1=.999999...?

[Lohrenswald]

How could there not be?

come with something better

 

[brennan]

Right. Because 'no there isn't' is a quantitatively better retort than 'yes there is'.

 

[onejayhawk]

In what sense Q is fully discrete? Every neighbourhood of a rational contains infinitely many rationals.

The idea of Q as discrete and R\Q as "the continuous part" comes from R\Q making "the bulk" of the reals. You'd think that Q is just separate points every here and there, and in between is "the continuous part". That's a misunderstanding, at least in the sense that there are no connected segments in the R\Q.

Also, you seem to think that the popcorn function is constant in some neighbourhood of an irrational point. Otherwise you wouldn't hold this as an evidence. That's also a misconception that rises easily, it's the exact reason that makes the function amazing.

The popcorn function is similar to the function that is x on irrational x:s and 0 elsewhere, and which is continuous only in 0: The continuity is a result of the "dampening" of the non-constant parts of the function near the point examined.

A valid point. I will rephrase.

It is discrete in the sense of the popcorn function, previously specified. Q is not continuous, but it is dense. It is easy to mistake one for the other in many contexts. Indeed, I think that is the point of all this discussion.

Right. Because 'no there isn't' is a quantitatively better retort than 'yes there is'.

That was no a 'no there isn't.' That was a, 'Lame. Try harder.'

For practical purposes, I am not sure this matters.

J

 

[Lohrenswald]

Right. Because 'no there isn't' is a quantitatively better retort than 'yes there is'.

you're the one doing that

 

[dutchfire]

This is the third (fourth?) time I'm asking.

I would like to field a second question.

Does "0.999...8" also equal 1? Why or why not?

The construction of real numbers [usually] is via (Cauchy) sequences. The "number" 0.99999... is given by the sequence
a_n = 1 - 10^-n

What is the sequence you propose for 0.9999....8?

 

[Borachio]

a_n = 1 -10^-(n-1) + 0.8/10^(n-1) ?

Or something?

Now, I'm not claiming that that's a valid sequence.

 

[El_Machinae]

That's what I was trying to get at earlier. All numbers can be expressed multiple ways. There's always a function that can be used to create that number. 0.999...8 doesn't seem to have this ability.

 

[Hygro]

Why is this not valid?
a_n = (1 - 10^-n) + (1 - 10^-n - 1)

actually this even weirder to me. Are you all also implying there can be no such thing as 0.000...1? i.e. a unit of infinite smallness?

 

[trader/warrior]

If 0.999...=1 wouldn't 0.000...1=0?



So that clearly means all numbers actually =0 because they all consist of infinite numbers of 0.000...1's! NOTHING IS REAL TO ME ANYMORE!

Horribly math impaired brain out.

 

[Lohrenswald]

Why is this not valid?
a_n = (1 - 10^-n) + (1 - 10^-n - 1)

actually this even weirder to me. Are you all also implying there can be no such thing as 0.000...1? i.e. a unit of infinite smallness?

There are infinitessimally small numbers called dx (I think)
but you can't have infinitely many digits before a last digit

 

[Hygro]

ok so then a_n = 1 - 10^-n - dx

 

[Lohrenswald]

Ask one of the guys who have finished a math dege I guess
I didn't think of integrals as sums until like 6-7 months ago

 

[Borachio]

Yeah. That big crazy "S" sign didn't give it away to you, then?

 

[Lohrenswald]

no

 

[Atticus]

WAre you all also implying there can be no such thing as 0.000...1? i.e. a unit of infinite smallness?

There is no such. What would you get if you divide it with 2? (You can't accept 0, since then the rule a/b=c => a=bc would break down).

There are infinitessimally small numbers called dx (I think)

No there aren't, not in real numbers. Some people just think that's an easier way to think integrals etc.

 

[Leifmk]

If 0.999...=1 wouldn't 0.000...1=0?

That latter thing is not a thing. You cannot have an infinite number of digits and then a last one.

 

[mrt144]

There are infinitessimally small numbers called dx (I think)
but you can't have infinitely many digits before a last digit

Why not?

 

[Akka]

I have the feeling this thread is turning into a parody of itself. I'm not sure who is serious and who is just joking to keep the dead horse moving now.

 

[Terxpahseyton]

You people are infuriating.

 

[Agent327]

I'm sure. I noticed another problem with the 'proof' that 0.999...=1.

1+1=2

It follows that 0.999...+0.999...=2 (since 0.999...=1)

1x1=1.

It follows that 0.999...x0.999...=1 (since 0.999...=1).

However, if one actually adds or multiplies 0.999... to or by 0.999... you get something like 1.999...8 or 0.999...8.

Now you will say, an infinite number doesn't have a last number. Which would obviously be correct.

However, no matter how far you add 9s, the addition (or multiplication) simply won't add up to 2 (or 1).

I haven't tried, but I would assume that it is equally possible to prove 0.999... does not equal 1.

What I'm actually more interested in: what is the point of 0.999...? Since it has been proved to be 1. And, since equality goes both ways: what is the point of 1=0.999...? In what calculations would this be useful? (I tried adding/multiplying 0.999... with 0.999... by calculator, but the calculator can't handle it.)