POLL: Do you believe that .9999... = 1? ** UPDATED WITH MORE PROOFS **

All right, it does. I do have references to proofs made by mathematicians and teachers. But some people are stubborn and I'll give some proofs. Read the proofs, and then vote.

PROOF 1:
1/3 = .3333...
2/3 = .6666...
THEREFORE 3/3 = .9999...

PROOF 2:
LET x = .9999...
THEREFORE:
10x = 9.9999...
9x = 9
x = 1

There are more proofs. But the point of the poll is... are you convinced?

It is not a Yes No question. It's all about significant figures. IN other words, it depends on the data you're dealing with.

i think if the number of nine's are infinite, then yes, .999.... = 1
it is just about limit

If .9999 = 1 then .9999 +.9999 should = 2

Well I make it 1.9998

Like Saaby says it depends on the data and how you interpret it.

Originally posted by x-ray:
If .9999 = 1 then .9999 +.9999 should = 2

Well I make it 1.9998

Like Saaby says it depends on the data and how you interpret it.

Click to expand...

<font size="2" face="Verdana, Arial">It is a yes/no question. If you interpret "..." to mean that the digits repeat forever, then 0.9999... does in fact equal 1.

If you don't then you are adding 0.99990000... not 0.99999999...

It's the same thing as
9/10 + 9/100 + 9/1000 + etc. which also adds up to 1.

Or how about looking at it this way...

1/9 = 0.1111...
What is 0.9999.../9 ?

1/3 doesn't equal an infinite number of .3's. It APPROACHES 1/3, but isn't equal. Same with .9 with an infinite number of nines. It will never be EXACTLY 1.0 with an infinite number of 0's, but it will get so close it doesn't really matter. Like they said, it is all about limits and tolerances.

Spud

Originally posted by Tater Rocket:
1/3 doesn't equal an infinite number of .3's.

Click to expand...

<font size="2" face="Verdana, Arial">Actually, it does.

If you say it doesn't, then let's call the difference x.

So, 1/3 = x + 0.3333...

Show me a non-zero value for x that makes that equation true.

Originally posted by Max:
</font><blockquote><font size="1" face="Verdana, Arial">quote:</font><hr /><font size="2" face="Verdana, Arial">Originally posted by x-ray:
If .9999 = 1 then .9999 +.9999 should = 2

Well I make it 1.9998

Like Saaby says it depends on the data and how you interpret it.

Click to expand...

<font size="2" face="Verdana, Arial">It is a yes/no question. If you interpret "..." to mean that the digits repeat forever, then 0.9999... does in fact equal 1.

If you don't then you are adding 0.99990000... not 0.99999999...

It's the same thing as
9/10 + 9/100 + 9/1000 + etc. which also adds up to 1.

Or how about looking at it this way...

1/9 = 0.1111...
What is 0.9999.../9 ?</font><hr /></blockquote><font size="2" face="Verdana, Arial">Ah my mistake, didn't notice the "..." (infinite 999's)

in our limited human mind filled with the theories and constructs we call mathemathics it is 1.
in reality though ... if an objective reality even exists ... who knows? maybe not even the Vorlons know ...
but for practical reasons it is 1.

and most important, to reduce our horrible fear of things we cannot capture with our mind, such like eternity in this case, it is 1. it has to be. call it religion.

bernhard

Originally posted by Tater Rocket:
1/3 doesn't equal an infinite number of .3's. It APPROACHES 1/3, but isn't equal. Same with .9 with an infinite number of nines. It will never be EXACTLY 1.0 with an infinite number of 0's, but it will get so close it doesn't really matter. Like they said, it is all about limits and tolerances.

Spud

Click to expand...

<font size="2" face="Verdana, Arial">A good point: Give me a non-zero number that is between .9999... and 1.

Originally posted by logicnerd411:
A good point: Give me a non-zero number that is between .9999... and 1.

Click to expand...

<font size="2" face="Verdana, Arial">Zero isn't between .9999... and 1 either.

No. .999... Approachs 1 as a limit. There can be only 1 answer to any one math question. However, most math problems in the practical world are poised incorrectly.

Do the problem below in your head.

3 guys check into a motel for the night. The desk clerk is new at his job and charges each guy ten dollars. Later he relises he over charged them, he should have charged them nine dollars. So, he takes five dollars out of the till to pay them back. On the way to the rooms he gets greedy and pockets two dollars for himself. Then he gives each guy back a dollar, now they have each paid nine dollars. 3x9=27 dollars plus the two in his pocket =29 dollars. Where did the other dollar go?

This is the reason I was given. Given two real numbers x and y,x is not equal to y iff there exists another real number z not equal to zero such that x+z=y. Thus if .99999... is not equal to 1, there exists a positive real number z such that .9999... + z = 1. But there is no such z, since the difference between .9999 and 1 is smaller than any positive real number. The difference is in fact zero. This is basically what Max said.

Originally posted by logicnerd411:

PROOF 2:
LET x = .9999...
THEREFORE:
10x = 9.9999...
9x = 9
x = 1

Click to expand...

<font size="2" face="Verdana, Arial">I get 9x = 8.9991

An nice one I've always liked is showing that there are more real numbers between 0 and 1 than natural numbers, even though there are an infinite number of either. The natural numbers are 1,2,3,... and the reals between zero and one are any positive decimal expansion without any natural number to the left of the decimal point (and excluding .99999999... since that is 1). Examples of reals between zero and 1 are .1 , .3333333.... , e-2 = .718281828 followed by a bunch of stuff. Two sets are of equal size when we can (putting this very loosely) match each element of one set with one element of the other. More technically if there is an into and onto mapping from one set to the other. Now, imagine we have constructed a list of all the reals between 0 and 1 and we have matched them up with the naturals. This would look something like the following:

Natural Real
1 .xyz....
2 .abcd...
... ....

Well, there is always a way to construct a number not in the list (and therefore there is no possible matching up). This new number would be constructed by choosing a number that differs from the first real number in the first term to the right of the decimal, from the second real in the second digit etc.

x = (1 + 2 + 4 + 8 + 16 + ....)
x-1 = (2 + 4 + 8 + 16 + ....)
(x-1)/2 = (1 + 2 + 4 + 8 + 16 + ...)
(x-1)/2 = x
x-1 = 2x
x-1-x = 2x-x
-1 = x
-1 = (1 + 2 + 4 + 8 + 16 + ....)

As a calibration technician I would call this a resolution problem. Displayed on a 4 digit display, .9999 would appear as .9999, if rounded up to 3 digits you end up with 1.00.
If you look at .9999 on a 8 digit display you will get .99990000 which is definitely not equal to 1.
Its rather like the crazy specifications of some of the test equipment I have to calibrate!

[No message]

Originally posted by EMPOWERTORCH:
As a calibration technician I would call this a resolution problem. Displayed on a 4 digit display, .9999 would appear as .9999, if rounded up to 3 digits you end up with 1.00.
If you look at .9999 on a 8 digit display you will get .99990000 which is definitely not equal to 1.
Its rather like the crazy specifications of some of the test equipment I have to calibrate!

Click to expand...

<font size="2" face="Verdana, Arial">I chose 4 9's randomly. The "..." signifies repeating.

Originally posted by GJW:
x = (1 + 2 + 4 + 8 + 16 + ....)
x-1 = (2 + 4 + 8 + 16 + ....)
(x-1)/2 = (1 + 2 + 4 + 8 + 16 + ...)
(x-1)/2 = x
x-1 = 2x
x-1-x = 2x-x
-1 = x
-1 = (1 + 2 + 4 + 8 + 16 + ....)

Click to expand...

<font size="2" face="Verdana, Arial">Thus you have shown an instance of why the axioms of the real numbers apply to the real numbers (and not to infinite ammounts). There is something call "nonstandard analysis" that extends the real numbers to include infinitesimals and by the fact that the operation of the reals are valid on the extension, 1/an infinitesmimal or what you are calling x. But, the the axioms in this extension are chosen very carefully. An example of the problem of treating nonreals as reals is the hand waving in integral calculus that the integral is a summing of boxes of height f(x) with width dx, with area f(x)dx. dx is not a real number so the sort of problem you may encounter with misapplying the axioms of the real numbers and using the handwaving notion of an integral is that it is possible to demonstrate that a cylinder of fixed height and radius has infinite surface area.