Poll: 0.999... = 1

[Athinira]

So 0.333... is 0.0666... less than 0.4.

So 0.999... is less than 1?

If not, you're saying 0.4 x 3 = 1.

Again: You're confusing 0.333... with 0.3999....

0.333... < 4
0.3999... = 4

0.333... * 3 = 1
0.3999... * 3 = 0.4 * 3 = 1.2

 

[Piflik]

Of course there are infinitely small non-zero numbers.

So, how should we represent these numbers?

How bout 0.0000...0xx (where xx is any combination of decimals you can imagine)?

]

So, Piflik, have you come up with a reasonable explanation as to why the sum of ALL negative integer powers of two is not one? Why can I not replace all values of one with the sum of this series? What errors would this cause?

Yes, and just for you I will post it a third time

The point of converging infinite sums is, that you can always add another term and still not reach the limit. That's why it is called a limit. It will never go above this limit, no matter how many of these terms you add, and thus it can also never reach that limit, because then the next term you add will take it above the limit. And there will always be a next term, since we are talking infinity here.

Yes, but just because it never actually reaches that value, does not mean we cannot do maths with said value. We can work out what the value is, and in the case of 0.9 + 0.09 + 0.009 + 0.0009 + ..., it turns out to be 1.

Mathematicians use infinities all the time. Integral calculus uses it to find the area under graphs, geometric series can be used to solve Zeno's paradox. Just because in practice it never reaches that limit, does not mean it will do the same in theory.

You can do math with it all you want...I do it myself...it is useful. Inaccurate, but useful and when dealing with such small inaccuracies not working with it would be exceptionally stupid. I am not arguing that...for all practical intents and purposes 0.99999... = 1, but when you want to delve into the unambiguous beauty of maths, these two numbers have to be different.

Can you provide an example of where 0.999... = 1 causes a logical contradiction?

I would say the statement itself is a logical contradiction, but as far as day to day math goes, there is no situation I can think of, where such a degree of accuracy would be needed. I am not arguing that. I just say it is inaccurate.

 

[Maze1125]

So 0.333... is 0.0666... less than 0.4.

So 0.999... is less than 1?

If not, you're saying 0.4 x 3 = 1.

No 0.333... < 0.4

3 x 0.333... < 3 x 0.4

0.999... < 1.2

That's not a contradiction at all, as 1 is also less than 1.2.

So 0.999... = 1 < 1.2, everything is fine.

0.3999... = 0.4

But 0.3999... x 3 > 1, so there is still no problem.

 

[Addendum_Forthcoming]

If .9~ = 1, then both would be interchangeable. They aren't.

Yes they are.

The can't substitude .9~ for one because it is not properly divisible by 1.

Yes it is.
0.999... / 1 = 0.999... = 1

Does 12 x .9~ = 12? No.

Yes it does.

Really? How?

Logic dictates that a number shoulkd be perfectly (as represented by the figure '1') divisible by itself.

Logic should also dictate that the only way to divide .9~ is by itself to get 1, not 1.

Frankly this discussion is merely a mathematical equivalent of an idiotic wordgame.

Another mathematical equivalent would be saying the square root of 36 is -6, therefore -6 = 6.

If we are to treat numbers as concrete, then .9~ is not 1 ... for the simple fact that it isn't 1.

 

[Maze1125]

Of course there are infinitely small non-zero numbers.

So, how should we represent these numbers?

How bout 0.0000...0xx (where xx is any combination of decimals you can imagine)?

Well that's no good, as if you do a ... then another number, that means the ... terminates finitely. For example, the set of all positive integers up to n would be represented {1, 2, 3, ... n-1, n}

And if those ... do terminate finitely then that is just another non-infinitesimal real number.

 

[Vanaron]

If .9~ = 1, then both would be interchangeable. They aren't.

Yes they are.

The can't substitude .9~ for one because it is not properly divisible by 1.

Yes it is.
0.999... / 1 = 0.999... = 1

Does 12 x .9~ = 12? No.

Yes it does.

Really? How?

Logic dictates that a number shoulkd be perfectly (as represented by the figure '1') divisible by itself.

Logic should also dictate that the only way to divide .9~ is by itself to get 1, not 1.

Frankly this discussion is merely a mathematical equivalent of an idiotic wordgame.

Another mathematical equivalent would be saying the square root of 36 is -6, therefore -6 = 6.

If we are to treat numbers as concrete, then .9~ is not 1 ... for the simple fact that it isn't 1.

The square root of 36 is 6 not -6.

 

[Athinira]

Yes it does.

Really? How?

Logic dictates that a number shoulkd be perfectly (as represented by the figure '1') divisible by itself.

Logic should also dictate that the only way to divide .9~ is by itself to get 1, not 1.[/quote]

1 and 0.9r IS the same number, so you ARE dividing it with itself.

0.9r / 0.9r = 1
0.9r / 1 = 1
0.9r / (0.5 + 0.5) = 1
0.9r / (300 / 5 / 5 / (6 * 2) ) = 1

The representation is different, but the number is the SAME. 0.9r represents 1 in the same way that 1/2 represents 0.5 or 3/2 represents 1.5.


So 3.999... x 2 = 8?[/quote]

Yes.

Or 7.999... if you like, but since they both represent the same number, thats irrelevant. You could also write 8/1 or 16/2 if you like. Different representation, same number.

 

[Coldie]

If we are to treat numbers as concrete, then .9~ is not 1 ... for the simple fact that it isn't 1.

That is exactly why the numbers of the form 0.(9) are forbidden in the canonical representation of real numbers. If you allow 0.(9) to be a canonic number, then 1 will have two canon forms, but they must be unique (i.e. each Real number has exactly one canon form).

If you're not using the formal system of canonical real numbers, then 0.(9) can be used (as well as 0.9999... and 5/5 and whatever) to represent the number 1. Although it's still a bad idea, as you can see from this thread. It's a mathematical oddity that's completely alien to the layman.

 

[Addendum_Forthcoming]

If .9~ = 1, then both would be interchangeable. They aren't.

Yes they are.

The can't substitude .9~ for one because it is not properly divisible by 1.

Yes it is.
0.999... / 1 = 0.999... = 1

Does 12 x .9~ = 12? No.

Yes it does.

Really? How?

Logic dictates that a number shoulkd be perfectly (as represented by the figure '1') divisible by itself.

Logic should also dictate that the only way to divide .9~ is by itself to get 1, not 1.

Frankly this discussion is merely a mathematical equivalent of an idiotic wordgame.

Another mathematical equivalent would be saying the square root of 36 is -6, therefore -6 = 6.

If we are to treat numbers as concrete, then .9~ is not 1 ... for the simple fact that it isn't 1.

The square root of 36 is 6 not -6.

All positive numbers have two square roots, one positive and one negative. -6 x -6 = 36, in the same method as 6 x 6 = 36.

 

[Ishnuvalok]

SNIP

So 3.999... x 2 = 8?

I am going to post this again for you, since you seem to have not seen it because A) You missed it or B) You ignored it because it does not fit your conclusion.

Alright. Lets try this again.

We will define 0.999... as an infinite series.

0.999...=0+9(1/10) + 9(1/10)^2 + 9(1/10)^3 + 9(1/10)^4 + ...

Now we use a convergence theorem.

ar+ar^2+ar^3+...=ar/(1-r)

So

0+9(1/10) + 9(1/10)^2 + 9(1/10)^3 + ... = 9(1/10)/1-(1-10)

Which of course = 1.

 

[Maze1125]

Logic should also dictate that the only way to divide .9~ is by itself to get 1, not 1.

What?

0.999... / 0.999... = 1
0.999... / 1 = 0.999...

It also happens that 0.999... = 1.

Where's the problem?

Another mathematical equivalent would be saying the square root of 36 is -6, therefore -6 = 6.

No that's a quadratic problem, so it has two potential solutions.

1 - x = 0 is a linear problem, so it only has one solution for x. It just so happens that that solution can be represented in two different ways.

If we are to treat numbers as concrete, then .9~ is not 1 ... for the simple fact that it isn't 1.

Why do you believe that it factually isn't 1?

 

[Maze1125]

SNIP

So 3.999... x 2 = 8?

Yes, exactly.

 

[Piflik]

Of course there are infinitely small non-zero numbers.

So, how should we represent these numbers?

How bout 0.0000...0xx (where xx is any combination of decimals you can imagine)?

Well that's no good, as if you do a ... then another number, that means the ... terminates finitely. For example, the set of all positive integers up to n would be represented {1, 2, 3, ... n-1, n}

And if those ... do terminate finitely then that is just another non-infinitesimal real number.

As far as I know my terminology has not yet been introduced into the accepted maths, so I can decide what the '...' stands for, right? Just because finite numbers are represented that way, doesn't mean that would not work for infinite numbers...actually if you let n-> infinity, then the '...' in your example would also be an infinite number...

 

[Athinira]

Really? How?

Logic dictates that a number shoulkd be perfectly (as represented by the figure '1') divisible by itself.

Logic should also dictate that the only way to divide .9~ is by itself to get 1, not 1.

But you are dividing by itself, because the numbers are the same, just with different representation.

0.9r represents 1 in the same way that 3/2 represents 1.5, 2/8 represents 0.25, or 8/8 also represents one. Same number, just a different way of writing it.

And yes it also applies to 0.3r which is another way to write 1/3. Same number, different representation.

Your square-root example is terrible, because it's well-known that all square-roots of a positive number have TWO results. Completely unrelated to this.

Again: Leading mathmaticians over the world disagrees with you. You just don't understand the rules of real numbers.

SNIP

So 3.999... x 2 = 8?

Yes. Or 16/2. Or 12 - 4. Or 32/4. Or 7.999...

It doesn't matter how you decide to represent it, it's still the same number written in a ton of different ways.

 

[Addendum_Forthcoming]

That is exactly why the numbers of the form 0.(9) are forbidden in the canonical representation of real numbers. If you allow 0.(9) to be a canonic number, then 1 will have two canon forms, but they must be unique (i.e. each Real number has exactly one canon form).

If you're not using the formal system of canonical real numbers, then 0.(9) can be used (as well as 0.9999... and 5/5 and whatever) to represent the number 1. Although it's still a bad idea, as you can see from this thread. It's a mathematical oddity that's completely alien to the layman.

No I get what he's saying, it's not like it's hard to envisage .9~ as being '1', but It's still just an idiotic playing around with established ideals concerning the usage of numbers.

It would be like writing an essay in phonetic English rather than with actual words. One could argue it's still English, it's just a stupid thing to do.

 

[Maze1125]

As far as I know my terminology has not yet been introduced into the accepted maths, so I can decide what the '...' stands for, right?

No. It has to be unambiguous with existing terminology.
If your was accepted and I wrote 0.000...01, then it wouldn't be clear if I was talking about a very small and unknown, but still finite, power of 10. Or an infinitesimal power of 10.

It'd be like me suddenly deciding that |x| meant "multiply by -1". If someone wrote |6|, people wouldn't know if we were talking about the original sense or my sense, which would be the difference between the answer being 6 or it being -6.

actually if you let n-> infinity, then the '...' in your example would also be an infinite number...

But you wouldn't do that. You'd either know if you were talking about a finite or infinite set, or you'd represent the infinite possibility by writing {1, 2, 3, ...} separately.

 

[Athinira]

As far as I know my terminology has not yet been introduced into the accepted maths, so I can decide what the '...' stands for, right?

Sure, if you like. But you still perfectly well understand what '...' stand for when WE (or anyone who creates a thread about this) use it, so if you decide that it stands for something else, then you are just deliberately misunderstanding us, aka. trolling

 

[Vanaron]

All positive numbers have two square roots, one positive and one negative. -6 x -6 = 36, in the same method as 6 x 6 = 36.

No they don't the square root of a positive number is the positive number which squared equals the first.

(-6)^2 = 36, yes.

but

sqrt(36) = 6, and that's that.

The confusion comes from the fact that when the teacher tells you that

if x^2 = 36 then x = 6 or x = -6, and that's right, but the math isn't complete because

x^2 = 36

does not imply

x = sqrt(36),

it implies

|x| = sqrt(36) = 6

which implies

x = 6 or x = -6.

 

[crudus]

Every math major I have talked to and showed that to has described that as "shady".

Well, that is how my dad explains it, and he's a math professor, so I'd rather take his word over that of some math majors who can't come up with anything better than calling it 'shady'.

Well, they were nice enough to supply me with a more satisfying answer using infinite series. I am not arguing against the result; I am arguing against the proof used.

 

[Coldie]

No I get what he's saying, it's not like it's hard to envisage .9~ as being '1', but It's still just an idiotic playing around with established ideals concerning the usage of numbers.

It would be like writing an essay in phonetic English rather than with actual words. One could argue it's still English, it's just a stupid thing to do.

Yup, you got that right. There's no purpose to this other than math wizards mocking the living daylight out of mathematically challenged individuals. 0.(9) is not used for anything else, because it is nothing more than an idiotic way to write down 1.

Math has other wondrous things, ones that actually have meaning and purpose. Such as Euler's Identity that you could see in my avatar, for instance.