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-   -   .9 Repeating (https://www.techwarelabs.com/community/showthread.php?t=485)

Jason425 10-05-2003 05:46 PM

ok let's make this simple.. you have a big thing of grapes, a truckload, no a highway of trucks full of grapes. Say there are 1,000,000,000,000,001 grapes. Take one grape away and you have 1,000,000,000,000,000 grapes. Now tell me how there is the same number of grapes before and after? Think of it this way and you can see how rediculous it is to just move to another number for kicks.. 8)

eviltechie 10-05-2003 07:40 PM

i like being a scientific man so i will share my view the scientific way

due to our limitations of experiments, our instruments can only record accurately up to 6 significant figures

therefore, 0.99999999999999999 and repeating
would round up to be 1.00000

Jason425 10-05-2003 09:02 PM

yes, but rounding isn't exact... and since we're talking about .x1 , shouldn't it be exact? since we don't know exactely, we call them 2 seperate numbers and end it there

james 10-05-2003 09:58 PM

Re: .9 Repeating
 
Quote:

Originally Posted by vee_ess
yeah, 9.9~ - .9~ is not 9 but 9.0~9 cuz if you were to actually do it you'd have to keep the same number of sig figs.

NOOO. that is SOO wrong. Sorry. You are totally misleading people. This is an abstract mathematical problem where sig figs DO NOT MATTER. infinite is infinite. therefore, muliplying 0.9~ by 10 does NOT shift everything to the left one space. There are the same number of 9's to the right of the decimali as there were before and after. therefore
a = { 10n = 9.9~
b = { n = 0.9~
a-b = { 9n = 9 => n = 1 but by by n = 0.9~ therefore 1 = 0.9~ is quite legitimate.

Some people here have definitely not taken calculus, and certainly haven't taken calc w/ theory behind it. If you haven't, i'd seriously consider STOP misleading people b/c of "simple" logic, that unfortunately doesn't hold shit on Newton. As others have indicated, you can do this in a much more elegant proof involving limits, but the above is enough. Now there is some confusion about how to deal with addition and subtraction of "infinity". THe problem here is, we have no infinity. We have infinitely many, so yea, we can add and subtract all day long. Very nice, if you DO want to learn how to add and subtract and compare the sizes of different kinds of infinity (among other things that are seemingly impossible w/ "simple" logic), I suggest learning advanced Set Theory.

For those who think that a number can't be two things at once, I suggest learning some more quantum physics. The problem of Schrodenger's cat, once you can wrap your mind around it, can be very enlightening. (hint: if your head doesn't hurt, you don't understand anything any quantum physicist has ever said. hint 2: if your head does hurt, you probably still don't understand most of what any quantum physicist has ever said.)


this

Jason425 10-05-2003 10:50 PM

well now i'm just upset and frustrated.. i'm in pre-cal.. grrr.. :mad:
oh well.. you can have your one number = 2 diff number bs.. :rolleyes:

lithiumdeuteride 11-05-2003 03:10 PM

I will now sum up (no pun intended) and refute the various invalid arguments people have made, then offer three valid proofs, in order of complexity, that zero-point-repeating-nine really does equal one.

Aoshi: Your arguments are flawed because we are not dealing with real life here. In laboratory measurements, there are error bands, written as "+/- 4%", for example. There is no uncertainty when dealing with pure math. It is valid to say that 6 +/- 2 is the same measured value as 7 +/- 2, but it's NOT valid to say that 6 = 7. Repeating numbers do have actual values, and CAN be multiplied. Consider 4/3. It is NOT an irrational number, because it CAN be expressed as a ratio of real, whole numbers, and therefore can be multiplied or divided or whatever.

Grinnin Reaper: Get used to the idea that common sense sometimes fails in the realm of math and science. Also, rounding does not simply things, it makes them more difficult to keep track of, because you have introduced error bounds where none previously existed.

vee_ess: There is little similarity between pi and 0.99999999~ because the former is a non-repeating decimal, and the latter has a simple pattern. Any ratio of two whole numbers can be expressed as a decimal with a repeating pattern. Since there are infinite combinations of numbers to make ratios with, it follows that any repeating decimal can be expressed as a ratio of two whole numbers. A ratio that is equal to 0.99999999~ is 1/1, or 1.

evil techie: You are certainly quite wrong about the accuracy of our instruments. Experiments are dependant upon the accuracy of instruments, and NOT the other way around (unless you count breaking the equipment). We also know the speed of light to at least eight significant digits. I therefore conclude that you have no idea what you're talking about, as I had previously suspected.

Here are those 3 proofs I promised:


PROOF 1:

1/3 = 0.33333333~
multiply both sides by 3:
1 = 0.99999999~


PROOF 2:

let x = 0.99999999~

multiply both sides by 10:
10x = 9.99999999~

subtract x = 0.99999999~ from 10x = 9.99999999~:
10x = 9.99999999~
- x = 0.99999999~
9x = 9.00000000~

divide both sides by 9:
x = 1


PROOF 3:

the series SUM(c*r^n,n,m,infinity) converges when |r|<1. read this as: "the riemann sum of c times r to the nth power where n equals m to infinity".

SUM(c*r^n,n,m,infinity) converges to this value: (c*r^m)/(1-r)
the proof that this what the series converges to is found in many calculus books.

setting variables:

let r = 1/10:
SUM(c*(1/10)^n,n,m,infinity) = (c*(1/10)^m)/(1-(1/10)) = c*(1/10)^m)/(9/10) = (10/9)*c*(1/10)^m

let c = 9:
SUM(9*(1/10)^n,n,m,infinity) = (10/9)*(9)*(1/10)^m = 10*(1/10)^m

let m = 1:
SUM(9*(1/10)^n,n,1,infinity) = 10*(1/10)^1 = 1

kakomu 11-06-2003 02:02 AM

it's you Primarscources!!!!

Ok, here's the refutation.

First of all, the use of a Calculator is invalid. It does not give exact answers when it comes to infinity (as it would infinitly loop and never come up with an answer). It actually uses series and sequence to calculate infinite numbers.

So, I'm going to explain this with Summation and series:

Sigma from 1->Infinity of 9 * .1^x

If you were to continually add this up, you will find that there is an asymptote of 1 (as a limit of 1). This clearly shows that repeating 9 does NOT equal 1.

If you were to try and do the integral of 9*.1^x from 1 to infinity, you will find that just plugging in infinity does NOT work.

Also, on the subject of Limits, if we have the Limit as X->infinity of 1/x, it equals 0, however, it's pretty obvious that there is NO number you can plug into 1/x to equal 0. With that being said, it's pretty simple to make the assessment that 0 is the ASYMPTOTE, much like 1 is the ASYMPTOTE of repeating 9.

I rest my case :D

lithiumdeuteride 11-06-2003 12:03 PM

don't rest just yet
 
Let's assume you are correct for a moment, and that 0.99999999~ actually IS a different number than 1, and get right down to the fundamentals of numbers in general. If the numbers are different, then there either ARE or ARE NOT other numbers between them. If there ARE other numbers between 0.99999999~ and 1, I challenge you to find one. If there ARE NOT any numbers between 0.99999999~ and 1, then 0.99999999~ must be the PRECEDING number, that is, it comes right before it. If it does come right before it, then a basic property of the Real Number System is invalid. That property is that there IS NO NEXT NUMBER. The Real Numbers are continuous, and often referred to as a Continuum.

I quote from the definition of Real Numbers from www.whatis.com :

"If x and z are real numbers such that x < z, then there always exists a real number y such that x < y < z."

This means that for any two numbers x and z where one of them is actually larger than the other, and thus different, as we are supposing 0.99999999~ and 1 to be, there is another number y inbetween them. Find that number y, I dare you!

Jason425 11-06-2003 07:14 PM

both arguments are invalid, as you two have proven, any argument involving the word infinity can not be proven with certainty (at least this one) both arguments are correct, but then who's right? Nobody, because it's infinity. (assuming that silly number rule is correct, or perhaps we just proved it wrong?)

james 11-07-2003 12:36 AM

Quote:

Originally Posted by Jason425
both arguments are invalid, as you two have proven, any argument involving the word infinity can not be proven with certainty

what are you talking about?? Mathmaticians deal with infinity in very precise ways all the time. the people here are using the word infinity in imprecise ways because they don't know math from a hole in the wall, hence all the confusion. People have dealt with this issue in a variety of legitimate ways. Why do people keep dealing with this in such ignorant ways? I don't understand, if you've had any math at all, you know this answer, if you don't, why even bother TRYING to make an explanation


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