# Thread: On the Equality of infinite Decimals, MMIX

1. ## On the Equality of infinite Decimals, MMIX

Greetings,
Ever since I was a child I have found my peers to be too slow in class so I now do my own research. However like Godel I have reached a point where I am now ready to reject much of the classical mathematical theory.

It is often said that 0.99999 recurring = 1

But 0.99999 is always less than 1 no matter how many nines are written therefore I do not accept this hypothesis.

If I may say so, this is perhaps the first time something has been disproven from the literature. I am looking for other advanced mathematicians to aid in the writing of this paper. ** PLEASE only respond if you are a gifted and talented individual. I do not want to waste time explaining my theorems to morons **

2. Originally Posted by boygenius112
Greetings,
Ever since I was a child I have found my peers to be too slow in class so I now do my own research. However like Godel I have reached a point where I am now ready to reject much of the classical mathematical theory.

It is often said that 0.99999 recurring = 1

But 0.99999 is always less than 1 no matter how many nines are written therefore I do not accept this hypothesis.

If I may say so, this is perhaps the first time something has been disproven from the literature. I am looking for other advanced mathematicians to aid in the writing of this paper. ** PLEASE only respond if you are a gifted and talented individual. I do not want to waste time explaining my theorems to morons **
$\displaystyle x = 0.9999999...$

$\displaystyle 10x = 9.999999...$

$\displaystyle 10x - x = 9.9999999... - 0.99999...$

$\displaystyle 9x = 9$

$\displaystyle x = 1$.

A simple, rigorous and infallible mathematical proof that yields your 'paper' worthless. I think you need to research more into the concept of infinity.

Your fallacy is on the basis of 'no matter how many nines you add, it's always less than one'. That implies that we are adding a finite number of descending powers of 9. We are not. We are adding an infinite amount.

Methinks thou is a troll.

3. nice try subtracting two infinite numbers

Using infinite peano arithmetic:
1 - 0.999999 ... = 0.000 ... 001 =/= 0

your method of proof is clearly flawed .. by your clownish argument, if we let x = 99999 ...
then 10x = x so x = 0 (contradiction)

4. Originally Posted by boygenius112
nice try subtracting two infinite numbers

Using infinite peano arithmetic:
1 - 0.999999 ... = 0.000 ... 001 =/= 0

your method of proof is clearly flawed .. by your clownish argument, if we let x = 99999 ...
then 10x = x so x = 0 (contradiction)
First of all, peano arithmetic is restricted to operations on NATURAL NUMBERS ONLY. Infinite decimals are quite irrational. Unless you're willing to accept the fact that we are dealing with the number 1 here, and hence peano axioms hold .

But in order to address your illegality of subtractin infinite numbers, allow me, then, to express it, not as an infinite number, but as the sum of infinitely many finite numbers.

$\displaystyle x = \displaystyle \sum_{n = 1}^{\infty} 9 \times 10^{-n}$

$\displaystyle = 9 \times \sum_{n = 1}^{\infty} 10^{-n} = 9(10^{-1} + 10 ^{-2} + 10^{-3}...)$

$\displaystyle 10x = 10 \times 9 \times \sum_{n = 1}^{\infty} 10^{-n}$

$\displaystyle = 9 \times \sum_{n = 1}^{\infty} 10 \times 10^{-n}$

$\displaystyle 9 \times \sum_{n = 1}^{\infty} \times 10^{1-n}$

$\displaystyle = 9 ( 10^0 + 10^{-1} + 10 ^{-2} + 10^{-3}...)$

$\displaystyle 10x - x = 9 \times \sum_{n = 1}^{\infty} \times 10^{1-n} - 9 \times \sum_{n = 1}^{\infty} 10^{-n}$

$\displaystyle = 9\big( 10^0 + 10^{-1} + ^{-2} + 10^{-3}...\big) - 9\big( 10^{-1} + 10^{-2} + 10^{-3} ...\big)$

$\displaystyle 9x = 9 \times 10^0$

$\displaystyle x = 10^0 = 1$

Now I'm not subtracting infinite numbers, I'm subtracting an infinite amount of finite numbers .

The fact of the matter is, that the infinite sum, given by $\displaystyle \displaystyle \sum_{n = 1}^{\infty} 9 \times 10^{-n}$ converges, and its finite value is 1.

And no. My proof doesn't apply to $\displaystyle 9999999$. The sum you proposed there is $\displaystyle \displaystyle \sum_{n = 0}^{\infty}9 \times 10^n$, which does not converge.

A simple integral convergence test is all that is necessary to prove that.

5. ## Confusion

Dear Mush,

I'm not a mathematician, but I'm interested in the logic of your position.

The query that I have regarding it is related to the Law of Contradiction: A is *not* Not-A.

Using that pattern, one would normally say .999... (A) is not 1 (Not-A). But what we seem to wind up with in the case that you present is .999... (A) *is* 1 (Not-A).

I mean that ordinarily one can say that, "Mary is either in her bedroom (A) or somewhere else (Not-A)". But if A can be Not-A, following the pattern of what you propose mathematically, it seems that one could say, "Mary is in her bedroom (A) *and* somewhere else (Not-A).

Confusing.

What's the solution?

Regards,

Jack Owens

6. Originally Posted by jackowens
Dear Mush,

I'm not a mathematician, but I'm interested in the logic of your position.

The query that I have regarding it is related to the Law of Contradiction: A is *not* Not-A.

Using that pattern, one would normally say .999... (A) is not 1 (Not-A). But what we seem to wind up with in the case that you present is .999... (A) *is* 1 (Not-A).

I mean that ordinarily one can say that, "Mary is either in her bedroom (A) or somewhere else (Not-A)". But if A can be Not-A, following the pattern of what you propose mathematically, it seems that one could say, "Mary is in her bedroom (A) *and* somewhere else (Not-A).

Confusing.

What's the solution?

Regards,

Jack Owens
Your theory assumes that, regardless of whether or not 0.999... is equal to 1, 0.999... is NOT equal to 1.

This is an invalid theory.

I'm not saying that 0.999... is equal to 1 AND equal to something else. I am stating that 0.999... is equal to 1, and only equal to 1.

7. ## Puzzle

Dear Mush,

In regard to your post of 1/29/09 (#6):

I'm still struggling.

If we take your set of equations from your post #2, when I get to 9X = 9, I'm looking for a referent.

In other words, 9X refers to 9 0.9999999...'s. Then, 9 0.9999999...'s divided by 9 gives 1 0.9999999....

Make sense?

Regards,

Jack:

8. Originally Posted by jackowens
Dear Mush,

In regard to your post of 1/29/09 (#6):

I'm still struggling.

If we take your set of equations from your post #2, when I get to 9X = 9, I'm looking for a referent.

In other words, 9X refers to 9 0.9999999...'s. Then, 9 0.9999999...'s divided by 9 gives 1 0.9999999....

Make sense?

Regards,

Jack:
We start with the equation $\displaystyle x = 0.999...$

And we logically arrive at the conclusion:

$\displaystyle 9x = 9$

There is one unique solution to this equation, and that is x = 1. This is what you get when you divide both sides of the equation by 9.

$\displaystyle \frac{9}{9}x = \frac{9}{9}$

$\displaystyle x = 1$

Now we started with x = 0.999... and now we have x = 1, but like I said, there is one unique solution to that equation. The conclusion that ends the dilemma is that we do not have 2 solutions to an equation which only has one solution. We have 1 solution which can be expressed in two difference ways, but are exactly the same thing.

x = 0.999...
x = 1

0.999... = 1

9. The www has discussion ad nauseam on this topic. There are several threads in MHF that also bang on with this topic.