1. ## 0.999... = 1

In mathematics, the recurring decimal 0.999, which is also written as or , denotes a real number equal to 1. In other words, the notations "0.999" and "1" represent the same real number. The equality has long been accepted by professional mathematicians and taught in textbooks. Various proofs of this identity have been formulated with varying rigour, preferred development of the real numbers, background assumptions, historical context, and target audience.

The non-uniqueness of real expansions such as 0.999 is not limited to the decimal system. The same phenomenon occurs in all integer bases, and mathematicians have also quantified the ways of writing 1 in non-integer bases. Nor is this phenomenon unique to 1: every non-zero, terminating decimal has a twin with trailing 9s, such as 28.3287 and 28.3286999. For simplicity, the terminating decimal is almost always the preferred representation, contributing to a misconception that it is the only representation. Even more generally, any positional numeral system contains infinitely many numbers with multiple representations. These various identities have been applied to better understand patterns in the decimal expansions of fractions and the structure of a simple fractal, the Cantor set. They also occur in a classic investigation of the infinitude of the entire set of real numbers.

Algebraic

Fractions

One reason that infinite decimals are a necessary extension of finite decimals is to represent fractions. Using long division, a simple division of integers like 1⁄3 becomes a recurring decimal, 0.333, in which the digits repeat without end. This decimal yields a quick proof for 0.999 = 1. Multiplication of 3 times 3 produces 9 in each digit, so 3 × 0.333 equals 0.999. And 3 × 1⁄3 equals 1, so 0.999 = 1.

Another form of this proof multiplies 1/9 = 0.111 by 9.  An even easier version of the same proof is based on the following equations: Since both equations are valid, by the transitive property, 0.999 must equal 1. Similarly, 3/3 = 1, and 3/3 = 0.999. So, 0.999 must equal 1.

http://en.wikipedia.org/wiki/Proof_t...99..._equals_1  Reply With Quote

2. Its fun isnt it! I like all the 3s and 9s. 9 has some strange properties it seems.

But really, 0.9999 recurring is only ever =1 when taken "to the limit" - but that must mean infinty surely --- which is never going to be possible. (??) Which means surely , that in fact it will never be the equal to 1. These proofs must be "slights of mathematics".

But then, what are facts?  Reply With Quote

3. Digit manipulation

Another kind of proof more easily adapts to other repeating decimals. When a number in decimal notation is multiplied by 10, the digits do not change but the decimal separator moves one place to the right. Thus 10 × 0.999 equals 9.999, which is 9 greater than the original number.

To see this, consider that subtracting 0.999 from 9.999 can proceed digit by digit; in each of the digits after the decimal separator the result is 9 − 9, which is 0. But trailing zeros do not change a number, so the difference is exactly 9. The final step uses algebra. Let the decimal number in question, 0.999, be called x. Then 10x − x = 9. This is the same as 9x = 9. Dividing both sides by 9 completes the proof: x = 1. Written as a sequence of equations, The validity of the digit manipulations in the above two proofs does not have to be taken on faith or as an axiom; it follows from the fundamental relationship between decimals and the numbers they represent. This relationship, which can be developed in several equivalent manners, already establishes that the decimals 0.999 and 1.000... both represent the same number. (A second version is as follows: 1/11=0.09090909..... 10/11=0.90909090..... (1/11)+(10/11)=.99999999..... But,(1/11)+(10/11)=1 Therefore,by substitution, 1=.999999999..... )  Reply With Quote

4. That has some correlation to the old elemental difference as we learnt it in Physics, doesn't it? The charge difference between electrons and protons is so minimal that it cannot be denoted by any real number, yet is enough of a difference to charge one positively, and the other one negatively. In a way the difference in charge between the two is about the difference between 0.999... and 1 - there is a very real difference, but cannot be denoted because it is "zero without being zero". Following me there? ----------------------------------------

Other proofs which lead the expressability of mathematics ad absurdum are for example that which I call the "zero sham proof" which equates 1 to a number close to 0 - plus what ever constant may be used there. Let us take the number 1.
This number can also be written as 0^0.
If we differentiate 0^0 in the manner of f(x) = nx^(n-1), then f(x) = 0^(-1).
0^(-1) = ∞ for all intents and purposes, because x^(-1) is always 1/x. 1/x however then would be 1/0, and division by zero gives one infinity.

Now if we integrate this number ∞ to reach back to the original, then

∫f(x)dx = 1/(x^(n+1) + C ... f. ex. ∫4x dx = 4/2x^(1+1) + C = 4/2x^2 + C= 2x^2 + C.

In this case however ∫∞ dx = 1/∞x + C ... 1/∞ ≈ 0 ... thus ∫∞ dx ≈ 0 + C

Quod erat demonstrandum?   Reply With Quote

5. ## Servus, Sigurd...

I didn't get any of that....a blede Gschicht.  Reply With Quote

6. Originally Posted by Sigurd Let us take the number 1.
This number can also be written as 0^0.
If we differentiate 0^0 in the manner of f(x) = nx^(n-1), then f(x) = 0^(-1).
0^(-1) = ∞ for all intents and purposes, because x^(-1) is always 1/x. 1/x however then would be 1/0, and division by zero gives one infinity.

Now if we integrate this number ∞ to reach back to the original, then

∫f(x)dx = 1/(x^(n+1) + C ... f. ex. ∫4x dx = 4/2x^(1+1) + C = 4/2x^2 + C= 2x^2 + C.

In this case however ∫∞ dx = 1/∞x + C ... 1/∞ ≈ 0 ... thus ∫∞ dx ≈ 0 + C

Quod erat demonstrandum? I've seen quite a few "proofs" utilizing ∞ that "prove" all kinds of craziness. However, I'm sure we all know that the lemniscus cannot generally be used in standard mathematical operations outside of integration and summation. What is interesting is Cantor's transfinite mathematics that does allow us to manipulate infinite sets arithmetically.  Reply With Quote

7. The simple fact is surely that you cant do mathematics like this using infinite numbers of any kind. It is indeed sham - it is an unacceptable procedure.

One can never actually carry out any of these operations , let alone move on the next stage, since one can't definitively conclude an infinite expression. That includes Infinity itself ( here I ) and any expression, such as 0.999 recurring. You tell me the difference here, without using shorthand cop outs! You would never be able to conclude the expression , its full statement without the short hand would take an infinity of time to express!

Besides,

I x 2 = I , I x 3 = I , I - anything expressable = I
.....................   Reply With Quote

8. So what happens to π (Pi) when 1. is considered absolutely as .9~, does it have a square root? ;-P  Reply With Quote

9. This only works if you round off, just as Pi is not simply 3.14. Wouldn't a rational function with an asymptote at 1 (x or y) disprove this possibility (because the function becomes undefined at 1, but continues on at .9~ into infinity)?  Reply With Quote

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