what wrong in this prove? 0.999=1.000. let x=0.999, 10x=9.999 10x-x=9x, 9.999-0.999=9 , 9x=9 then x=1?

Answer 1

As mentioned by others, 10*.999 = 9.99, not 9.999, so your proof is wrong. Now, if your decimal has a truly infinite number of nines, then it would be equal to 1. And speaking of which:

questor_2001, just saw your comment on the other question. I still haven't "settled" with Tom - mainly because we don't know each other and it's hard to hold a conversation on Y!A. Anyway, I find interesting the conjunction of statements: "At any given point in the decimal sequence, the 1 has a zero, the other has a 9. Rounding the subsequent 9's to 10 begs the question." "More to the point, 2/3 = 0.666666666666666......7, meaning that wherever you have stopped to look at the final digit, it is a 7, not a 6, and even when you look another 10,000 digits past it, the terminal digit is a 7, not a 6." Tell me, why is it that whenever you look at the final digit of an n-digit approximation to .999... you truncate, but when you look at the final digit of an n-digit approximation to 2/3, you round? Isn't that begging the question?

In fact, in your argument, you actually present (quite inadvertently, I'm sure) a fairly convincing proof that .999... is equal to 1. You state: "1.0000000000000000000000000000... is different than 0.99999999999999999999999, to any degree of accuracy demanded." Let's test that:

To within 1/10, is 1 different from .9999...? No, because .9999... + .1 > 1.
To within 1/100, is 1 different from .9999...? No, because .9999... + .01 > 1.
To within 1/1000, is 1 different from .9999...? No, because...

In fact, we can show quite easily that there is NO positive rational ε such that |1 - .9999...| ≥ ε. Which means that these numbers are the same to any degree of accuracy, which means they must be the same, period.

If we want to formalize this, we can note that by the archimedian property, the nonexistence of positive rational ε such that |1 - .9999...| ≥ ε implies the nonexistence of positive real ε such that |1 - .99999...| ≥ ε (the archimedian property basiaclly says that there are no positive real numbers smaller than every positive rational number). But if both of these are real numbers, then so is their difference, and the only nonnegative real number smaller than every positive real number is zero. Therefore 1 - .9999... = 0, which means 1 = .9999....

Also, as I note in my answer to http://answers.yahoo.com/question/index;_ylt=Aptc2Yy1m1to3M3w2ZrbVOfsy6IX?qid=20060904113823AA1iALg , in anything like the decimal system you're going to have identities like .9999... =1. This doesn't even depend on the archimedian property, but simply on the fact that the reals are dense - there is a real number strictly between any two distinct real numbers, but there is no decimal between .9999... and 1 (no, appending a 1 to the end of the decimal doesn't work, because there is no end to append it to). This is why having multiple decimals that give the same number doesn't surprise professional mathematicians - they usually understand that this sort of thing has to happen in order to get the decimal system to work. And why having different decimal expansions, while a necessary condition of two numbers being unequal, is NOT a sufficient condition.

- Pascal

Answer 2

If you're going to copy a question, at least COPY it correctly. The original is subtle though incorrect, but your version lacks even that charm.

let x = 0.999

then 10x=9.990, not 9.999

10x-x=9x = 9.990-0.999 = 8.991

therefore, x = 0.999 which was to be proved

Sorry, but 0.999 <>1.000

....and it wasn't when the more interesting version was asked last week either!

Answer 3

If x=0.999 then 10x=9.99 thus not 9.999

Answer 4

Here x is not equal to 0.9999999999999..............
you have taken up to three decimal
0.999=1.000. let x=0.999, 10x=9.999 10x-x=9x, 9.990 -0.999=8.991 , 9x=8.991 then x=0.999 therefore x is not equal to 1
Hence x should be equal = 0.999999999999999..........upto infinity then x =1

Answer 5

that is wrong, because if x = 0.999, than 10x = 9.99 and not 9.999

Unless you mean 0.999... :P

Answer 6

Nothing wrong with it, you can do it that way. There are so many proofs...

You can also model an approximation to n decimal places as the sum of the terms of the geometric series with a=0.9 and r=0.1 so that the summation to infinity is a/(1-r)=0.9/0.9 = 1