Matter /CAN/ be created!

[Olas]

Saying that a is equivalent to b means that a=b if and only if b=a. Equivalence thus implies equality).

So equivalence equals equality? You just blew my mind. I suppose mathematically that would be written as. ===

 

[oktalist]

.000_1 = 0
But .000_1 is such a messy way of typing that number, so let us clean it up. It basically means an infinitely small number so using & as a symbol for an infinitely large number:

1/& = 0

Now we all know any number multiplied by 0 equals 0, even an infinitely large number.

If that was true then you could do this:

1/& = 0

Multiply both sides by &

1 = 0

Regardless, the math for proving .999r=1 is still wrong, considering the equations are just proving that x=x

It's not wrong, it's just not a proof of .999... = 1. That doesn't mean that no proofs of .999... = 1 exist. Several do exist.

 

[Lukeje]

Saying that a is equivalent to b means that a=b if and only if b=a. Equivalence thus implies equality).

So equivalence equals equality? You just blew my mind. I suppose mathematically that would be written as. ===

Nah, it would be ===>= (using => for implies and == for equivalence).

 

[Shadowkire]

An infinitely large number multiplied by 0 is NOT 0, it is undefined.

Sorry, you are correct, so let me redo that:

.000_1 = 0

Subtracting or adding 0 from/with any number equals that number

So .999_ - .000_1 must equal 1 or .999_...

.999_8 So .999_8 = 1

.999_8 - .000_1 = .999_7

Let us try this an infinite number of times, checking our answers along the way. Let us try it with addition as well. Every time it shows 1 equals a new number.

In fact, adding or subracting .000_1 an infinite amount of times will eventually return every single number as equaling 1. Which is false.

 

[Makhiel]

Regardless, the math for proving .999r=1 is still wrong, considering the equations are just proving that x=x

That's the point isn't it? They are one and the same number.

My previous argument still stands and you don't even have to multiply.
0.9 can be rewritten as 0.9000 (plus arbitrarily more zeros if you want) but you cannot slap more zeros to 0.9_ no matter what.

edit:

An infinitely large number multiplied by 0 is NOT 0, it is undefined.

Sorry, you are correct, so let me redo that:

.000_1 = 0

Subtracting or adding 0 from/with any number equals that number

So .999_ - .000_1 must equal 1 or .999_...

.999_8 So .999_8 = 1

.999_8 - .000_1 = .999_7

Let us try this an infinite number of times, checking our answers along the way. Let us try it with addition as well. Every time it shows 1 equals a new number.

In fact, adding or subracting .000_1 an infinite amount of times will eventually return every single number as equaling 1. Which is false.

How did 0.999_ suddenly became 0.999_9 ?

 

[Wyes]

Regardless, the math for proving .999r=1 is still wrong, considering the equations are just proving that x=x

Except that we are trying to prove x=x, that's the point of the exercise.
0.999... = 1 is a widely accepted mathematical fact. This is not a bunch of people on a forum going 'Look I'm a genius and I proved this new thing', it's a guy going 'Hey I learnt this and thought it was super cool it (which it is) but I overestimated its significance'

Personally I'm a much bigger fan of the simpler approach (which has been shown here before).

1/3 = 0.333...
1/3 * 3 = 0.333... * 3
3/3 = 0.999...
1 = 0.999...

It's really just as simple as that.

Now, I'm not going to go and call anyone an idiot for disbelieving this, it doesn't seem right, in a way it's nice to see so many people try to apply critical thinking to the problem, and not just assuming its correct (though let's face it, that's really coming about because they heard it from a guy on the internet).

 

[Shadowkire]

.000_1 = 0
But .000_1 is such a messy way of typing that number, so let us clean it up. It basically means an infinitely small number so using & as a symbol for an infinitely large number:

1/& = 0

Now we all know any number multiplied by 0 equals 0, even an infinitely large number.

If that was true then you could do this:

1/& = 0

Multiply both sides by &

1 = 0

That was my point, until someone pointed out infinity doesn't work with normal math, that if 1 = .999_ then it follows that 1 = 0. I used that to point out the ridiculousness of the OP's proof.

 

[Shadowkire]

An infinitely large number multiplied by 0 is NOT 0, it is undefined.

Sorry, you are correct, so let me redo that:

.000_1 = 0

Subtracting or adding 0 from/with any number equals that number

So .999_ - .000_1 must equal 1 or .999_...

.999_8 So .999_8 = 1

.999_8 - .000_1 = .999_7

Let us try this an infinite number of times, checking our answers along the way. Let us try it with addition as well. Every time it shows 1 equals a new number.

In fact, adding or subracting .000_1 an infinite amount of times will eventually return every single number as equaling 1. Which is false.

How did 0.999_ suddenly became 0.999_9 ?

the underscore is my way of saying "repeating", the number after the underscore is my way of saying "at the end of this infinitely repeating number is an 8, or a 7."

Yes I am aware of how odd it is to try an place a number at the end of an infinite.

 

[Wyes]

.000_1 = 0
But .000_1 is such a messy way of typing that number, so let us clean it up. It basically means an infinitely small number so using & as a symbol for an infinitely large number:

1/& = 0

Now we all know any number multiplied by 0 equals 0, even an infinitely large number.

If that was true then you could do this:

1/& = 0

Multiply both sides by &

1 = 0

That was my point, until someone pointed out infinity doesn't work with normal math, that if 1 = .999_ then it follows that 1 = 0. I used that to point out the ridiculousness of the OP's proof.

The OP's proof is not a good one, I feel it over complicates it, but one can replicate the results using other, much simpler processes.

EDIT: Actually the wiki article [http://en.wikipedia.org/wiki/0.999...] really covers this pretty well.

 

[Makhiel]

the underscore is my way of saying "repeating", the number after the underscore is my way of saying "at the end of this infinitely repeating number is an 8, or a 7."

Yes I am aware of how odd it is to try an place a number at the end of an infinite.

I didn't ask what the underscore meant.
And yes, it's so odd that it is impossible. As such none of your proofs in which you've used it is valid.

 

[oktalist]

Yes I am aware of how odd it is to try an place a number at the end of an infinite.

It's not just odd, it's completely backwards and meaningless.

You are wasting your time trying to disprove this "proof". The "proof" presupposes the proposition that it is trying to prove, a proposition that happens to be true, so the only way to disprove it is with a "disproof" that presupposes the contradictory proposition.

 

[irishda]

Except that we are trying to prove x=x, that's the point of the exercise.
0.999... = 1 is a widely accepted mathematical fact. This is not a bunch of people on a forum going 'Look I'm a genius and I proved this new thing', it's a guy going 'Hey I learnt this and thought it was super cool it (which it is) but I overestimated its significance'

Personally I'm a much bigger fan of the simpler approach (which has been shown here before).

1/3 = 0.333...
1/3 * 3 = 0.333... * 3
3/3 = 0.999...
1 = 0.999...

It's really just as simple as that.

Now, I'm not going to go and call anyone an idiot for disbelieving this, it doesn't seem right, in a way it's nice to see so many people try to apply critical thinking to the problem, and not just assuming its correct (though let's face it, that's really coming about because they heard it from a guy on the internet).

That's the point isn't it? They are one and the same number.

The OP's original equations don't show .999r=1 because the math is off. It's like that riddle about the waiter splitting a $30 bill three ways and losing $3. The logic is just slightly off.

9(.999r) =/= 9

The assumption is made in the original equation that 9x=9, but 9(.999r) is only 8.99r. 9(.999r) will equal 9 only if you already go into the equation with the assumption that .999r=1. If we ACTUALLY go through the equation by replacing the x with .999r (since that's what x is supposed to equal) we get the TRUE result of the equation.

.999r=.999r
10(.999r)=10(.999r)
9.99r=9.99r
9.99r-.999r=9.99r-.999r
9=9
9/9=9/9
1=1

OR substitute the opposite

x=x
10x=10x
10x-x=10x-x
9x=9x
x=x

Either way you don't get .999r=1

As for the fractions, I understand you need to know fractals in order to refute that one, but it's still refutable.

 

[Makhiel]

The assumption is made in the original equation that 9x=9, but 9(.999r) is only 8.99r.

But 8.99r is 9.

 

[Truth Cake]

Ummmmmmmmmmmmm.

1/3 = 0.3 rep

2/3 = 0.6 rep

3/3 = ?

Once again, this is dealing with infinitesimals, which cannot be measured, so there's bound to be stupid little errors like that when you try to measure them (or leave out measuring them when they're supposed to be included, as the case may be).

I'll be blunt- 1/3 does not QUITE equal .333... (and by proxy, 2/3 doesn't quite equal .666...), that's just as close as we can come to measuring it since we can't write something to the infinity decimal place. (unless you've found a way that I don't know about, which no offense, but I highly doubt)

And obviously 3 divided by 3 is 1, plain and simple.

 

[Wyes]

Either way you don't get .999r=1

As for the fractions, I understand you need to know fractals in order to refute that one, but it's still refutable.

As with most mathematical proofs (particularly in mathematical induction), you presuppose that what you're trying to prove is true. Then, if it is true, it's all internally consistent; if it's not true, there is some contradiction which breaks the internal consistency.

In the proof using fractions, we are working under the assumption 1/3 = 0.333...
This is the key step. Most people do not question this step, because it is elementary. If you disagree with it, then we'll never agree.
If you are interested in other proofs, there's a geometric series proof here [http://www.purplemath.com/modules/howcan1.htm], though you'll only accept this one if you can accept that what we know about geometric series is correct.

As for refutability; these are not 'refutable' because they are facts, it doesn't matter how much you know about infinitives, transinfinitives or fractals, they remain true. Maybe some day we will discover we were wrong, but it hasn't happened yet, and I doubt that anybody here is going to be the person to disprove it (but hey, stranger things have happened).

 

[Wyes]

Once again, this is dealing with infinitesimals, which cannot be measured, so there's bound to be stupid little errors like that when you try to measure them (or leave out measuring them when they're supposed to be included, as the case may be).

I'll be blunt- 1/3 does not QUITE equal .333... (and by proxy, 2/3 doesn't quite equal .666...), that's just as close as we can come to measuring it since we can't write something to the infinity decimal place. (unless you've found a way that I don't know about, which no offense, but I highly doubt)

And obviously 3 divided by 3 is 1, plain and simple.

I must I must double post; 0.333... and their ilk are not approximations. If you try to take them to some amount of decimal places, then yes, they are approximations, but they go on to infinity! There is no 'error'.

 

[irishda]

The assumption is made in the original equation that 9x=9, but 9(.999r) is only 8.99r.

But 8.99r is 9.

Again, only if you already believe that .999r=1. But that's changing the math to suit your belief. It doesn't change the fact that 9x=8.999r. If you divide that by 9, then oh look at that, you get x=.99r. Not x=1

 

[Truth Cake]

I must I must double post; 0.333... and their ilk are not approximations. If you try to take them to some amount of decimal places, then yes, they are approximations, but they go on to infinity! There is no 'error'.

You could've just edited your first pose to include the quote and response...

And yes, they ARE approximations, all non-terminating decimals are approximations since a precise amount in decimal form would have infinite numbers, which obviously cannot be written out or measured.

Put it this way: if you just have 3/3 by itself, as in 3 divided by 3, as in how many groups of 3 marbles can you make out of a whole set of 3 marbles? 1 group. Not .999... group(s), 1 group.

 

[oktalist]

Once again, this is dealing with infinitesimals, which cannot be measured

It's got nothing to do with measurement, and there is no infinitesimal involved.

I'll be blunt- 1/3 does not QUITE equal .333... (and by proxy, 2/3 doesn't quite equal .666...), that's just as close as we can come to measuring it since we can't write something to the infinity decimal place. (unless you've found a way that I don't know about, which no offense, but I highly doubt)

We have found a way to write infinity decimal places. That's what the ... on the end of .333... means.

 

[Maze1125]

But that confounds the problem because the only way of creating a recurring decimal is the decimal approximation of a fraction.

This seems to be the root of your confusion, your assumption that infinite decimals are just approximations.

They are not. An infinite decimal is defined as the limit of a series, and limits are precise. Specifically, if we're defining decimal 0.x[sub]1[/sub]x[sub]2[/sub]x[sub]3[/sub]x[sub]4[/sub]..., then the series that defines it is sum[sub]n=1[/sub][sup]m[/sup](x[sub]n[/sub]/10[sup]n[/sup]) and the limit of that series is, of course, sum[sub]n=1[/sub][sup]infinity[/sup](x[sub]n[/sub]/10[sup]n[/sup]).

Now, with the decimal 0.999..., x[sub]n[/sub]=9 for all n.
Therefore the sequence that defines the decimal is sum[sub]n=1[/sub][sup]m[/sup](9/10[sup]n[/sup]), and the limit of that is sum[sub]n=1[/sub][sup]infinity[/sup](9/10[sup]n[/sup]), which can be proven to be precisely 1 (I hope you don't need me to do that bit, but I can if you wish me to).

Further, the exact same method can be used to prove that recurring decimal notation of fractions are not approximations either, but are also precise.