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0.9 repeater = 1

I guess you're right, 1^infinity does raise some issues. For example, as n -> infinity (1 + 1/n)^n does not equal 1, but rather equals e.

So, let's try using the L'Hopital rule and see if it makes things any clearer:

lim (x -> infinity) 1^x
lim (x -> infinity) der(1^x)
lim (x -> infinity) (ln(1))*(1^x)
lim (x -> infinity) 0 * (1^x)

OK, I guess that didn't help. How about this: The infinite sum as n goes from 1 to infinity of (1/n)^2 is not infinity. That means that as n approaches infinity, the value of (1/n)^2 must approach 0. So, that means that lim (n -> infinity) (1/n)^2 = zero. Now, for every value of n, the value of (1/n)^2 is greater than the value of (1/10)^n. So, if lim (n -> infinity) (1/n)^2 = zero and lim (n -> infinity) (1/n)^2 >= lim (n -> infinity) (1/10)^n, than lim (n -> infinity) (1/10)^n = 0.

Or, in other words, I hope that if I bury you in math, you won't notice any errors that I make.
 
BillyJoe said:
(abc...n(GETS THERE AT INFINITY = NEVER)/(999...9). So 0.9 repeating(GETS THERE AT INFINITY = NEVER) = 9/9 = 1.

regards,
BillyJoe
Click to expand...

I'm confused BillyJoe. What, in this context, does your phrase "gets there at infinity" mean? .999... is a number, it doesn't change, it doesn't 'get there' or go anyplace.... can you explain to me what you mean?
 
SquishyDave: Ok 0.9 repeater equals one, thats standard stuff, so who can show me a proof for it?
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Several people have already posted some of the usual demonstrations of this fact, but they are not considered formal proofs in mathematics. rwald was on the right track.

One of the standard proofs uses the geometric series theorem, which says that the sum of the infinite series <nobr>a*(k^n)</nobr> where n goes from 0 to infinity, is equal to <nobr>a/(1-k)</nobr>.

Another way of expressing 9.999... is to use the above infinite series with a=9 and k=1/10. Since <nobr>-1<k<1,</nobr> this is a convergent series and according to the theorem the sum is equal to 9/(1-(1/10)) = 10.

Thus we have proved that 9.999... = 10. Now subtract 9 from both sides. End of proof.

BillyJoe: In other words. 0.999999999999999...... NEVER equals 1
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Sorry, BiilyJoe (and everyone else who said the same thing), but if you gave that answer on a university exam, it would be marked wrong.

0.999... is exactly equal to 1. And the proof of this is airtight.

Another infinite series that is exactly equal to one is:

1/2 + 1/4 + 1/8 + 1/16 + ...

And for exactly the same reason.
 
For those still not convinced that .999... is eactly equal to one, I have a question -

What is 1 - .999... ?

Here's another way to ask the same question -

If 1 > .999... then name a number between the two.
 
Heh, I just did it on my calculator (only not with the infinite progression) - I thought it was going to be 0.111..., but it was 0.000...1! And, as someone pointed out earlier, an infinite progression of one digit ending with another doesn't make sense. So yeah, it's essentially 0. I've said it before and I'll say it again, I am not a mathematician.

edited to take out my wrong maths!

I'm confused!
 
Right, I'm not going to sound very clear here, and that reflects the fact that I'm not very clear, but...

it looked to me like 1 - 0.999... was going to equal 0.111... because each of those 9s in the progression needs a 1 above it to take it back up to the next decimal place up, if you see what I mean. So why isn't the answer 0.111...?
 
xouper said:
For those still not convinced that .999... is eactly equal to one, I have a question -

What is 1 - .999... ?

Here's another way to ask the same question -

If 1 > .999... then name a number between the two.
Click to expand...

actually Xouper, just to show you that there's always a quibble, try reading around that topic of "non-standard analysis". Its a very powerful tool in mathematics - much more powerful when talking about limits, infinitesmals etc, than standard analysis (epsilon/delta type stuff). Until a few years ago the field was somewhat controversial, but is now AFAIK generally accepted....
 
Surely its 0.999 repeater tends towards one.

The proof of 1/3 + 1/3 + 1/3 = 3/3 = 1, yes, but 1/3 cannot be accurately described in a decimal system.

for all intents and purposes, yes it can be given as 1, but it isn't, so there, the problem lies with the use of the decimal system.

use hex, you know it makes sense.
 
Kullervo said:
Yeah me I have a question. Speaking of powers of infinity, this one caused some amusement a couple months back.

evaluate (1 + 1/n)^n as n approaches infinity.

Intuitively, 1/n approaches 0 as n approaches infinity, so 1 + 1/n becomes closer to 1, and 1 to any power is 1.

So, the limit has to be 1, right?
Click to expand...

The "intuitive" approach fails because (1 + 1/n) ^ n is NOT the same as (1^n) + (1/n)^n. 1/n approaches 0 as n approaches infinity, as does (1/n)^n. But (1+x)^n is always greater than (1 + x), no matter how small x is. As n gets larger, 1/n gets smaller, but (1 + 1/n) gets multiplied by itself more times and gets larger each time...

Consider the result for a few values of n (to 2 decimal places):
When n = 1, result = 2
n = 2, 2.25
n = 3, 2.37
n = 4, 2.44
n = 100, 2.70
 
Captain_Snort: The proof of 1/3 + 1/3 + 1/3 = 3/3 = 1, yes, but 1/3 cannot be accurately described in a decimal system.
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Yes it can. There is no "problem" with decimal representation.

.333... is exactly equal to 1/3

.333... is shorthand notation for the infinite series <nobr>3*(1/10^n)</nobr> as n goes from 1 to infinity.
 
Tez: actually Xouper, just to show you that there's always a quibble, try reading around that topic of "non-standard analysis".
Click to expand...
Are you suggesting that non-standard analysis has a quibble with .999... = 1?
 
BillyJoe

I think Xouper nailed it with his proof, one that is actually provable.

As ceptimus said, it is just two different ways of expressing the same number. That's all.

1/3 = 3/9 = 0.3 repeated to infinity

3/3 = 9/9 = 1 = 0.9 repeated to infinity

It's all the same number just written differently.
 
xouper said:
Are you suggesting that non-standard analysis has a quibble with .999... = 1?
Click to expand...

Not really with the conclusion, however, if you define 0.99.. through the standard limit mechanism, then NSA will have a problem with "how many" numbers are between 0.99... and 1. i.e. in standard analysis there'd be *no* numbers between the two (which is the essence of your post and normally what we mean by equal!) my (limited) understanding of NSA is that there'd still be an infinite number of NS numbers between the two...

I've never had the chance to chat with an expert on NSA - all I know about the subject is basically what i've learnt from others who dabble in it for physical reasons...
 
People seem to have an easier time accepting that 0.3333... = 1/3 than that 0.9999... = 1. The explanation is the same for both, though.

The value of the infinite decimal 0.3333... is defined to be the limit of the infinite sequence 0.3, 0.33, 0.333, 0.3333, ... . The limit of that sequence is 1/3 because no matter how close you want to get to 1/3, you can get that close to 1/3 by going far enough in the sequence and you'll stay that close or closer no matter how much further you go in the sequence. The same cannot be said of any number other than 1/3. (A sequence can have at most one limit.)

An exactly analogous argument shows why 0.9999... = 1.

You can ask why a particular definition was chosen, or how a particular result follows from a particular definition; but it makes no sense to claim that something is false, if it's true by definition.
 
Good point, 69dodge. You're going by a strict definition-of-limits proof, which is equally as valid as any any other method thus far used. So, is there anyone who still doesn't understand this concept?
 
Kullervo said:
Yeah me I have a question. Speaking of powers of infinity, this one caused some amusement a couple months back.

evaluate (1 + 1/n)^n as n approaches infinity.

Intuitively, 1/n approaches 0 as n approaches infinity, so 1 + 1/n becomes closer to 1, and 1 to any power is 1.

So, the limit has to be 1, right?
Click to expand...

No, it doesn't work that way. Just like when you calculate derivatives using limits, you can't take the limit of one part first (say, the numerator going to zero) and then the second part (say the denominator) and expect to get the right result (which for derivatives would give you zero every time). You really need to take the WHOLE thing together.

Someone can correct me if I'm getting this wrong, but I believe what you get is:

(1+1/n)^n = 1 + (n!/(n-1)!1!)(1/n) + (n!/(n-2)!2!)(1/n)^2 + (n!/(n-3)!3!)(1/n)^3 + ....
(n!/0!n!)(1/n)^n

(this is complete, no limits yet)

The first term after 1 becomes,
(n!/(n-1)!1!)(1/n) = (n/1)(1/n) = 1

The second term becomes
(n!/(n-2)!2!)(1/n)^2 = (n*(n-1)/2)(1/n)^2 = (n-1)/n*(1!/2!)

In the limit of n approaching infinity, this becomes
just (1^2)/2!
Now if we take the limit of the whole thing, it should be at least believable that you get:

1 + 1/1! + 1^2/2! + 1^3/3! + ... = e^1

Make any sense? I kept the 1^x because you can do the same thing with (1+x/n)^n and get e^x.
 
I suppose there is a name for numbers that can be written exactly in decimal notation with a finite number of digits?

I don't know what the name is, but I'll call such numbers, 'r-numbers' for now.

Clearly, any irrational or trancendental numbers, like e, pi etc. are not going to be r-numbers.

There are also an infinity of rational numbers like 1/7, 1/3 etc. that are not r-numbers either.

Now, with just one exception, any r-number can also be written in a different form, terminating in an infinite series of 9s. for example, -12.3456 can be written as -12.3455999999...

Note that if trailing zeros are ignored, then there are just two ways of writing an r-number - the normal way, and one with an infinite number of 9s.

The interesting exception - an r-number that can be written in just one way instead of two is the number zero.

ceptimus.
 
The word you're looking for is "terminating decimals." It's true to say that all rational numbers can be represented by a repeating decimal or a terminating decimal. Of course, if you consider the 0 af the end of a terminating decimal to be repeating indefinitely, than all terminating decimals are actually repeating decimals.

In other words, 12.345 = 12.345000000000...

Irrational numbers (such as pi, e, sqrt(2), etc.) are those which cannot be represented as either repeating or terminating decimals.
 
scribble said:
I'm confused BillyJoe. What, in this context, does your phrase "gets there at infinity" mean? .999... is a number, it doesn't change, it doesn't 'get there' or go anyplace.... can you explain to me what you mean?
Click to expand...
My phrase was GETS THERE AT INFINITY = NEVER meaning that you never get there.

.999... means that you keep tacking on another 9......no, don't stop, keep tacking on another 9......yep, keep going......KEEP GOING.....
 
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