Is Dividing 1 by 3 an Invalid Problem?

[Hurkyl]

So it was chosen 0.999... was equal to 1.

Effectively yes.

FYI, historically, we used the real numbers long before we ever had the idea to represent them with decimals. I'm not sure what the original idea for decimals was, but I would imagine the value of a decimal was originally meant to be an infinite sum. And the corresponding infinite sum that computes the value of 0.999... is a geometric series whose sum is 1.

 

[Drakkith]

So it was chosen 0.999... was equal to 1.

It was not chosen that specifically 0.999... should equal 1. It is simply the way our number system is set up that makes it that way.

 

[tedima]

Irrelevant. Your example has nothing to do with what we are talking about.

Yeah it kinda moved off topic

 

[tedima]

Effectively yes.

FYI, historically, we used the real numbers long before we ever had the idea to represent them with decimals. I'm not sure what the original idea for decimals was, but I would imagine the value of a decimal was originally meant to be an infinite sum. And the corresponding infinite sum that computes the value of 0.999... is a geometric series whose sum is 1.

That explains everything. Thanks.

 

[Mensanator]

When you divide 1 by 3, you get .33333... repeating forever of course. My question is whether this operation could ever be considered to end. It looks to me like it's an invalid problem since you could never get a final answer, but simply keeping adding threes to the end of it when you try to solve. Does this make any sense?

No, it makes no sense. You are confusing a number with its representation. Try using base 3.

 

[disregardthat]

It was not chosen that specifically 0.999... should equal 1. It is simply the way our number system is set up that makes it that way.

We decided that 0.9999... = 1 when we decided that 0.999.. makes sense in the particular way it does today.

 

[Oriako]

Just read this:
http://en.wikipedia.org/wiki/0.999...

 

[Caramon]

1 = 0.999...

Proof:

x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1

:)

 

[Char. Limit]

1 = 0.999...

Proof:

x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1

:)

fify

 

[HallsofIvy]

No, it wasn't "chosen". 0.9999... means .9+ .09+ .009+ .0009+ .00009+ ...= .9(1+ .01+ .001+ .0001+ .00001+ ...)

That last is a "geometric series" which is taught in any good "precalculus" or "algebra II" class.
a(1+ r+ r^2+ r^3+ r^4+ ...) has sum a/(1- r) as long as |r|< 1.

Here, a= 0.9 and r= .1. a/(1- r)= .9/(1- .1)= .9/.9= 1.

 

[disregardthat]

No, it wasn't "chosen". 0.9999... means .9+ .09+ .009+ .0009+ .00009+ ...= .9(1+ .01+ .001+ .0001+ .00001+ ...)

That last is a "geometric series" which is taught in any good "precalculus" or "algebra II" class.
a(1+ r+ r^2+ r^3+ r^4+ ...) has sum a/(1- r) as long as |r|< 1.

Here, a= 0.9 and r= .1. a/(1- r)= .9/(1- .1)= .9/.9= 1.

The point is that it was chosen to mean exactly that, not that we chose to do our calculations correctly...