New Proof that 1 = .999999999 Repeating

[mchammer187]

Originally posted by: feelingshorter
infinity is not a bumber, so 1 is still more than .99999999 just by a very small amount

that small amount is not a number unless it is zero

 

[Locut0s]

Originally posted by: feelingshorter
infinity is not a bumber, so 1 is still more than .99999999 just by a very small amount

Infinity is indeed not a number but it is a perfectly valid mathematical concept. In fact all of calculus and much of the rest of mathematics besides is based on the notion of infinite series and sequences of numbers that either add to or reach a finite limit. Within any such sequence or series it is correct to say that the sequence is not equal to its limit as long as we are looking at any finite number of terms. However if we may be allowed to approach this limit arbitrarily close simply by evaluating the sequence or series up to some given n, then we say that the limit of the series is equal to this finite number. This concept is at the very foundation of all of Calculus.

 

[aplefka]

Shit man, start the presses, alert all major news headquarters. And is the world really round?

Venk is a fvcking genius!

Die in a very hot fire. :|

 

[nolson1]

Originally posted by: feelingshorter
infinity is not a bumber, so 1 is still more than .99999999 just by a very small amount

It must be sad to not know the truth that .9 repeating = 1 exactly

 

[Wadded Beef]

alert! new proof that:

you = teh suck!!1

 

[DrPizza]

Originally posted by: TuxDave

Originally posted by: DrPizza
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.

Dude... that scares me. But on the other hand, does the limit exist? What happens if you have x approaching from the negative side?

Shame on me! It's a right handed limit... as x -> 0+

incidentally, it bugs me that I can't remember exactly how to prove it.
As I think about it, I recall something like
Let L = sqrt(x + sqrt(x + ...
add x to both sides
L + x = x + sqrt(x + sqrt(x + ...
sqrt both sides
sqrt(L + x) = sqrt(x + sqrt(x + ...
since the right side is L,
sqrt (L+x) = L
(I'm leaving out the negative roots, since it should be obvious that both are positive numbers)

square both sides, L + x = L^2
and
L^2 - L = x
There are 2 values of L that satisfy >^2 - L = 0
1 and 0
I can't remember how to demonstrate that it has to be 1 though.
anyone?

 

[Locut0s]

Originally posted by: DrPizza

Originally posted by: TuxDave

Originally posted by: DrPizza
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.

Dude... that scares me. But on the other hand, does the limit exist? What happens if you have x approaching from the negative side?

Shame on me! It's a right handed limit... as x -> 0+

incidentally, it bugs me that I can't remember exactly how to prove it.
As I think about it, I recall something like
Let L = sqrt(x + sqrt(x + ...
add x to both sides
L + x = x + sqrt(x + sqrt(x + ...
sqrt both sides
sqrt(L + x) = sqrt(x + sqrt(x + ...
since the right side is L,
sqrt (L+x) = L
(I'm leaving out the negative roots, since it should be obvious that both are positive numbers)

square both sides, L + x = L^2
and
L^2 - L = x
There are 2 values of L that satisfy >^2 - L = 0
1 and 0
I can't remember how to demonstrate that it has to be 1 though.
anyone?

I'm not sure this really works but here's another possible way of showing this. If you play around with this on a calculator you can probably convince your self that this infinite sequence is bounded above by 1. So lets just assume that it is then for any given term in the sequence we have:

a(n+1) = sqrt(x+a(n)) < sqrt(x+1) but as x->0 sqrt(x+1) -> 1

So this sequence is bounded above by 1 as x->0.

However we also know that this is a monotonically increasing function for any arbitrarily small value of x. So because it is an increasing function bounded above by 1 as x->0 the limit of the sequence as x approaches 0 =1.

 

[DrPizza]

Originally posted by: Locut0s

Originally posted by: DrPizza

Originally posted by: TuxDave

Originally posted by: DrPizza
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.

Dude... that scares me. But on the other hand, does the limit exist? What happens if you have x approaching from the negative side?

Shame on me! It's a right handed limit... as x -> 0+

incidentally, it bugs me that I can't remember exactly how to prove it.
As I think about it, I recall something like
Let L = sqrt(x + sqrt(x + ...
add x to both sides
L + x = x + sqrt(x + sqrt(x + ...
sqrt both sides
sqrt(L + x) = sqrt(x + sqrt(x + ...
since the right side is L,
sqrt (L+x) = L
(I'm leaving out the negative roots, since it should be obvious that both are positive numbers)

square both sides, L + x = L^2
and
L^2 - L = x
There are 2 values of L that satisfy >^2 - L = 0
1 and 0
I can't remember how to demonstrate that it has to be 1 though.
anyone?

I'm not sure this really works but here's another possible way of showing this. If you play around with this on a calculator you can probably convince your self that this infinite sequence is bounded above by 1. So lets just assume that it is then for any given term in the sequence we have:

a(n+1) = sqrt(x+a(n)) < sqrt(x+1) but as x->0 sqrt(x+1) -> 1

So this sequence is bounded above by 1 as x->0.

However we also know that this is a monotonically increasing function for any arbitrarily small value of x. So because it is an increasing function bounded above by 1 as x->0 the limit of the sequence as x approaches 0 =1.

works for me.
Another way to convince myself (or rather, others),
Going back to where I had L = the root,
then eventually, L^2 - L - x = 0, plug in a variety of small values of x and view their graphs. There are 2 roots: one positive and one negative. As x -> 0, one root approaches 0 from the left, and the other root approaches 1 from the right. Clearly, the root is not a negative value, therefore it most be the positive value... 1.

 

[newParadigm]

1 repost does not = .9999.... reposts

 

[Bacardi151]

Originally posted by: 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

but nobody ever implied it until he brought it up

 

[Locut0s]

Here's one for you guys who can't seem to wrap your mind around the concept of infinity. Which is bigger the set of all integers or the set of all odd integers? Both sets obviously contain an infinite number of elements. The answer is that they are the same size! You might think that the set of all integers has to be larger, after all the set of all odd integers is a subset of the set of all integers. Also the set of all odd integers doesn't include the even integers which are in the set of all integers. However if we create a one-to-one and onto mapping, or function, 2*n+1 we now have a method of mapping each and every element in the set of integers onto exactly one element in the set of odd integers, therefore they have to be the same size.

 

[acadia11]

Originally posted by: mchammer187

Originally posted by: acadia11

Originally posted by: mchammer187

Originally posted by: acadia11

Originally posted by: amoeba

Originally posted by: acadia11
You are all gay, and fcuking nerds, I got a better one for you though ... try this one

a = b

a^2 = b^2

a^2 = b * b

a^2 = a * b

a^2 - b^2 = a * b - b^2

(a + b)(a - b) = (a - b)b

a + b = b

b + b = b

2 * b = b

So,

2 = 1 .......... oh my f'cking god 2 = 1!!!!!!!



1 = 0 , so that means that everything should be free in society. since every number actually equals 0!!!!!!!

nice trick....

unfortunately the

(a+b)(a-b) = (a-b)b

and then

a+b = b

is where it unravels.

You could have done it much quicker with

2*0 = 1* 0

2= 1

OMG!!!

Yeah, but 0/0 is undefined, so it isn't quicker, it blatently false.

On the otherhand,

(a-b)(a+b) = (a-b)b

a^2 -ab + ab - b^2 = ab - b^2

a = b , so

a^2 - b^2 = aa - b^2

a^2 - b^2 = a^2 - b^2

....

so going back, (a-b)(a+b) = (a-b)b

so, 2 = 1, OMG !!!!!! .... looks good to me, down with capitalism!!!

Who's not paying taxes with me.

when you divide both sides of (a-b)(a+b) = (a-b)b by (a-b) you have to assume that a != b or you get an indeterminate for beause it is division by zero otherwise thus amoeba is correct

I'm not saying the proof is correct.

I said 0/0 is undefined, or indeterminate more correctly, and blatently wrong.


However x,y,z... looks good to me.


On another note?

I can't assume a != b, I arleady just assumed
a = b, how can a = b and a != b, that's a contradiction.

Man, you people are really nerdy, it was much funnier when capitalism was coming to an end.

Thanks for taking the jam out of my donut.

i was saying in order to divide both sides by (a-b) you must assume a != b

if you do not make that assumption than you are dividing by zero

you are contradicting yourself when you divide by (a-b)

because in order to do so you must ASSUME you are not dividing by zero otherwise the equation blows up

because you are dividing by (b-b) or (a-a) or (b-a) or (a-b) all are zero

therefore if a=b then you cannot perform the division at all because it is performed on the basis that a!=b


so I was saying either a!=b or dont do the division at all

Which amounts to the proof is bullshit and 2 !=1, which was the joke, but somewhere the humor in the proof got lost, because, you all are a bunch of nerds who are to busy trying to be correct or correcting someone instead of just laughing.

Atleast one guy go it, the guy who called me a communist.

That, that was funny.

Lighten up, life, is too short.

 

[acadia11]

Originally posted by: Locut0s
Here's one for you guys who can't seem to wrap your mind around the concept of infinity. Which is bigger the set of all integers or the set of all odd integers? Both sets obviously contain an infinite number of elements. The answer is that they are the same size! You might think that the set of all integers has to be larger, after all the set of all odd integers is a subset of the set of all integers. Also the set of all odd integers doesn't include the even integers which are in the set of all integers. However if we create a one-to-one and onto mapping, or function, 2*n+1 we now have a method of mapping each and every element in the set of integers onto exactly one element in the set of odd integers, therefore they have to be the same size.

Ah but the set of all real numbers is bigger than the set of integers or set of odd integers!!!

Muahahahahahaha!!!!!

 

[Locut0s]

Originally posted by: acadia11

Originally posted by: Locut0s
Here's one for you guys who can't seem to wrap your mind around the concept of infinity. Which is bigger the set of all integers or the set of all odd integers? Both sets obviously contain an infinite number of elements. The answer is that they are the same size! You might think that the set of all integers has to be larger, after all the set of all odd integers is a subset of the set of all integers. Also the set of all odd integers doesn't include the even integers which are in the set of all integers. However if we create a one-to-one and onto mapping, or function, 2*n+1 we now have a method of mapping each and every element in the set of integers onto exactly one element in the set of odd integers, therefore they have to be the same size.

Ah but the set of all real numbers is bigger than the set of integers or set of odd integers!!!

Muahahahahahaha!!!!!

Hehe. In fact the set of the real number line from [0..1] is larger than the set of all integers.

 

[zerocool1]

Originally posted by: DannyBoy

Originally posted by: MrChad
Oh
My
God
:roll:

You would think something as well known as this on the forums couldn't POSSIBLY get reposted, it's been discussed over and over so many god damn times now.

yea really

 

[DrPizza]

Originally posted by: Locut0s

Originally posted by: acadia11

Originally posted by: Locut0s
Here's one for you guys who can't seem to wrap your mind around the concept of infinity. Which is bigger the set of all integers or the set of all odd integers? Both sets obviously contain an infinite number of elements. The answer is that they are the same size! You might think that the set of all integers has to be larger, after all the set of all odd integers is a subset of the set of all integers. Also the set of all odd integers doesn't include the even integers which are in the set of all integers. However if we create a one-to-one and onto mapping, or function, 2*n+1 we now have a method of mapping each and every element in the set of integers onto exactly one element in the set of odd integers, therefore they have to be the same size.

Ah but the set of all real numbers is bigger than the set of integers or set of odd integers!!!

Muahahahahahaha!!!!!

Hehe. In fact the set of the real number line from [0..1] is larger than the set of all integers.

Don't you mean, the set of all real numbers from [0,0.999...]

 

[darbius]

Originally posted by: Lizabath
no you are wrong 91ttz,

.999... = 1

Take some advance math courses and you will find out why.

I've taken advanced math courses and that is just one of the few rules of math I'm not willing to accept. No matter which way you spin it or how many proofs you try to do, .999 repeating will never, ever equal 1.

Edit: And this is all crap anyway, because anyone who's taken advanced math knows that highly advanced math is more entrenched in philosophy than numbers.

 

[DanTMWTMP]

Originally posted by: darbius

Originally posted by: Lizabath
no you are wrong 91ttz,

.999... = 1

Take some advance math courses and you will find out why.

I've taken advanced math courses and that is just one of the few rules of math I'm not willing to accept. No matter which way you spin it or how many proofs you try to do, .999 repeating will never, ever equal 1.

hehe, created a new account to hide your other one when the internet ALREADY provides you with anonymity??

Are you that scared that people will regard you as a dumbass after making that statement? What is so hard to grasp the idea of infinity, and that yes, the limit to the sequence 0.99999~ is 1 (ONE)? So you'll disagree with millions of math professors out there that worked their arses off for their math degrees? You'll disagree with millions of engineers out there? You'll disagree with millions of teachers out there, just because you will not accept a factual concept?

ya..

 

[blahblah99]

Originally posted by: acadia11
You are all gay, and fcuking nerds, I got a better one for you though ... try this one

a = b

a^2 = b^2

a^2 = b * b

a^2 = a * b

a^2 - b^2 = a * b - b^2

(a + b)(a - b) = (a - b)b

a + b = b

b + b = b

2 * b = b

So,

2 = 1 .......... oh my f'cking god 2 = 1!!!!!!!



1 = 0 , so that means that everything should be free in society. since every number actually equals 0!!!!!!!

Except, a-b = 0, and you're dividing by 0. Nice try.

 

[MySoS]

In a math were 0.9999R != 1, calculus isn't valid. Since we know calculus is valid, then 0.9999R is equal to 1.

 

[maziwanka]

Originally posted by: aplefka
Shit man, start the presses, alert all major news headquarters. And is the world really round?

Venk is a fvcking genius!

Die in a very hot fire. :|

go fvck yourself asshat

 

[MySoS]

According to mathematics all the points in the universe could be contained in the tip of a needle. For many this hard to accept, but if you think about it, then makes sense. Another math concept with many find hard to believe like 0.999R = 1.

 

[acadia11]

I''m putting an end to this debate ..999999999R != 1, .999999999999R is approximately 1.

End of debate.

Semantics are important, sort of like 1 / infinity is not equal to 0, it is approximately 0.

And dependant upon the granularity of the comparison determines the outcome. Even computers work this way when comparing two numbers.

 

[dornick]

you says you get the final word? you werent even the OP...

Read the whole thread moron, everything you just said has been disproven in multiple places.

 

[Matthias99]

Originally posted by: darbius

Originally posted by: Lizabath
no you are wrong 91ttz,

.999... = 1

Take some advance math courses and you will find out why.

I've taken advanced math courses and that is just one of the few rules of math I'm not willing to accept. No matter which way you spin it or how many proofs you try to do, .999 repeating will never, ever equal 1.

Ah, these are always fun.

I'd check out that Dr. Math link above for more information, but here's the proof. An honest-to-god actual proof that the number defined by "0.999..." is exactly equal to the number defined by "1.000...".

First off, what you really mean by saying "0.999..." is "the limit as x goes towards infinity of (sum from 1 to x of (9 * 10^(-x)))" -- that is, "0.9 + 0.09 + 0.009 + ...". Start from there.

(Those of you who have taken calc or even pre-calc should recognize that this is a geometric series, and it is possible to prove that all geometric series involving real numbers obey certain rules, and the sum of this series must be (9/10 + 1/10 = 10/10 = 1). But proving that is more involved, and so I am going to stay along these lines.)

LEMMA: "0.9 + 0.09 + 0.009 + ..." is a real number.
Since each term in the series is a real number (they are actually rational numbers, of the form 9/10, 9/100, 9/1000, ..., but rationals are a subset of the reals), and any two real numbers added together also produces a real number (this is a basic property of addition on reals), therefore the sum of this series must also be a real number. QED.

"1.000..." (AKA 1/1) is also a rational/real number.

Now let's talk about equality of real numbers. There are a number of ways to define equality on real numbers, but one of the more useful ones is to say that "for two real numbers a and b, a = b if and only if a - b = 0" (I'm not gonna delve further into this, but basically, these things are just defined as part of the properties of real numbers). So our question then becomes:

"Does '1.000...' - 'the sequence referred to as "0.999..." ' = 0?"

Taking the sequence above, you can see that the difference between "1" and "0.999..." for a finite number of terms is (sum from 1 to x of (1 - (9 * 10^-x))). Let's look at the values of this sum:

x = 1, value = 1 - .9 = 0.1 = 1/10 = 1 / (10^1)
x = 2, value = 1 - .9 - .09 = 0.01 = 1/100 = 1 / (10^2)
x = 3, value = 1 - .9 - .09 - .009 = 0.001 = 1/1000 = 1 / (10^3)
x = 4, value = 1 - .9 - .09 - .009 - .0009 = 0.0001 = 1/10000 = 1 / (10^4)
x = 5, value = 1 - .9 - .09 - .009 - .0009 - .00009 = 0.00001 = 1/100000 = 1 / (10^5)
...
x = n, value = 1 / (10^n)

The actual difference we are looking for here is whatever the limit of this sum is as x goes to infinity.

The limit as x goes to infinity of (1 / (10^x)) is exactly zero. Thus, "1.000..." - "0.999..." is also exactly zero.

The typical challenge back to this is that "the limit's not zero, it's an infinitely small positive number -- 'an infinite number of zeroes followed by a one' ". No such number can exist in the real number system; the only number that can be the answer to this limit is zero. Attempting to define such a number via sequences, series, and limits -- or any other mathematical method -- is impossible. Intuitively, "an infinite number of zeroes followed by x" (where x is anything that's not "0") cannot exist, since then the sequence of zeroes would not be infinite.

Basically, "0.999..." = "1.000...", no qualms about "logic" or other things like that. When you're working with real numbers, they are identical. The sum of a converging infinite sequence can be exactly equal to a real number -- not just approximately equal. Frankly, that's the entire point of calculus.