filtherton

Doesn't 1/infinity equal 0? Certainly the limit of 1/x as x approaches infinity is 0. If you were to slap limit notation before the above proof it would make more sense. I agree though, that as a proof, it ain't exactly proper.

daking

No

It wouldnt make any more sense. You cannot start with A = B-lim(x->infinity) (1/x) to prove that A equals B.

filtherton

What does it equal?

daking

It doesnt equal anything, infinity isnt a number its a concept to represent unboundedness.

filtherton

What is the limit of 1/x as x-> infinity?

daking

ofcourse the limit is zero.

filtherton

So, wouldn't it follow what 1/infinity is, for all intents, zero?

molloby

No, because infinity only makes sense when it is used in the context of a limit or an infinite sum.

It cannot be used as a stand alone number.

daking

What molloby said

filtherton

That makes sense, i guess when i see 1/infinity i just assume the limit is implied.

molloby

I see where you're coming from but there is a difference between saying:

1/infinity = 0
and
1/x gets arbitrarily close to 0 as x gets arbitrarily large

Rigour goes a long way in maths.

a-j

daking is correct the limit is 1.
The limit of y=sin(x)/x = 0/0, using l'hopital's rule the limit of y=cos(x)/1 as x->0 = cos(0)/1 = 1/1


Wow, how is the limit of sin(x)/x as x->inf obviously 1?
This limit is obviously 0, since |sin(x)|<=1. Limit of |sin(x)|/x as x->inf = 0. Take a look at the epsilon delta definition of limit why this works.

xepherys

Hmmm, there seem to be two approaches to this that have been semi-overlooked, and both disprove this (AFAICT):

1/infinity IS NOT a real number because infinity is not a real number. If this is the case, you cannot logically prove the theory using a non-existing number (not sure why "i" is used...)

1/infinity IS a real number because it has SOME positive value (however infintesimally small). If this is true, and it has value, then 1.0 != 0.9r but rather 1.0 = 0.9r + 1/infinity, whatever value that may be.


In either case it's a quirk of human logistics rather than mathematical fact.

molloby

Infinity is not a numerical value full stop. Attempting to use it as a number simply doesn't work.

1=0.9r: There are some nice proofs earlier in this thread- espescially the geometric series proof.

Remember, any mathematical statement that includes infinity as a value is meaningless- it is as simple as that.

daking

Yea it was late (4 am) when i was typing, for some reason was thinking about x to 0. I am aware the limit as x tends to infinity is 0.

The function i should of used was x sin(1/x) whose limit is 1 as x->inf.

filtherton

I was refering to the limit as x goes to infinity because that is what he asked at the end of the particular post he was referring to.


Yeah, on second thought, i should have read things more carefully before i opened my vitual mouth.

TheBrit

let x = 0.125r
1000x = 125.125r
(Subtract x from both sides, we have defined x as 0.125r. 125.125r - 0.125r must = 125)
999x = 125
x = 125/999
I assure you that's correct.

Using the same maths:
x = 0.9r
10x = 9.9r
9x = 9
x = 9/9
x = 1

Or think about it this way. If two numbers are different, you prove this by subtracting them, and getting a difference. 0.9r continues into infinity, therefore there is no difference between 0.9r and 1, therefore 0.9r = 1.

AntoineMartinez

for all those who don't understand a thing these people are saying, try subracting .9r from 1

1.0000000000000000000
- .9999999999999999999
you would just keep borrowing from the previous, making 9's forever

daking

Yea thats kind of intuitive, 1-0.999...= 0.0000....

And if they are all 0's well, the answer has to be 0

Munku

<~ Idiot. r = repeating.

kutulu

.9r is not equal to one. It is always less than one. The limit of 1/x as x->infinity IS zero, however function will always be greater than zero, no matter how small the value it. Just because the limit of a function is equal to zero it does not mean the that value of the function will ever be equal to zero.

kutulu

No, 1-0.999.... = 0.0000......1 which will always be greater than 0.

Slavakion

But this thread has proven that .9r does equal one. Many times, many ways.

Although which is technically larger than zero, it is infinitely small.

It's like you took the universe, un-orthoganized all of the theoretical multiverses, raised the whole mess to the google power and added one. Then set it to an exponential growth model that continued until the end of time. At which point you compared it [the whole universe mess] to a quark. Where's that quark? Oh, I can't find it. It's so ridiculously small compared to everything else around me.

Even imaginary numbers are bigger than 1-.9r and they don't exist (or so we've been told).

kutulu

It doesn't matter if it is infinitely small or infinitely large, it still isn't equal to zero. It's like if you have to move an object 1 foot. You start by moving it half a foot. Then 1/4 ft, then 1/8 ft..... You will never move it the full 1ft, even if there is no way to measure the distance between the object and the 1ft barrier, there is still distance to be covered.

filtherton

From a physics perspective, you're never really touching anything, because no object has a clearly defined boundary on the atomic level. For all intents and practical purposes though, .9r equals one. The properties of numbers don't always transfer directly to the properties of physical dimensions. You can't disprove a logic based mathematical proof using physical dimensions.

adam

Here's another way to think about why .9999999999999 (repeated indefinitely) = 1

Suppose it doesn't: then 1 - .9999r must be > 0. But I can demonstrate that for any difference you pick, I can get it closer to zero than that. Ergo, the difference must be zero.

I have an undergrad degree in math. A lot of it comes down to the fact that things involving infinity (in this case, an infinite sequence) don't behave in ways that are particularly intuitive. That's what homework was for.

daking

Incorrect, there is no termination to the result, you cant do "00...1" on the end.

Imaginary numbers "exist" to the same extent that Real numbers "exist". Further more there is no orderedness in complex (imaginary) numbers compared to purely real numbers, so you cant really say they are "bigger" without using a different metric |a+bi| .
Firstly, you are getting confused with the partial sum and the sum itself, its true to say that


for any N one chooses, but the fact is that the sum to infinity does equal 1.



So if you mean by "infinitely large" you mean the sum to infinity then you are wrong and the sum does equal 1. However if you mean by "infinitely large" "for any choice of N, the sum is less than 1 then you are correct. Its not clear which one you ment.

The problem occurs where you say "..it still isn't equal to zero." i presume you mean the result of 1-0.9r. Well this does equal zero, as it is the sum to infinity :



not a partial sum to N.

molloby

daking, what program did you use for those equations?

Could be useful.

daking

Maple 9.5, you should be able to find it on torrent sites.

molloby

Thanks, will look that up.

phukraut

Maple is overkill for this, try looking for MathType. It can export to gif.

daking

Maple is a great all round math utility tho.

molloby

I like, I like.

Nothing like a bit of Gauss's law to start the day:

Slavakion

Yeah, that was supposed to be a humorous remark. I should've known that a math joke was doomed to fail. I know that imaginary numbers do exist, they were only termed imaginary because the first people to work with them weren't really comfortable (square root of a negative number? WTF!). I hadn't really thought about the fact that you can't compare imaginary and real numbers. I suppose the opposite names should have clued me in.

On another note, my teacher got fed up with the proof, and crossed it out on the grounds that you can't define x twice.

KnifeMissile

I'm sorry, I don't mean to harp on this but, as anyone who has seen me in this forum before can attest, I just can't let this go.

Do you still believe that 0.999... (that r notation really is too confusing) is not exactly equal to 1?

Stompy

To sum up what someone said earlier: it just is. It works.

The same goes for 1/infinity. It's an idea that can be proven (saying it equals 0, or gets insanely close to 0, as x approaches infinity), but used by itself makes no sense.

It's like comparing the integral of 1/x and 1/(x^2) as x goes from 1 to infinity. By itself, and looking at it, you'd think they're the same, but they aren't.

The first diverges to infinity whle the second converges to 0.

Think about it for too long and you start going crazy and end up believing things that aren't so

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bendsley

In anything, especially math, there are assumptions. People above are saying that .99999999999999 = 1. It doesn't, .99999999999999 = .99999999999999. But for arguments sake, .99999999999999 ~1. So, basically, its weird.

Thing with infinity is that no number is definite when referring to infinite possibilities, because you can always add one (1).

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daking

Noone is saying that 0.99999999999999 equals one. People are saying that "0.999..." or "0.9r" equals one and they are correct to think so. It doesnt approximately equal to one or asymptotic to 1 it exactly equals one.

The second converges and equals 1 not 0, but yes your right the first integral diverges.

Stompy

D'oh! Yeah, kinda left off a step there at the end in forgetting to account for the 1 to infinity

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Rangsk

Take any two distinct, real numbers. Any two, doesn't matter which two. Ok, so what makes them distinct? What makes the distinct is the fact that there are an infinite number of real numbers inbetween them. Can you name even one number between .999.. and 1? I can't. If you can, then you've proven they're distinct. If you cannot, then they are not distinct, and are thus equivalent.

Of course, I would have to prove that you cannot, but there are plenty of valid proofs already listed in this thread. It's not "voodoo math".. it is perfectly valid.

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