Poll: 0.999... = 1

[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.

 

[Addendum_Forthcoming]

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.

From what I can remember of year 6 maths, all positive numbers have two square roots. Besides, as I said .. example of idiotic math games not unlike .9~ = 1

 

[Maze1125]

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.

Yes, that proof is a bit shady, it's perfectly correct, but it implies a lot of complicated mathematics that isn't explicitly said. And in that it is a bit shady.
If you were talking to a mathematician. That's a fine proof to use, as they should be able to fill in the gaps themselves. But if you're talking to someone who doesn't know the maths, then you're essentially keeping half of what you're doing "under the table" to avoid confusion which, even though all the stuff you're hiding is perfectly correct, is a bit shady.

 

[Ishnuvalok]

Man, screw maths. It's 0430 and I am tired, lol. I give up. I either don't get this at all, or maths is broken.

It's the former option. It's that you don't understand this.

Never assume because you don't understand something, that everything that has to do with that field is wrong. That's ignorant.

 

[Agayek]

That doesn't make sense. You're saying that 10x-x is a number with an infinite number of decimal places occupied by nines, but the last decimal place is occupied by a one. Which means that you're essentially saying that 10x-x has both an infinite number of decimal places and a finite number of decimal places. That's a contradiction.

You cannot perform mathematical operations on an infinitely repeating number. Therefore, you must at some point terminate the string. At that point, you can then multiply it by 10 and proceed.

However once you do that, 9.9999...999 will have shifted to the left, so 0.999...999 will have one more significant digit. Thus, you get 8.999...991.

Edit:

But it goes on to infinity, so technically there's no 1.

Same answer to you too.

 

[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.

From what I can remember of year 6 maths, all positive numbers have two square roots. Besides, as I said .. example of idiotic math games not unlike .9~ = 1

I agree that the 0.999... = 1 maybe useless in actual math (although it may reveal problems on basic understanding of math).

But square roots of positive numbers are by definition positive... Or else:

if sqrt(36) = 6 and sqrt(36) = -6

then

sqrt(36) * sqrt(36) = -36

or worse

sqrt(1) = 1 and sqrt(1) = -1

then 1 = sqrt(1) = -1

which implies 1 = -1 and that's just wrong.

And that's not useless or a idiotic math game.

 

[Maze1125]

From what I can remember of year 6 maths, all positive numbers have two square roots. Besides, as I said .. example of idiotic math games not unlike .9~ = 1

These aren't games. These are facts.

Square roots are factually defined to always be positive.
The equation x[sup]2[/sup] = 36 has two solutions 6 and -6.
But that is not the same thing as saying sqrt(36) = -6.
sqrt(36) = 6, always. This is done in order to ensure that "sqrt(x)" is a valid function.

Equally, 0.999... = 1, that is a fact. We're not playing games, it just an interesting fact. Just like e[sup]i*pi[/sup] = -1. It's a very interesting fact that is very unintuitive the first time you see it. But that doesn't make it wrong.

 

[Addendum_Forthcoming]

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.

The only maths I ever had to do since the end of school was pearson's co-efficient for psych experimentations when measuring and testing relationships between psychological concepts such as happiness, optimism and extroversion using three tests and a lie scale test to eliminate untruthful test subjects.

Of course all that came on a CD you put into a computer an write down the results in the boxes ... hey presto ... computer works out RSS values for all of them <.<

 

[Maze1125]

That doesn't make sense. You're saying that 10x-x is a number with an infinite number of decimal places occupied by nines, but the last decimal place is occupied by a one. Which means that you're essentially saying that 10x-x has both an infinite number of decimal places and a finite number of decimal places. That's a contradiction.

You cannot perform mathematical operations on an infinitely repeating number. Therefore, you must at some point terminate the string. At that point, you can then multiply it by 10 and proceed.

However once you do that, 9.9999...999 will have shifted to the left, so 0.999...999 will have one more significant digit. Thus, you get 8.999...991.

Edit:

But it goes on to infinity, so technically there's no 1.

Same answer to you too.

Here's a more rigorous proof for you then:

An infinite decimal is defined to be:
lim(as n->infinity)sum(from k=1 to n) (a[sub]k[/sub] * 1/10[sup]k[/sup])
where a[sub]k[/sub] is the kth digit of the decimal.

Therefore, 0.999... is defined to be:
lim(as n->infinity)sum(from k=1 to n) (9 * 1/10[sup]k[/sup])
So all we need to do is show that that is equal to one.
Which is true iff for all e>0 there exists an N such that for all n>N |1 - sum(from k=1 to n) (9 * 1/10[sup]k[/sup])| < e

Now sum(from k=1 to n) (9 * 1/10[sup]k[/sup]) is a finite sum, and so we can calculate that
|1 - sum(from k=1 to n) (9 * 1/10[sup]k[/sup])| = |1/10[sup]n[/sup]|

So we need to show that for all e>0 there exists an N such that for all n>N |1/10[sup]n[/sup]| =1 then |1/10[sup]n[/sup]| e>0, then let N = 1/e and then |1/10[sup]n[/sup]| N

Hence the claim that, for all e>0 there exists an N such that for all n>N |1 - sum(from k=1 to n) (9 * 1/10[sup]k[/sup])| < e, is true.
So, by the definition of a limit, lim(as n->infinity)sum(from k=1 to n) (9 * 1/10[sup]k[/sup]) = 1
Therefore, by the definition of infinite decimals, 0.999... = 1

QED

 

[Athinira]

From what I can remember of year 6 maths, all positive numbers have two square roots. Besides, as I said .. example of idiotic math games not unlike .9~ = 1

I think his point was that while the square root of a positive number CAN have two solutions, in reality it only has one (aka. it's one or the other). Which one it is isn't necessarily something we can deduce, which is why he wrote x = 6 or x = -6

 

[Addendum_Forthcoming]

From what I can remember of year 6 maths, all positive numbers have two square roots. Besides, as I said .. example of idiotic math games not unlike .9~ = 1

These aren't games. These are facts.

Square roots are factually defined to always be positive.
The equation x[sup]2[/sup] = 36 has two solutions 6 and -6.
But that is not the same thing as saying sqrt(36) = -6.
sqrt(36) = 6, always. This is done in order to ensure that "sqrt(x)" is a valid function.

Equally, 0.999... = 1, that is a fact. We're not playing games, it just an interesting fact. Just like e[sup]i*pi[/sup] = -1. It's a very interesting fact that is very unintuitive the first time you see it. But that doesn't make it wrong.

As far as I remember, all positive numbers have 2 sq roots. A square root is just that. I hardly see how you're able to debate semantics after committing yourself to idiotic number games.

An analogous argument if I were to use similar semantics would be to simply say that .9~ isn't 1 because it doesn't use the same figure. A faultless argument on the basics of semantics (and what I still think is a logical one to make regardless in response to such foolish games)

 

[Maze1125]

From what I can remember of year 6 maths, all positive numbers have two square roots. Besides, as I said .. example of idiotic math games not unlike .9~ = 1

I think his point was that while the square root of a positive number CAN have two solutions, in reality it only has one (aka. it's one or the other). Which one it is isn't necessarily something we can deduce, which is why he wrote x = 6 or x = -6

No, he wasn't.

If you have the equation x[sup]2[/sup] = 36 then it has two solutions x = 6 or x = -6.
Equally, if you have the equation sqrt(36) = |x| then x = 6 or x = -6.

But if all you have is sqrt(36) = x, then that equation only has one solution, x = 6.
x = -6 would be wrong as the sqrt function always give a positive answer.