r^2 + r^3 + ... is equal to
r / (1 - r) = (1/2) / (1 - (1/2)) = (1/2) / (1/2) = 1


But, my geuss is that you've had very little experience in that area seeing how you recently denied the definition of multiplication.

I'm done here