Math problem

[Farbror]

Uh, yeah, I kinda need help with a math problem. Listen closely now...

Let's say we have a number, 0.999999.... and so on, with an infinite number of nines. Let's call that number X.

X=0.99999...

If we multiply that by ten....

10X=9.999999999...

With me so far? Okay... now take 10X minus X.

10X-X=9X=9.99999....-0.9999999....=9

9X=9

X=1

But that's not true! We earlier said that X=0.99999....! That means that 1=0.99999999........., which we know isn't true!

Where have I done wrong?

[Ryam BaCo]

infinite - infinite could be zero.

9.999999... has infinite numbers after the comma, 0.999999999... too

in fact the difference between infinite and infinite is something. it could be 8. it could be 12. it could be 0. nobody knows. because nobody knows how infinite infinite is.

that was your error.

got all the austrian cleverness?

[ysbreker]

something like that is called a rounding error

[ragnar]

Uh, yeah, I kinda need help with a math problem. Listen closely now...

Let's say we have a number, 0.999999.... and so on, with an infinite number of nines. Let's call that number X.

0.99999... with an _infinite_ number of nines is _equal_ to 1.

[Erkki]

0.99999... with an _infinite_ number of nines is _equal_ to 1.

No it isn't. It is nearing 1. (don't know if "nearing" is right in english) 0.99999... -> 1.

---

Your mistake was that you assumed 0.9999.... and 0.999999 times 10 have the same ammount of 9's after the period, while in fact 0.99999 times 10 should have one less (i'm not 100% sure though).

[Ryam BaCo]

0.99999... with an infinitive number of nines can't be the same as 1 because 1 is 1.00000... with an infinitive number of zeros...

you can't subtract infinitive numbers, because an infinitive number is not equal to another infinitive number.

example: infinitive - 8 is still infinitive

so infinitive - infinitive could be everything.

and 9.9999999... - 0.999999... could be something like 9.00000000...0001 which is not 9

to clear my first post up.

[Farbror]

Yeah, I agree that you cannot use normal mathematics on infinity.

But we haven't got infinity! We just have an infinite number of decimals! It's not the same thing. And we know that the first decimal in both number is always nine, the second one is always nine, we know that all decimals are identical in the both numbers, and the same number minus the same number equals zero. Infinity minus infinity doesn't equal zero, beacuse normal mathematics doesn't work with infinity. But we haven't got infinity!

Your mistake was that you assumed 0.9999.... and 0.999999 times 10 have the same ammount of 9's after the period, while in fact 0.99999 times 10 should have one less (i'm not 100% sure though).

...no...

No I don't think so, they both have an infinite number of decimals. If we multiply pi by ten, we get 31.41.... but we still have an infinite number of decimals, not one less in Pi times ten.

And likewise, Ryam, if you take Pi minus Pi, you should definately get zero, beacuse they're the same number, even though they both have an infinite number of decimals.

This is tricky...

[Wormsie]

I think it's a butterfly.

[Esseb]

9.99999....-0.9999999....=9

is false

Note that if you multiply 0.99999... with 10, the result has one less digit to the right of the decimal. One less than infinite, basically. (But still infinite, if you want to split hairs about it)

Doing the math with a finite number of digits to the right of the decimal (there's a word for those, I'm sure), makes it easier to see:

X = 0.999

10X = 9.99

10X - X = 9X = 0.999 - 9.99 = 8.991

9X = 8.991

X=1.123875

Edit: It seems Erkki and others already pointed this out, never mind then.

[Farbror]

I find it hard to grasp the concept "Infinity times ten is more than infinity times one", but I guess it's the only explanation. Thanks.

[Twilo]

I proved a similar problem to this before with an infinite series, I'll get back to you on that

Update:

OK, here it is:

I don't have the fonts for proper notation here, so I'll try and approximate the characters. The function is sum of all the terms in a sequence from n to p.

In this case, n is 1 and p is infinity, so this is the sum of all the terms of the form:

9(10^(-n)) or 9 times 1/10^n

for various n's, this might be 0.9, 0.009, 0.00000000009 etc

Now if we add all of these up we get 0.99999.... recurring.

The function in standard notation looks sort of like this:

infinity

S (9*10^-n)

n=1

"S" is sigma, meaning sum of. I don't know if there's ascii for that.

We will call this function X as you have.

now, 10X =

infinity

10*(S(9*10^-n))

n=1

(ie, 10 times the Sigma Function)

And then if we substract X from that, we get:

infinity

9*(S(9*10^-n))

n=1

(ie, 10 times the sigma function, minus 1 times the sigma function)

or

infinity

S(81*10^-n)

n=1

which produces, rather nicely, 8.99999999999...

(ie: 8.1 + 0.81 + 0.081 + 0.0081 etc)

which is exactly what we should expect.

[Farbror]

Neat! Thanks!

[Erkki]

Yeah, that seems right.

[ragnar]

0.99999... with an infinitive number of nines can't be the same as 1 because 1 is 1.00000... with an infinitive number of zeros...

It _is_ the same. Why? Because there is _no_ number so small that you can subtract it from 1 and get a larger number than 0.999... (with those infinite number of nines).

[ragnar]

To make the correct proof of the original problem:

                 inf          
                 ---   9      
                 \    ---
0.9999999.... ==  /      k     
                 ---  10      
                 k=1          


inf              inf      
---   9          ---   9  
\    --- * 10 =  \    --- 
/      k         /      k 
---  10          ---  10  
k=1              k=0      

inf          inf           0     
---   9      ---   9      ---   9       9       9 
\    ---  -  \    ---  =  \    ---  =  ---  =  ---  =  9
/      k     /      k     /      k       0      1
---  10      ---  10      ---  10      10 
k=0          k=1          k=0     

This says that we have 10X-X=9, i.e. 9X = 9, i.e. X=1

Q.E.D.

[Twilo]

this is aberrant because the series has a limit at 1 as k goes to infinity

This is quite different to being equal to it.

[ragnar]

this is aberrant because the series has a limit at 1 as k goes to infinity

Yes, k goes to infinity. That's precisely the point of my proof and it has a limit at 1. It _is_ 1.

Try read a math book about limit theory.

[ragnar]

You might want to read this page:

http://www.math.utah.edu/~carlson/teaching/calculus/series.html

[Twilo]

The whole point of my proof (which is not an incorrect proof) is to illustrate to farbror how you get an 8.99.. as he expects.

8.99.. is, of course, limited at 9.

Is 0.999... equal to 1? No. It's a series that approximates it. My own favourite "book about limit theory" illustrates this finely as part of its definition of a limit:

"..in finding the limit of f(x) as x approaches a, we never consider x = a." -Calculus - J. Stewart

The series can't consider 0.999... as it can't consider k = (infinity), instead we consider a limit, which is "arbitrarily close" to 0.999... (and as we don't know of any number > 0.999... and <1, then 1 is as close as we can get.)

Less pretentious please.

[Huz]

Everyone shut up.

[ptdc]

I have a math problem. My problem is that it's maths and not math.

[Twilo]

If you want something between 0.999... and 1, try:

  inf
  ---
  \
1+ /    (10^-k) - (10^-(k+1))
  ---
  k=1

the series converges on 0 as k tends to infinity, but the sum is always greater than it.

Adding this series to 0.999... gives 1