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Does 0.999999999........ = 1?

Does 0.999999999........ = 1?

Posers and Puzzles

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Originally posted by obvek
so am i right, wrong, or in-between? I'm not talking just about pesos, of course, but about all types of money, like the franc, which is no longer used, or the euro, which has replaced the franc and many other currencies, too.
Sorry, I was having a laugh, because the point of the story is that he can buy beer indefinitely because of the tangled loop in the exchange rate. I'm not sure I quite understand what you said at first. Could you explain again for my benefit?

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What I said was, if the pesos or some other form of money equaled another form of money, could it possibly equal .9999999..., and would this be another way to answer the question originally asked in this forum?

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Originally posted by obvek
What I said was, if the pesos or some other form of money equaled another form of money, could it possibly equal .9999999..., and would this be another way to answer the question originally asked in this forum?
I don't see specifically why it has to be money, but in fact the Argentinian unit of currency (I forgot what it's called) was worth $0.999... for quite a while IIRC.

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of course it isnt, you cannot say that one number is equal to another number. that is like saying "does 1=2?". 0.999999... does not equal 1.

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Originally posted by schaef
of course it isnt, you cannot say that one number is equal to another number. that is like saying "does 1=2?". 0.999999... does not equal 1.
I'd recommend reading the rest of the thread.

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Originally posted by THUDandBLUNDER
YES or NO?

.
0.999999999999.........=1

only if:

a)2=3,which isnt
b)you are nuts,which is true.

hence true!

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Originally posted by fisher154
0.999999999999.........=1

only if:

a)2=3,which isnt
b)you are nuts,which is true.

hence true!
I recommend reading the post before yours (and doing what it says) 😀

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Originally posted by fisher154
0.999999999999.........=1

only if:

a)2=3,which isnt
b)you are nuts,which is true.

hence true!
a) Proof given in post is true, hence 0.999... = 1

b) As result you are a nut.

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Originally posted by TheMaster37
a) Proof given in post is true, hence 0.999... = 1

b) As result you are a nut.
here what i think is the absolute final proof...

,99... can be rewritten as:

[9/(10^1) + 9/(10^2) + 9/(10^3) + ... + 9/(10^n-1) + 9/10^n]

where n APPROACHES infinity.

The LIMIT of .999... as n approaches infinity is 1.

HOWEVER!!! n never can actually reach infinity because a number n+1 can always be found... so while .999 becomes infinitesimally close to 1, it never can actually equal 1.

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Originally posted by Gambitzoid
here what i think is the absolute final proof...

,99... can be rewritten as:

[9/(10^1) + 9/(10^2) + 9/(10^3) + ... + 9/(10^n-1) + 9/10^n]

where n APPROACHES infinity.

The LIMIT of .999... as n approaches infinity is 1.

HOWEVER!!! n never can actually reach infinity because a number n+1 can always be found... so while .999 becomes infinitesimally close to 1, it never can actually equal 1.
actually i was wrong...

.99... repeating =

the limit as n approaches infiniity of

[9/(10^1) + 9/(10^2) + 9/(10^3) + ... + 9/(10^n-1) + 9/10^n]

that limit equals 1 which is equal to .999...repeating

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Originally posted by Gambitzoid
actually i was wrong...

.99... repeating =

the limit as n approaches infiniity of

[9/(10^1) + 9/(10^2) + 9/(10^3) + ... + 9/(10^n-1) + 9/10^n]

that limit equals 1 which is equal to .999...repeating
*bows*

There you are, you understand 🙂

The proofs in this thread are somewhat more cumbersome, as some people do not like basic proof involving limits.

You could have saved yourself alot of typing by writing 0.999... as 1- 10^(-n) with n tending to infinity 😛

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Well, after reading the original post, and a few there after it. I decided to explain my reasonings here.

Note:I did not read many posts(8 pages was a bit to tedious)

This whole problem derives from the assumption that 1/3 =0.333333...So on and so on. But the problem is that .33333... is not equal to 1/3. It is merely as close as possible. There is not finite number that can represent 1/3.

The problem is that attempting to multiply the answer to 1/3 by 3 will give you .99999999...ect. ect. Many represent this number as 3/3, which equals one. As before, this number in NOT equal to 1, but is the closest number possible to being one.

I have my own logic that I work through, I just merely attempted to explain the workings of my head to you. If it didn't make sense, I'm sorry, but I can't really put that into clearer words.

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You're close 🙂

From what i've heard of it, Non-standard analasis allows for infinitely small numbers to exist. In that NS-analasis you are completely right.

However, in the standard analasis those kind of numbers do not exist. You said it yourself; 0,999... is the closest number to 1. The difference is smaller then anything positive you can write down, hence the difference must be 0.

The point where you make a 'mistake' in standard analasis it that 1/3 is exactly 0,333... The reason for that is the dots. They say that the row of 3's is infinite. No matter how many you write down, the actual number has more 3's. An approximation for 1/3 is 0,33 or even 0,33333333, however 0,333... is the exact value.

If you have problems with what i said about the difference being 0, look back to one of my earlier posts, i gave a nice proof about it 🙂

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no it is the greatest number lesser than one

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Oh really? How about 1/2 then?