Poll: 0.999... = 1

[Sebenko]

Depends.

What degree of accuracy am I recording to?

If it's an integer, and I'm using the round function, then it's one. If it's not using round, then 0.9 = 0. Yay truncation!

Short? Long?
Single? Double?

Stop thinking like a computer scientist. We're recording to infinite accuracy. There is no truncation and there is no rounding. We're not claiming that 0.999... is approximately 1, we're claiming that it IS 1.

Uh... no.

I'm a computer scientist, it's what I do.

Do you really need infinite accuracy? I doubt it.

It's all very exciting as a hypothetical question, but I don't think there's ever been a real world situation where recording to infinite accuracy has been the only option.

 

[Rubashov]

So you agree with my prof that 0 = 1? Because if you want to do traditional maths with infinity, you would have to...

Infinity is an integral part of the so-called "traditional" math. Infinity and infinite numbers are also very prominent in the Set Theory, look it up.

You seem to have no understanding of how math actually works, so would you kindly post your alleged "proof" of 0 = 1?

I find you lack of math disturbing.

I did in the first post you quoted, but here it is again:

1-1 = 0
(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1) = 0
= 0
1+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+..... = 0
1 = 0

Summation((-1)^n) from n = 0 to infinity does not converge; it's value is undefined. That is to say (1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+..... does not equal zero and 1+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+..... does not equal 1. Neither equals anything because they are both divergent.

Summation(9/(10^n)) from n = 1 to infinity does converge, however; in fact, it converges to 0.999... or, in more familiar notation, 1. So your example is irrelevant.

 

[Rubashov]

Depends.

What degree of accuracy am I recording to?

If it's an integer, and I'm using the round function, then it's one. If it's not using round, then 0.9 = 0. Yay truncation!

Short? Long?
Single? Double?

Stop thinking like a computer scientist. We're recording to infinite accuracy. There is no truncation and there is no rounding. We're not claiming that 0.999... is approximately 1, we're claiming that it IS 1.

Uh... no.

I'm a computer scientist, it's what I do.

Do you really need infinite accuracy? I doubt it.

It's all very exciting as a hypothetical question, but I don't think there's ever been a real world situation where recording to infinite accuracy has been the only option.

What's needed in the real world is irrelevant because the question at hand is a conceptual one. I'm a computer scientist as well. However, I recognize that the limitations that reality imposes on computers don't apply to questions about mathematical concepts. The mathematical concept 1/2 is never 0, even though the value a computer returns when integer 1 is divided by integer 2 is 0.

 

[Maze1125]

Depends.

What degree of accuracy am I recording to?

If it's an integer, and I'm using the round function, then it's one. If it's not using round, then 0.9 = 0. Yay truncation!

Short? Long?
Single? Double?

Stop thinking like a computer scientist. We're recording to infinite accuracy. There is no truncation and there is no rounding. We're not claiming that 0.999... is approximately 1, we're claiming that it IS 1.

Uh... no.

I'm a computer scientist, it's what I do.

Do you really need infinite accuracy? I doubt it.

It's all very exciting as a hypothetical question, but I don't think there's ever been a real world situation where recording to infinite accuracy has been the only option.

The issue of accuracy is irrelevant.

0.999... = 1 if you round and also if you don't. The statement is true both ways. So pick what ever accuracy you want.

 

[Soraryuu]

Ok, blank slates now. New angle of attack. So, 0.999_ is a number that has "infinite" nines? You're using infinity in a number?

Infinity is not a number.

Since it's not a number, you can't multiply with it.

Since you can't multiply with it, you can't count with it.

Therefore, any repeating number is false. Incomplete.

1/3 is impossible to make into a complete number, same for all other fractions that lead to repeating. All of this 0.999_ stuff is a mathematical mistake on par with 1/0.

In other words, bad math. Don't drag forth your equations before you disprove this.

 

[Rubashov]

Hey, let's try something.

Let n = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... and so on, forever, involving all negative integer powers of 2.

Now, 2n = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... and so on.

This is 1 more than n.

2n = n + 1

Subtract n...

n = 1

So what you're saying is that n is actually infinitesimally smaller than 1, huh?

Would that make it equal to 0.999..., since that is also infinitesimally smaller than 1?

But I've just shown it's one.

What am I doing wrong? Where is the flaw that shows that when n + 1 = 2n, n =/= 1?

Again your geometric series...I said it before and will say it again. A limit is not a value. It is a limit and will never reach that value.

The limit of the sum 1/(2^n) for n -> infinity is 1. That much is true, but the function will approach this value asymptotically and never reach it.

The value of an infinite series is defined as the limit of the sequence of partial sums. Even a cursory study of infinite series will show that this is true. I suggest you get on that. [http://tutorial.math.lamar.edu/Classes/CalcII/SeriesIntro.aspx]

 

[Sebenko]

The issue of accuracy is irrelevant.

0.999... = 1 if you round and also if you don't. The statement is true both ways. So pick what ever accuracy you want.

If it's 1 in any case, why is it being discussed?

 

[Optimystic]

Therefore, 10x - x = 9

I sense a disturbance in the Force...

 

[emeraldrafael]

Snip

Of course you're not going to believe me, cause its different then whats being siad. BUt thats what he said. You're working on two different plains of mathematical reality. Seems pretty straight forward to me. When you give a number a value, you cant have it go to infinity doesnt have value. Besides, the guy's been teaching and doing math for more then 50 years of his life, I'm pretty sure he knows what he's saying. How do you argue what he says? What is a value of infinity on a value system?

Snip.

Yes, but those all have values. Thats what he's saying. He's saying that as long as the number is real and has value, it cant equal another without rounding. So whether its an interger, natural, whatever, it has value, so that human mind can comprehend. He's not saying that .999...! cant exist, just that when you give it value, like that because its a written out number, it can not be 1 because one as a value is more then .999...! will be.

 

[Delta342]

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.

Please tell me you did not just say that square roots are always defined to be positive..

I enjoy how this thread has blossomed out of control. Maybe I should start one about when parallel lines meet at infinity (projective space), when triangles can have 3 right angles (on a sphere) when the angles of a triangle can all be 0 (hyperbolic plane) when a rational point P = (x,y) usually when x,y are in Q on a specific curve (an elliptic curve) and P + infinity = P.. I love Mathematics.

Oh and to reiterate myself. Generally 0.99999etc is usually taken to be one. It is in fact the limit but is just written as equality. If you look into non-standard analysis this is very much NOT true and if you go into much much more complicated and pure Mathematics then the concept of equality doesn't really mean a huge amount anyway.

 

[quhouv vhqwiufv qwiu]

Here's the key to the whole thread:

Some people want to define infinity as "infinity plus rounding at the end towards an arbitrarily choosen reference" (how does the number know? Must be the mathematician)

Other people instead think, that infinity means just infinity, and that if one wants to do something on top of it, one needs to do something on top of it.

It doesn't matter how people think of infinity because the reals do not permit infinitesimally small numbers.

 

[quhouv vhqwiufv qwiu]

2003 called, it wants its math meme back.

This has nothing to do with any meme. I'm not some 4chan idiot. I want to see how many people reject the concept.

Rule 50

OT: I guess this makes sense, but I'm going to show my geometry teacher this just to see what she says.

What is rule 50?

 

[Maze1125]

Please tell me you did not just say that square roots are always defined to be positive..

Yes, I did say that, because it is true.

It is useful to have the concept of a square root be a function, which can't happen if it can be both negative and positive. Because it would be then one to many.

Oh and to reiterate myself. Generally 0.99999etc is usually taken to be one. It is in fact the limit but is just written as equality.

0.999... is defined as a limit. So if the limit is 1, then it is 1, as they are the same thing.

If you look into non-standard analysis this is very much NOT true

Okay, please show your working.

and if you go into much much more complicated and pure Mathematics then the concept of equality doesn't really mean a huge amount anyway.

Yeah, I'm going to have to disagree with you there too.

 

[Maze1125]

If it's 1 in any case, why is it being discussed?

Because not everyone can see why it's 1.

Snip

Of course you're not going to believe me, cause its different then whats being siad. BUt thats what he said. You're working on two different plains of mathematical reality. Seems pretty straight forward to me. When you give a number a value, you cant have it go to infinity doesnt have value. Besides, the guy's been teaching and doing math for more then 50 years of his life, I'm pretty sure he knows what he's saying. How do you argue what he says? What is a value of infinity on a value system?

I'm not arguing with him, I'm arguing with what you claim he says.

And I honestly want you to show him the proof I gave you, and for you to return and tell me what he says. Of course, you'd have to show him a print out of it, because the specific details of it matter quite a lot.

 

[Maze1125]

Ok, blank slates now. New angle of attack. So, 0.999_ is a number that has "infinite" nines? You're using infinity in a number?

The point is that the 9s continue forever, you can choose to call that concept infinity, or to just call it "continuing forever".

Infinity is not a number.

Actually, in some number systems, such as the Riemann Sphere [http://en.wikipedia.org/wiki/Riemann_sphere] infinity is a number. But that's fairly irrelevant here.

1/3 is impossible to make into a complete number, same for all other fractions that lead to repeating. All of this 0.999_ stuff is a mathematical mistake on par with 1/0.

0.333... and 0.999... aren't defined using infinity, but using limits as they tend to infinity. So there is absolutely no need to manipulate infinity when discussing them.

See the following proof:

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]

Snip.

Yes, but those all have values. Thats what he's saying. He's saying that as long as the number is real and has value, it cant equal another without rounding. So whether its an interger, natural, whatever, it has value, so that human mind can comprehend. He's not saying that .999...! cant exist, just that when you give it value, like that because its a written out number, it can not be 1 because one as a value is more then .999...! will be.

Again, this is incorrect. In the real number system, if you have two values (i use the term "value" here because a number can be expressed in many ways, including decimal form), then there must exist a value between those two for them to be different. For X < Y to be true in the real number system, there must exist a number Z, so that X < Z < Y. Since there doesn't exist a value between 0.9r and 1, it's the same number.

Going back to your post again, this part stuck out to me...

4: .999...! =/= 1 in the sense that they have values. 1 > .999....! because it is not the whole one when you look at it on the plain of values in numbers (compared it to have .999...! % of an apple. Though it maybe tiny and insignificant, its still not the whole apple.

As you know, you rounded off that part by saying "Though it maybe tiny and insignificant", but here is the thing: A number can't be "insignificant" in the real number system. It's either significant (meaning that you can find Z in between X and Y) or it's non-existant (or infinitesimal, which is equivalent to non-existant in the system). It must exist to be part of the series. And in this case, Z doesn't exist, which makes it impossible for X and Y to be two different values. 0.9r IS the whole apple. It's a representation of 1.

.

No they don't. The Extended Real Numbers only add the values of infinity, either both positive and negative infinity, or just infinity. They don't add infinitesimals. You have to go as far as the Surreal Numbers to get infinitesimals, and even then I've yet to see anyone prove that 0.999... =/= 1 even in the Surreal Numbers.

Oh i wasn't aware of that. My mistake and point taken.

Not quite, if you're working in the integers, then 30/8 doesn't have a solution, you don't just round.

Got a source for that, because my old math teacher disagress

So does computers executing code btw (i know you already discussed that computer contrictions doesn't necessarily fit with real-world concepts of math, but still wanted to mention it).

 

[chunkeymonke]

heres the thing though its not you can post all proofs you want but that is just because humans haven't fully mastered how to use infinites in math for example 1/3 does not equal .333 repeating that is just the closest approximent we can give because 1 can not be divided by 3 so no .99999 repeating is not equal to 1

 

[emeraldrafael]

If it's 1 in any case, why is it being discussed?

Because not everyone can see why it's 1.

Snip

Of course you're not going to believe me, cause its different then whats being siad. BUt thats what he said. You're working on two different plains of mathematical reality. Seems pretty straight forward to me. When you give a number a value, you cant have it go to infinity doesnt have value. Besides, the guy's been teaching and doing math for more then 50 years of his life, I'm pretty sure he knows what he's saying. How do you argue what he says? What is a value of infinity on a value system?

I'm not arguing with him, I'm arguing with what you claim he says.

And I honestly want you to show him the proof I gave you, and for you to return and tell me what he says. Of course, you'd have to show him a print out of it, because the specific details of it matter quite a lot.

why, the print out I mean? I emailed him exactly what you said, I'm just waiting for him to get back to me.

 

[Athinira]

heres the thing though its not you can post all proofs you want but that is just because humans haven't fully mastered how to use infinites in math for example 1/3 does not equal .333 repeating that is just the closest approximent we can give because 1 can not be divided by 3 so no .99999 repeating is not equal to 1

You're wrong. Simple as that.

Your own personal logic is nothing against the proofs, as well as the fact that it's the commonly accepted truth amongst highly educated mathmaticians.

In fact, your post here is even full of fallacies. You say that 1 can't be divided by 3, which can be summerized in one word: Bullsh*t.

Math is all about proofs. Every single concept in math - assuming it's true - can be proven, either in theory or in practice, so simply stating your logic isn't enough. It's going to take NOTHING less than a counter-proof to our proofs to show that you are right and we are wrong.

 

[Maze1125]

Not quite, if you're working in the integers, then 30/8 doesn't have a solution, you don't just round.

Got a source for that, because my old math teacher disagress

No, but I've got an argument:

If I'm working in the real numbers and try to calculate the sqrt(-1), I don't just call it 0 as that is the closest real number to i, I say I can't calculate it.
If I'm working in the positive integers and try to calculate 5 - 16, I don't just call it 1 because that's the closest positive integer to -11.

So why would it be any different if you get a fraction but you're working only in the integers?

What would you call it if you tried to calculated sqrt(2) when you were working only in the rationals?

Also, my maths teacher can beat up your maths teacher.