New Proof that 1 = .999999999 Repeating

[dornick]

i can't see it

 

[surreal1221]

WTFBBQ OMGHI2U2 IM OUTTIE!

 

[glenn1]

I agree that there's no smallest positive number. PROOF:

Assume there *IS* a smallest positive number.
A positive number divided by a positive number is still a positive number.
Divide the smallest number by 2... it's now half the original size. But wait... it it's half the original size, then it's smaller than the smallest positive number. Since it's also positive, it leads to a contradiction. Therefore, the assumption must be false.

Maybe it's an imaginary number. But wait, mathmatics already has those, so I guess you're familiar with the concept.

WHY IS THIS IMPORTANT????

It's not. It's an argument over an abstract concept and the conclusion doesn't matter in the slightest. Let me sum the argument: .9999~ = 1 because mathmatics logic requires it to be so. And since a mathematical proof will follow the rules of math logic it is impossible to show anything but the predestined conclusion. Practical application or interest to anyone not a mathmatician, infinity approaching zero.

 

[gwlam12]

Infinity is not a number. You cannot divide 1 by infinity.

 

[ugopk]

I don't think this would work.

Because if you can do it with 1 and .99..., you can do the same proof for .99 and .98 and so on...

basically saying all numbers = all numbers

 

[Bacardi151]

1/infinite != 0.


1. it is not written in such ways. it is written as 1/x, where x -> inifinite
2. infinite is not a number, if it is, tell me what it is. (it is a concept, a limit, used in mathematics to express something)
3. it is a proof to show that 1/x approaches 0 as x approaches infinite, but not equal. and through common sense we can write it as 0 because it's more or less insignificant to the yielding result

 

[spidey07]

Originally posted by: Anubis
you know i didnt read that HUGE other thread but we proved this in college, so i dont see what the big issue is

because there are still stupid people out there.

 

[spidey07]

Originally posted by: malak

Originally posted by: DanTMWTMP
whoever thinks that 0.9999... does not equal 1 is a total dumbass.

It's not about philosophy or anything. This is straight up math. 0.9999infinity = 1. end of discussion.

Mathmatically it's not possible. One day you will learn. Just like in elementary school they told you that the square root of a negative number isn't possible and then later they tell you it is, later they will tell you technically .99R does not equal 1.

no, no they won't.

 

[RelaxTheMind]

1 - .9999~ = .0000~1

Prove it wrong.

if you can explain it with a magic number i can explain it with a magic number. why would my number exist and not your .9999~

in theory .9999~ = 1 but in the real world where people already have an idea of what 1 is... 1=1 and .9999 = .9999

or else life would act much like this dammed thread.

 

[spidey07]

Originally posted by: RelaxTheMind
1 - .9999~ = .0000~1

Prove it wrong.

if you can explain it with a magic number i can explain it with a magic number. why would my number exist and not your .9999~

in theory .9999~ = 1 but in the real world where people already have an idea of what 1 is... 1=1 and .9999 = .9999

or else life would act much like this dammed thread.

there's this whole thing called a number system that just so happens to perfectly relate to the "real world" down the the sub-atomic level, I suggest you study it and learn what it means.

 

[mrSHEiK124]

Originally posted by: RelaxTheMind
1 - .9999~ = .0000~1

Prove it wrong.

if you can explain it with a magic number i can explain it with a magic number. why would my number exist and not your .9999~

in theory .9999~ = 1 but in the real world where people already have an idea of what 1 is... 1=1 and .9999 = .9999

or else life would act much like this dammed thread.

I was just about to post that
1=1 ; .999~1=.999~1
If you have .9 repeating for a set amount of 9's, make that N, then if you have N-1 0's (after a decimal point of course) followed by a 1, that makes 1, but .999~1 cannot be 1, it can be rounded to 1, because nobody gives a shiite about .0000000000000001, it won't make a difference, but it is still .999~1, only way to change it is by adding .001~1 to it.

 

[Howard]

Originally posted by: sheik124

Originally posted by: RelaxTheMind
1 - .9999~ = .0000~1

Prove it wrong.

if you can explain it with a magic number i can explain it with a magic number. why would my number exist and not your .9999~

in theory .9999~ = 1 but in the real world where people already have an idea of what 1 is... 1=1 and .9999 = .9999

or else life would act much like this dammed thread.

I was just about to post that
1=1 ; .999~1=.999~1
If you have .9 repeating for a set amount of 9's, make that N, then if you have N-1 0's (after a decimal point of course) followed by a 1, that makes 1, but .999~1 cannot be 1, it can be rounded to 1, because nobody gives a shiite about .0000000000000001, it won't make a difference, but it is still .999~1, only way to change it is by adding .001~1 to it.

Nobody is saying 0.99 (with a set amount of 9s) is equal to 1, they're saying 0.9 with an infinite number of 9s is equal to 1.

 

[element]

Originally posted by: RelaxTheMind
1 - .9999~ = .0000~1

Prove it wrong.

if you can explain it with a magic number i can explain it with a magic number. why would my number exist and not your .9999~

in theory .9999~ = 1 but in the real world where people already have an idea of what 1 is... 1=1 and .9999 = .9999

or else life would act much like this dammed thread.

0.0000~1 doesn't exist. When are you going to write that 1? Never. Do you understand the concept of never? that means it is 0.0000~ forever.

 

[mrSHEiK124]

Originally posted by: Howard

Originally posted by: sheik124

Originally posted by: RelaxTheMind
1 - .9999~ = .0000~1

Prove it wrong.

if you can explain it with a magic number i can explain it with a magic number. why would my number exist and not your .9999~

in theory .9999~ = 1 but in the real world where people already have an idea of what 1 is... 1=1 and .9999 = .9999

or else life would act much like this dammed thread.

I was just about to post that
1=1 ; .999~1=.999~1
If you have .9 repeating for a set amount of 9's, make that N, then if you have N-1 0's (after a decimal point of course) followed by a 1, that makes 1, but .999~1 cannot be 1, it can be rounded to 1, because nobody gives a shiite about .0000000000000001, it won't make a difference, but it is still .999~1, only way to change it is by adding .001~1 to it.

Nobody is saying 0.99 (with a set amount of 9s) is equal to 1, they're saying 0.9 with an infinite number of 9s is equal to 1.

yes I know, I thought I'd confuse people if I said ~-1 0's followed by a one.

 

[stupidusmaximus]

jeez...when would u really need to know that 0.999... = 1 ....just round the bloody thing off....u get 1...same thing...and less thinking....muahahahahah...

 

[beer]

What is the hard part of understanding 'infinite amount of 9s?" There is no way to truncate that. Period. Quit using that in your supposed 'arguements'

 

[nolson1]

Just another way to look at it...

Let .9 repeating = N
Then 10N = 9.9 repeating

Then:

10N = 9.99999999.....
1N = 0.99999999....

Subtract the two eqations...
Get

9N = 9

or N = 1
Therefore 1=0.9 repeating

 

[91TTZ]

This topic has become disinteresting- now's the time on Sprokets where we dance!

 

[GTPoompt]

this comes up all the time in every forum, but i haven't ever commented.

lim 1/x -> 0 as x -> infinity = 0. so then ok, 1/infinity = 0 because you can't have .000000 ..... 1 because 0's are infinite.

but then 1 = infinity * 0? it doesn't follow that basic rule of divison. infinity * 0 is an indeterminate form.

Both answers are correct. But don't say the mathematical one is more correct because math was created by logic to explain things.

Can anyone clarify 0 * infinity is indeterminate? It COULD be undefined but I am not sure.

another thing, why would 1/infinity be the smallest number? why wouldn't (1/A)^(A) / ((A)^(A))? but remember 1/A is indeterminate and A^A is also indeterminate. If A^A is indeterminate then 1/A must be indeterminate. this is where A -> infinity.

 

[MAME]

Originally posted by: ugopk
I don't think this would work.

Because if you can do it with 1 and .99..., you can do the same proof for .99 and .98 and so on...

basically saying all numbers = all numbers

you're making exactly no sense

.999...8 is not a number just like .000...1 is not a number (note the '...'s represent an INFINITE amount of numbers)

 

[BespinReactorShaft]

I think you just discovered one of the most potent post-generating algorithms out there.

 

[Bacardi151]

Originally posted by: MAME

Originally posted by: ugopk
I don't think this would work.

Because if you can do it with 1 and .99..., you can do the same proof for .99 and .98 and so on...

basically saying all numbers = all numbers

you're making exactly no sense

.999...8 is not a number just like .000...1 is not a number (note the '...'s represent an INFINITE amount of numbers)

you're the one who's not making any sense, if '...' represents an infinite amount of numbers, how can you have a 1 at the end? that means the amount of numbers in between .000...1 is not infinite, and what comes after 1? do you even know what the concept of infinite is? you can't add a number to infinite, which you did by saying infinite and then putting a 1 at the end of the infinite series of #s

 

[dornick]

Originally posted by: Bacardi151

Originally posted by: MAME

Originally posted by: ugopk
I don't think this would work.

Because if you can do it with 1 and .99..., you can do the same proof for .99 and .98 and so on...

basically saying all numbers = all numbers

you're making exactly no sense

.999...8 is not a number just like .000...1 is not a number (note the '...'s represent an INFINITE amount of numbers)

you're the one who's not making any sense, if '...' represents an infinite amount of numbers, how can you have a 1 at the end? that means the amount of numbers in between .000...1 is not infinite, and what comes after 1? do you even know what the concept of infinite is? you can't add a number to infinite, which you did by saying infinite and then putting a 1 at the end of the infinite series of #s

you just said the exact same thing he did. He was saying that those examples aren't numbers beacuse of the infinite block in between

 

[DrPizza]

Originally posted by: GTPoompt
this comes up all the time in every forum, but i haven't ever commented.

lim 1/x -> 0 as x -> infinity = 0. so then ok, 1/infinity = 0 because you can't have .000000 ..... 1 because 0's are infinite.

but then 1 = infinity * 0? it doesn't follow that basic rule of divison. infinity * 0 is an indeterminate form.

Both answers are correct. But don't say the mathematical one is more correct because math was created by logic to explain things.

Can anyone clarify 0 * infinity is indeterminate? It COULD be undefined but I am not sure.

another thing, why would 1/infinity be the smallest number? why wouldn't (1/A)^(A) / ((A)^(A))? but remember 1/A is indeterminate and A^A is also indeterminate. If A^A is indeterminate then 1/A must be indeterminate. this is where A -> infinity.

Congrats. You survived calculus without understanding what you were doing.

0 * infinity doesn't exist. Infinity is not a number.
If you have x*y
and you're wondering, what does x*y approach as x approaches 0 and y approaches infinity, then perhaps your calculus teacher told you to just "plug in 0 and infinity. If you get 0 * infinity then it's indeterminate." Stating it that way is a shortcut to get students to be able to do what they need to be able to do correctly most of the time. It's not appropriate, nor is it accurate.
The "0 * infinity" thing comes from limits. It's not actually 0, just a number that's approaching 0. It's not "infinity" either, as that's not a number. It's a number that's arbitrarily large.

When you have a limit that can be looked at in two parts,
the first part times the second part
and the first part is approaching zero, and the second part is not bounded (gets arbitrarily large), then that type of limit is called "indeterminate". In such cases, you can often use L'Hopital's Rule to find the actual limit.
0*0*0*0*0*0*0*0...
or, 0^(infinity) if you wish to express it that way, = 0.

However, a limit that takes the form of (approaching zero)^(approaching infinity) is an indeterminate form. I just checked through the 3 calculus texts in my class. None of them say "= 0/0" or "0^infinity" or "infinity/infinity". They all say "they are limits of the form 0/0" or "of the form infinity/infinity" or...

Here's where your understanding from calculus is going to break down.
Sqrt(0 + sqrt(0 + sqrt(0 + sqrt(0 + ....
infinitely nested. Obviously, it = 0.
Instead, sqrt(x + sqrt(x + sqrt(x + sqrt(x + ...
same thing, except an x instead of 0.
Now, find the limit as x approaches 0. Surprise! It's not 0 as you probably think. It's 1.

 

[DrPizza]

Also, while I'm at it, it's inappropriate to say
"the limit = infinity"
infinity is NOT a limit. It's a concept.
What is the limit of 2^x as x gets arbitrarily large?
Is written as Lim (x-> infinity) of 2^x
The answer is NOT infinity, the answer is "there isn't a limit" or "the limit does not exist"
However, it's common to refer to that limit as "= infinity" (or more appropriately, = +infinity) to denote that it increases without bound, which differentiates the solution from other types of problems that have no limit for other reasons.