More on 0.999~ vs 1: Comparing Numbers

[uart]

ok Robokapp. If you want a really unsophisicated argument then think about it like this. Let's just say for one moment that 0.999... and 1 are different ok. Then by what amount do they differ?

The usual punter will think about this for a while and then reply that they differ by 0.000...1

Now let's look at the "number" represented by 0.000...1, that's a decimal point followed by an infinite number of zeros with a one placed right on the end. Really think about what that means, the zeros go on for ever, they never ever end, and then when they end you want to place a one. Can you see the logical falacy there, the one on the end is pointless, if the zeros never end then you simple *cant* place a one on the end. In other words 0.000...1 is not a valid representation of a number, there's really no point in even writting it, but if we do write it then we must at least agree the it equals zero exactly.

So 0.999... and 1 are different and the amount that they differ by is precisely zero. Ok well maybe we should just say they're equal then. You know it makes sense.

 

[Robokapp]

Well...I was thinking about just that. I started typing a message but I erased it...but here's basically what my answer is:

1/2=0.5
2/3=0.666...
3/4=0.75
4/5=0.8
(...)
99999/100000=0.99999

So...the limit of (x-1)/x as x approaches positive infinity is...1 becasue it's an equal heavy rational equation.
Point is...how do you obtain a 0.9999...?
You go as close as possible to infinity for x in this equation: (x-1)/x

But why I'm showing all this: (x-1)/x will always be subunitray...never equiunitary. (I hope those words actaully exist. I never used them in english. I am taking about top < bottom vs top = bottom).

So...you do pull out a 0.9999... if you go far enough, but you do not pull out a 1 no matter where you go. Isn't that what an assymptote is?
I hope that makes any sense. This topic is getting personal for me because it doesn't take much complex math, yet it's...a curiosity if you will.

 

[uart]

Ok, so you don't understand the concept of infity or limits. You don't understand the difference between infinity and a very large but finte number. That's ok some people just don't have the mental capacity to ever grasp some concepts, no body here will really care if you accept your limitations and move on. Just don't try and use you personal mental limitations to "prove" that accepted mathematical fact is wrong.

 

[StatusX]

You agree that (x-1)/x has a limit of 1 as x goes to infinity. Let's say we use the notation [f(x)] to designate the limit of f(x) as x goes to infinity, which is just a number (if the limit exists). Then [(x-1)/x]=1. It doesn't "approach" 1, it just is 1. (x-1)/x approaches 1 as x gets large, but we have defined [(x-1)/x] to be the limit, which we all agree is 1. So, with this notation I've just invented, you would agree that [(x-1)/x]=1, right?

Well 0.999... is the same thing. We use the ... notation for the limit as more and more 9's are added. At any stage, a number with many 9's will be less than one, but 0.999... does not refer to any of these stages. It refers to the limit, and this is exactly equal to 1.

 

[Hurkyl]

Point is...how do you obtain a 0.9999...?

Obtain... from what? Do you mean the function f(x) = (x-1)/x? You cannot "obtain" 0.999... from f. In fact... if you do a little algebra, you will find that 1 is the only number you cannot "obtain" from f, and thus 0.999... = 1.

You go as close as possible to infinity

There is no "as close as possible to infinity". There are only two possibilities in the extended reals: you are either "at infinity", or there is something between you and "infinity". Or, symbolically...

If x \neq +\infty, then x &lt; 2|x| &lt; +\infty, no matter what x is.



Can we agree on the following two things?

(1) 0.\bar{9} \leq 1
(2) y &lt; 0.\bar{9}, whenever y is equal to zero followed by a finite number of 9's after the decimal point

If we can agree on these, then it's easy to prove that 0.\bar{9} = 1 as follows:

Let x be any number smaller than 1.
Then, 1 - 10^{-k} &gt; x for some positive integer k. For example, we could choose k = \lceil -\log(1 - x) \rceil.

However, 1 - 10^{-k} = 0.99\cdots9, where the r.h.s. has exactly k 9's. Therefore, x &lt; 0.\bar{9}.

This proves that 0.\bar{9} is bigger than any number smaller than 1... and therefore 0.\bar{9} cannot be less than 1. Therefore, it must equal 1.

 

[gnomedt]

Criticize this:

To convert a decimal with a repeating pattern of period R into a fraction, the standard procedure is to form a natural number with the digits of the pattern, divide said number by (10^(R)-1) = 999...^{{R times}}999, and reduce the fraction. For instance:

0.2857142857142857...
=285714/999999 (confirm it)
=(285714/142857)/(999999/142857) (dividing top and bottom by gcd)
=2/7
which is the correction fractional equivalent of the decimal above.

Hence,
0.99999999...
=9/9
=1

 

[HallsofIvy]

Criticize this:

To convert a decimal with a repeating pattern of period R into a fraction, the standard procedure is to form a natural number with the digits of the pattern, divide said number by (10^(R)-1) = 999...^{{R times}}999, and reduce the fraction. For instance:

0.2857142857142857...
=285714/999999 (confirm it)
=(285714/142857)/(999999/142857) (dividing top and bottom by gcd)
=2/7
which is the correction fractional equivalent of the decimal above.

Hence,
0.99999999...
=9/9
=1

Perfectly valid demonstration, but to be a "proof" that 0.999...= 1, you would have to show that the calculation done in, for example, writing 10(0.999...)= 9.999..., is valid. To do that (it is true, of course) you would have to appeal to the definition of base 10 representation of a number less than 1 and the definitions of multiplication and subtraction of such things. It is simpler to use that definition to state that 0.999... means the infinite series \sum_{n=1}^\infty (0.9)^n and then argue that that is a geometric series with sum \frac{0.9}{1- 0.1}= 1.

 

[Robokapp]

The demonstration I was presented with in my 7th grade I think is a little different.

let's say you want 20.05676767676767... the 67s repeating. I chose that example so it shows more than one single step.

you'd pick 200567-2005 (whole - the non-repeating part) and divide by as many 9s as the repeating decimals...in this case 99...and add as many zeros as the number of decimal spots non-repeating...in this case two zero.

so 20.05(67) can be written as...

198562/9900. I have never met this process anywhere outside my country in any class...and i have never had to learn or use it again. However...it's been fustrating me for a lnog time that...exactly like this topic's subject here, it shows that 0.9999=1

it's just...(9-0)/9
I guess I'm terribly wrong. The problem I'm facing is...the word "equal" to me signifies perfect identic things on both sides of the sign, even in different forms. Without going into the math, one would immediatelly assume that 0.999... comes on the number axis a little ahead of the number 1. However I've been proven wrong so...I'm wrong.

I'm sorry if i put too much effort in this topic...it has been on my mind for a while. I had some "personal" struggles with exact form answers of problems like 1/2+2.(9)*5 or something like that.

One final question. Can you really say that you can be 'at infinity'? Reason I'm asking is in my 8th grade, same math teacher told me you never write infinity with a ] and never consider it a solution because it is never going to be reached and if it is it is not applicable. A comment on this?

Thank you for your time.
~Robokapp

 

[d_leet]

198562/9900. I have never met this process anywhere outside my country in any class...and i have never had to learn or use it again. However...it's been fustrating me for a lnog time that...exactly like this topic's subject here, it shows that 0.9999=1

It definitely does not show that 0.9999=1, .9999 = 9999/10000 which is not equal to one, but if we take the notation that .999... represents the infinite geometric series with .9 as the first term and 1/10 as the common ratio then this is equal to one .999... = 1.

it's just...(9-0)/9
I guess I'm terribly wrong. The problem I'm facing is...the word "equal" to me signifies perfect identic things on both sides of the sign, even in different forms. Without going into the math, one would immediatelly assume that 0.999... comes on the number axis a little ahead of the number 1. However I've been proven wrong so...I'm wrong.

i don't understand this. I'm sure that at some point in your mathematical experience that you've had to solve equations such as
3x - 7 = 7x -5
As you can see these two expressions on each side of the equals sign are exactly equal if x = -3 but these two expressions look nothing like each other, so it seems to me that the sense of equality you are thinking of is more "cosmetic" than mathemtical. You also say even in different forms and that is pretty much what .999... is, it is another form of 1 since the two are exactly equal.

One final question. Can you really say that you can be 'at infinity'? Reason I'm asking is in my 8th grade, same math teacher told me you never write infinity with a ] and never consider it a solution because it is never going to be reached and if it is it is not applicable. A comment on this?

Definitely don't take me as the expert on this because I'm not entirely sure, but in my complex analysis class we did make use of a point at infinity and when doing mobius transforms actually mapped this point at infinity to some complex value in the plane, but that could just be an severe informatlity that we are allowed to use because of more advanced mathematics, or something like that.

 

[matt grime]

The reason you don't write [0,oo] is because there is no point on the real line that is (labelled) infinity: it is redundant to allow it as a possibility in the interval. So, in the real numbers, you are, in some sense 'never at infinity'. However, I would absolutely refuse to use such terminology. Just stick to mathematics, and the definitions of the objects at hand.

It is perfectly possible, even useful to include extra symbols such as oo, or i, and extend the reals to other sets which have other/extra properties. Indeed, one of the most useful objects in mathematics is the complex sphere, or the complex plane with the point at infinity added (this is P^1(C) as well), and we can do projective stuff with point(s) 'at infinity'. You just have to stop thinking of these things in the physical manner in which you evidently want to.

You must stop thinking about the real numbers, whatever they may be, as decimal expressions. Those are just representations of real numbers. 1/2=2/4 causes you no problems, does it? Why does 0.999...=1 cause issues?

 

[Robokapp]

It definitely does not show that 0.9999=1, .9999 = 9999/10000 which is not equal to one, but if we take the notation that .999... represents the infinite geometric series with .9 as the first term and 1/10 as the common ratio then this is equal to one .999... = 1.

I wanted to say 0.9999...
I always talked about 0.9999... but i forgot the ... because that's not the notation I learned.
I learned that if you want to write...let's say 0.6767 repeating you write
0.(67) in parenthesys like that. But you all use the ... instead so I tried to use yours but at times I forgot it.

0.9999 is a banality to convert...also an irrelevance.
However 0.(9) using the method I did with the long number as example would look like this:

(the whole number - the non-repeating number)/as many 9s as different repeating decimals followed by as many zeroes as the decimals not repeating.

So...

(09-0)/9 is just 9/9 or 1.