Uhhh....where IS everyeone?

[Shvegait]

If you were to apply .9r to a thing, could it really exist since the nines would continue forever?  Maybe...

This is a good point. .9 repeating is a worthless concept.

Think about it. How do you generate a repeating sequence of digits? Division.

8 divided by 9 becomes .888888888 on forever in long division. However, how can you generate .9999999 on forever by division? You can't. 9 divided by 9 gives 1.

But .9 repeating IS equal to the infinite series: Sum of 9/10^n from n=1 to infinity (0.9 + 0.09 + 0.009 + ...). And you can prove that this series converges to 1. It's been a year since I took Calculus B but... maybe I'll look through my old notes and let you know

[Isu]

Ugh, now I know why Computer code keeps to 1's and 0's.  

[Timballisto]

Lol.  That's another thing completely.

What if a device used 0s, 1s, and 2s?  If there were three possible states for a bit, then, we could do 1/3 more with a computer than we could now...interesting...now just to find how to add the two.  If zero is no current and one is current is on, maybe two could be inbetween somehow.

We might as well just go into threes while we're at it...

[Proxima]

It wouldn't be just one-third more. Nowhere near.

2^8 = 256
3^8 = 6561

2^16 = 65,536
3^16 = 43,046,721

2^24 = 16,777,216
3^24 = 282,429,536,481

2^32 = 4,294,967,296
3^32 = 1,853,020,188,851,841

I think we can stop there...... you see that if a bit could have three states instead of two, 32 bits could encode more than a quarter of a million times as much information as they currently do. A little bit more than a one-third increase!  

[guest]

I think that both sides are correct in different manners. &#A0;Even with infinite nines, when you think about it in pure logic, you never get to one. &#A0;At the same time, .9 repeating is infinitely closer to one, so it becomes one.

The whole confusion really is that people are confusing the value

0.9999........

with this infinite sequence of numbers

0.9, 0.99, 0.999, ...

It is true that all of the numbers in the sequence are less than 1.  However, this does not follow that the value 0.9999...... is less than 1, because the value 0.9999...... is itself not in the sequence.  Each number in the sequence above each only has a finite number of 9s, and so the number 0.9999...... would not be part of the sequence, but actually just outside of it.

So it is not a logical contradiction that 0.9999...... is equal to 1 even though every number in the sequence 0.9, 0.99, 0.999,... are less than 1.  The sequence and the value are two separate entities, despite being closely related.

[guest]

Ugh, now I know why Computer code keeps to 1's and 0's. &#A0;

Well actually, you'd get the same sort of thing no matter what base you use for the number.  If we are in binary, then instead of 0.99999..... being 1, we get 0.11111.... being 1.

Of course, one nice thing about computers as opposed to math is that computers never deal with infinite quantities in the way you would in math.  There would be no 0.11111... in a computer since you don't have an infinite amount of memory to hold all those 1s.

[guest]

Speaking of bits...

You guys know that the word "bit" is originally a contraction of the phrase "binary digit", right?

Now imagine what you'd get if we do ternary (base 3) instead of binary......

[Isu]

Ugh, now I know why Computer code keeps to 1's and 0's.  

This was really some kind of joke. I wasn't expecting a conversation to sprout over it.

[Shvegait]

Actually, guest...

The nth element of the sequence, starting at n=1 is (1 - 1/10^n)

The limit of (1 - 1/10^n) as n approaches infinity equals 1 - 0 = 1.

It is not correct to say that .9 repeating isn't in the sequence... It is just one of those concepts that people will feel uncomfortable about until they learn calculus...


The confusion is kind of interesting. People will accept that .3 repeating is 1/3. Don't they? You never see people saying that no matter how many 3's you write, you will always be less than 1/3 (you will, except at infinity). This is, I guess, because the concept of 1/3 isn't as strong in people as the concept of 1, which "clearly" any DECIMAL must be less than. It isn't as clear for other repeating decimals, although the same concepts are at work.

The other thing is that .9 repeating is never the result of long division, so there aren't really any concrete examples where people get .9 repeating as a result and must interpret it.

[guest]

Actually, guest...

The nth element of the sequence, starting at n=1 is (1 - 1/10^n)

The limit of (1 - 1/10^n) as n approaches infinity equals 1 - 0 = 1.

It is not correct to say that .9 repeating isn't in the sequence... It is just one of those concepts that people will feel uncomfortable about until they learn calculus...

I'm sorry, but you are actually not correct.  Your statement about limits is true, but generally in mathematics, the limit of a sequence is not considered to be in the sequence itself.  This is why for example most texts are careful to say the limit of such-and-such as n approaches infinity (or notationally, n -> inf), rather than to say "as n = inf".

I'm sure your calculus textbook must have examples like the function f(x) = (x^2-1)/(x-1) when they talked about limits.  Because division by 0 is undefined, they would draw the graph of f(x) with an open circle at x = 1, so that f(1) is undefined.  Yet the limit as x approaches 1, if you look at the graph, is clearly the value 2.  So the limit can exist even though it is not considered to be in the function's range.

The other thing is that .9 repeating is never the result of long division, so there aren't really any concrete examples where people get .9 repeating as a result and must interpret it.

That's true.  Of course, you can get .9 repeating by a nonstandard way of doing the long division:

Code: [Select]

   0.999...
 ---------
1 )1.000...
    9
  --------
    10
                    9
  --------
                    10
                     9
  --------
                     1...

[Timballisto]

This was really some kind of joke. I wasn't expecting a conversation to sprout over it.

Yes, I know, but I decided to make it a conversation, becuase that is an interesting thought, using  0s 1s and 2s...

[Shvegait]

Oh yeah, guest, you are right. Sorry, it's been a while. &#A0;X_X

But I doubt that's the reason people are confused about this whole thing. They see a 0.# and assume it must be less than 1.0. Although it's true that when people talk about it they inevitably say "0.99999999999999999", referring to that sequence.

[AstralLemming]

Speaking of bits...

You guys know that the word "bit" is originally a contraction of the phrase "binary digit", right?

Now imagine what you'd get if we do ternary (base 3) instead of binary......

I know exactly you're talking about. Streetlight administrator, You have competion for rudeness.

[drumnbach]

But what function would the third option in a ternary serve? Surely the on and off workings of binary is enough? You would have to devise an entirely new system of logic. The only purpose I can see a ternary serving is that it would throw uncertain values into the mixture, which inevitabley leads us to quantum mechanics. Would quantum mechanics actually serve any purpose in the world of computing? I can see it working for security applications, but only in a very obscure way.

I don't care what people say; quantum mechanics applies only to the subatomic level. Unless they shrink down circuit boards to the size of atoms, a ternary would be inefficient. Binary works so well BECAUSE there are so few options.

[Proxima]

What function it would serve is clear enough; in fact I already answered that.

Suppose you have a picture that takes up 1KB of storage space (8192 binary digits). Since 2^8192 ~ 3^5168, to store the same information in a ternary system requires just 5168 digits.

Mathematically, the fraction of digits needed by a ternary system compared to a binary is given by log 2 / log 3 = 0.631. A saving of 37%; equivalently, you could fit 58% more information (games, pictures, music, whatever) into the same amount of space.

Less efficent? I think not.