Poll: 0.999... = 1

[Ishnuvalok]

I'm not saying you're wrong; using your maths, you are right.

I'm saying our maths is wrong.

Just like our maps were wrong when they were missing countries and our globes when we 'knew' Earth was flat and was orbited by the sun.

The facts don't change. Only our way of seeing does.

I like how you ignore mine and many other's profs of 0.999...=1 and insist on living in your own little delusion where common sense is the base for all mathematics. It isn't.

 

[Maze1125]

Of course there are infinitely small non-zero numbers.

So, how should we represent these numbers?

 

[Redingold]

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?

 

[Maze1125]

SNIP

I was just reading Wikipedia (terrible source I know. But it is now... past 0400 and I'm too tired and lazy XD) and I found something I don't get. Can you help me understand this?

Why does 0.999... = 1, if 0.333... < 0.4?

Because 0.3999... = 0.4 and 0.3999... - 0.3333... = 0.0666.... So 0.333... is 0.0666... less than 0.4.

 

[Athinira]

Whatever. As I said in the post that infinitely small, it's still relevant

.9~ / 1 =/= 1

.9~ / 1 = .9~

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

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

Does 12 x .9~ = 12? No.

Therefore it's not 1 <.<

Your narrow mind is simply clouding your logic.

I'll repeat again: Numbers can be represented in more than one way. Maze 1125 states it very well: 0.9r and 1 is simply two different representations for the same value. And yes 12 x 0.9r = 12.

There have been mathematical proofs in abundance of this concept and it's accepted between the worlds leading mathmaticians. Now we just need the general (persistent) part of the population to understand it.

I'll repeat one last time: If x < y, then there exists a number z between the two. If you can't find z, then x = y. Period. Now show me which number lies between 0.999... and 1. And no, 0.0...1 is not a number that can exist in the Real Number system, because that system doesn't allow infinitesimals to exist.

SNIP

I was just reading Wikipedia (terrible source I know. But it is now... past 0400 and I'm too tired and lazy XD) and I found something I don't get. Can you help me understand this?

Why does 0.999... = 1, if 0.333... < 0.4?

Because you are confusing 0.333... with 0.3999...

 

[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.