Is 1 = 0.9999......

[MadRat]

So I can take that namecalling as your resignation from the argument?

 

[Darien]

Originally posted by: SilentRunning

Originally posted by: Darien

Q: How many mathematicians does it take to convince computer programmer types that 0.999... was indeed the same as 1 at Kyteland's workplace?


A: Three, but two of them must have PHD's and ignore that fact that wasn't the question that was asked.

You need to understand numbers to realize that .9999... = 1

Why are you replying to my tag line? Can't you read?

Nope

 

[spidey07]

LOL.

No more to add. I'm just smug that I understand basic science/math but still struggle with it's and its.

Hey, at least my smarts were in the "right" part of the brain.

 

[silverpig]

Originally posted by: MadRat


Unfortunately .999... cannot be represented on the number line meaning then it doesn't exist.

OMG. Any value can be represented on the number line (until you go into complex numbers, and then it's a plane, but that's beside the point). Any single value you can think of. If you give me any number, I can put it in between two numbers on the number line.

Originally posted by: MadRat


At the same time if you accept that .999... does exist then there must be a difference between 1 and that number.

Why are you saying there's a difference if you don't even think 0.999... exists?

Originally posted by: MadRat


To say that they are the same is a copout because to define .999... as anything but 1 means a difference exists. By the definition of .999... it does not equal 1 and therefore there must be a difference bewteen them.

How is that a copout? Are 4/2 and 2 not equal? Following your logic, by writing 4/2 not as 2 means a difference exists. There must then be some difference right?

Originally posted by: MadRat


Since you claim that no difference exists (because you say that .999...=1) then it also means the .999... value also does not exist.

Okay, so if there's no difference between 4/2 and 2, then 4/2 must not exist?

Originally posted by: MadRat


If you take any number and times it by a product of two you'll end up with an even result.

That is so wrong it's not even funny. If you multiply any INTEGER by 2, you will get an even number. Are you saying 2*pi is even? How about 2 * 1/5?

Originally posted by: MadRat


You've only shown how to make .333... into 1/3 with hocus pocus of using a limit as the definition of the product to gain an even numbered product, which cannot happen.

So you're saying that 1 is even now are you?

Originally posted by: MadRat


You cannot add three equal values to define an even product because that product does not exist; every product of 3 will equal an odd numbered result.

2 + 2 + 2 = 6

6 is even. I added 3 equal values and got an even um... product (product is usually reserved for the result obtained from a multiplication operation).

So 6 doesn't exist now? You're taking apart the entire bloody number line now.

Oh, and every product of 3 will NOT always equal an odd numbered result. 3 * (any even number) will ALWAYS be even.

Originally posted by: MadRat

The answer lies before the limit, not at the limit. The limit lay outside your product by definition.

0.999... = lim(n->inf) [Sum (i = 1..n) 9/10^i]

No. 0.999... is defined as the limit.

Originally posted by: MadRat

Its the arrogance of this whole argument that the limit=product that is absurd.

No one is saying that the limit equals the sum.

S = Sum(i= 1..n) 9/10^i is always less than 1 for any finite n. This, however, is not what 0.999... represents.

Originally posted by: MadRat

If .999... can exist then a number can exist between 1 and it... If you insist that .999... does exist then you have to accept the difference between 1 and .999... is possible, too.

Again, you can find a difference between 4/2 and 2? Please, show me.


You are obviously having some serious issues with understanding some very basic principles (even and odd numbers for starters). I would suggest you think about what you're posting, and perhaps try proving some of these statments you're throwing out from left field (ie, the 3 times a number will always be odd statment).

 

[RossGr]

There is nothing to argue you are simply wrong. As usual.

 

[SilentRunning]

Originally posted by: silverpig

No one is saying that the limit equals the sum.

Ummm, the proof does.

 

[silverpig]

No, it says the limit of the sum = 1.

 

[Viper GTS]

Originally posted by: silverpig
No, it says the limit of the sum = 1.

But it also states that the limit of 1x10^-x as x approaches infinity is zero, by which the converse must also be true - that one minus that number must be one.

Viper GTS

 

[silverpig]

Sum (i = 1..n) 9/10^i

The value of this sum depends on what you choose for n. It can be 0.9 or 0.999999, etc. No one is saying that these sums equal 1. The limit of this sum as n -> infinity is 1.

 

[silverpig]

Originally posted by: Viper GTS

Originally posted by: silverpig
No, it says the limit of the sum = 1.

But it also states that the limit of 1x10^-x as x approaches infinity is zero, by which the converse must also be true - that one minus that number must be one.

Viper GTS

That's a single value, not a sum.

 

[Zebo]

I never knew there was so much controversy about this. YES

I havent read the posts but here's simple math for those who don't like limits

you start with one like so: 1.0 / 3 = .3333333 right

and 3 x 1/3 or 3 x .333333 = .999999

Therefore 1=.999999

 

[MadRat]

Correction: "If you take any odd number and times it by a product of two you'll end up with an even result."

Correction: "You cannot add three equal odd values to define an even product because that product does not exist; every product of 3 and an odd number will equal an odd numbered result."

So I mistyped and left out that my point was referring only to products resulting from odd numbers. Just because I omitted the point doesn't make your argument any more valid. Nor does it make .999... any closer to being 1.

 

[SilentRunning]

Proof: 0.9999... = Sum___9/10^n
______________(n=1 -> m)___as m --> infinity

____________ = lim________ Sum__ 9/10^n
______________(m -> Infinity) (n=1 -> m)


first and second line of proof

 

[silverpig]

Perhaps nothing in my "post of quotes" added validity to my argument, but unlike yours, it didn't serve to discredit my argument either.

You have responded to 2 of my issues with your argument. I'll grant you that they were typos. Please explain the rest.

 

[NovaTone]

Sum (i = 1..n) 9/(10^i)

this is also an infinite geometric series. If you recall, when you have an infinite geometric series, the sum equals a1 / (1 - r) where a1 = the first value in the series, and r = the ratio.

Therefore, in this case, a1 = 9/10 and r = 1/10, and thus, 0.9999rep. = 1.

A proof without having to use limits.

 

[duke]

Originally posted by: Carbonyl
I never knew there was so much controversy about this. YES

I havent read the posts but here's simple math for those who don't like limits

1.0 / 3 = .3333333 right

and 3 x 1/3 = .999999

Therefore 1=.999999

I know what you're trying to say, but your explanation is technically incorrect.

 

[SilentRunning]

Originally posted by: silverpig
Perhaps nothing in my "post of quotes" added validity to my argument, but unlike yours, it didn't serve to discredit my argument either.

You have responded to 2 of my issues with your argument. I'll grant you that they were typos. Please explain the rest.

What are you trying to say. It lost something in the translation.

 

[Zebo]

Originally posted by: duke

Originally posted by: Carbonyl
I never knew there was so much controversy about this. YES

I havent read the posts but here's simple math for those who don't like limits

1.0 / 3 = .3333333 right

and 3 x 1/3 = .999999

Therefore 1=.999999

I know what you're trying to say, but your explanation is technically incorrect.

? Explain

I would agree with incomplete because I did'nt want to confuse some with a 10 multiplier and pople hate X's.

 

[MadRat]

Which points confuse you?

The idea that .999... doesn't fit on the number line? That just begs the question of what is the number closest to 0 which is not zero? I've claimed that if it exists then a difference between 1 and itself exists; if it does not exist then the argument is moot. The human mind cannot comprehend infinity therefore no man can define infinity in any objective terms. You can believe that the idea of infinity exists but it cannot be proven because it is a purely subjective value.

Saying that .999... is to 1 what 4/2 is to 2 is yet another absurdity. It has no relevance to the argument.

 

[SilentRunning]

Originally posted by: NovaTone
Sum (i = 1..n) 9/(10^i)

this is also an infinite geometric series. If you recall, when you have an infinite geometric series, the sum equals a1 / (1 - r) where a1 = the first value in the series, and r = the ratio.

Therefore, in this case, a1 = 9/10 and r = 1/10, and thus, 0.9999rep. = 1.

A proof without having to use limits.

Let k be any real number. Is it possible for k to be the limit for an infinite
geometric series? Are there values (r, a) so that:

(insert solution here)

Thus, k =a/(1 - r)


http://www.imsa.edu/edu/math/journal/volume1/articles/LinearFunctionsEtc.pdf

 

[MadRat]

Originally posted by: NovaTone
Sum (i = 1..n) 9/(10^i)

this is also an infinite geometric series. If you recall, when you have an infinite geometric series, the sum equals a1 / (1 - r) where a1 = the first value in the series, and r = the ratio.

Therefore, in this case, a1 = 9/10 and r = 1/10, and thus, 0.9999rep. = 1.

A proof without having to use limits.

Aww, contrare. Your proof again is like saying "1=1 but only up to .999... and no higher".

 

[ElFenix]

hey, heres an idea, subtract .999... from 1. tell me what you get. an infinite number of 0s. what is 0? its nothing. what do you have when you have an infinite amount of nothing? you still have nothing! its still 0! it doesn't matter how many decimal places you carry the answer out its still 0!

 

[silverpig]

Originally posted by: SilentRunning
Proof: 0.9999... = Sum___9/10^n
______________(n=1 -> m)___as m --> infinity

____________ = lim________ Sum__ 9/10^n
______________(m -> Infinity) (n=1 -> m)


first and second line of proof

I have 2 responses:

1.
I defined the sum as S = Sum (i = 1..n) 9/10^i
This is the sum to which I was referring. My sum implies that n is a single integer. Pick any integer for n and you will come up short of 1.

2.
I guess I should have been a little less exclusive when I said "no one" My bad. Please note however, that the sum you posted is slightly different from mine.

 

[MadRat]

Originally posted by: ElFenix
hey, heres an idea, subtract .999... from 1. tell me what you get. an infinite number of 0s. what is 0? its nothing. what do you have when you have an infinite amount of nothing? you still have nothing! its still 0! it doesn't matter how many decimal places you carry the answer out its still 0!

That would be a proof to say ".999... does not exist" but does not prove .999...=1.

 

[silverpig]

Originally posted by: SilentRunning

Originally posted by: silverpig
Perhaps nothing in my "post of quotes" added validity to my argument, but unlike yours, it didn't serve to discredit my argument either.

You have responded to 2 of my issues with your argument. I'll grant you that they were typos. Please explain the rest.

What are you trying to say. It lost something in the translation.

I was talking to MadRat.

I was referring to how he is now deconstructing the number line by saying that any two representations of a single number must mean that one of the representations must not exist.