
Re: It is a very bad idea and nothing less than stupid to define 1/3 = 0.333...
Posted:
Oct 7, 2017 10:29 AM


On 10/7/2017 4:06 AM, netzweltler wrote: > Am Samstag, 7. Oktober 2017 01:42:15 UTC+2 > schrieb Jim Burns: >> On 10/6/2017 6:03 AM, netzweltler wrote: >>> Am Freitag, 6. Oktober 2017 02:54:40 UTC+2 >>> schrieb Jim Burns: >>>> On 10/5/2017 3:12 PM, netzweltler wrote: >>>>> Am Donnerstag, 5. Oktober 2017 17:59:25 UTC+2 >>>>> schrieb Jim Burns: >>>>>> On 10/5/2017 10:00 AM, netzweltler wrote: >>>>>>> Am Donnerstag, 5. Oktober 2017 15:22:35 UTC+2 >>>>>>> schrieb Jim Burns:
>>>>>> [...] >>>>>>>> _We don't do what you're describing_ >>>>>>> >>>>>>> Nevertheless, >>>>>> >>>>>> "Nevertheless"? >>>>>> Do you agree that what you're describing >>>>>> is not what we're doing? >>>> >>>> *NETZWELTLER* >>>> DO YOU AGREE THAT WHAT YOU'RE DOING >>>> IS NOT WHAT WE'RE DOING? >>> >>> Let's say I agree. Doesn't mean that it is obvious to me >>> what *you* are doing. >> >> Great. Let's say you agree. Will you stop saying that >> "0.999... means infinitely many commands"? > > No.
When I showed you what 0.999... means, your reaction (paraphrased) was "But there are no 'infinitely many commands' there!"
Am I wrong about that paraphrase? Will you show me the infinitely many additions which I do? Which _I_ do, not which _you_ do.
If you can't, why do you keep saying that they're there?
> Because it is not obvious to me why the equation > 0.999... = 0.9 + 0.09 + 0.009 + ... > should be wrong.
By definition, 0.999... = LUB{ 0.9, 0.9 + 0.09, 0.9 + 0.09 + 0.009, ... }
If, by 0.9 + 0.09 + 0.009 + ... one means LUB{ 0.9, 0.9 + 0.09, 0.9 + 0.09 + 0.009, ... } then it's not wrong. But you don't mean that.
The answer to _why_ you're wrong is: it's not defined your way.
Now that you know why, will you stop saying that "0.999... means infinitely many commands"?
[...] > > You are not addressing what I mean what has to be proven. > How do you prove that 0.999... is pointing to a location > on the number line? To me you are just assuming that it does.
By the real number axiom that any bounded nonempty set of reals has a least upper bound, LUB{ 0.09, ... } represents a unique real number.
Suppose you have the set N of natural numbers. How would you prove that there is no predecessor of 0 in it? Wouldn't using the axiom (Ax)~(Sx = 0) be "just assuming" that there isn't?
> Then you are comparing its "location" to the locations of > the terminating decimals in (11/10^n)n?N and conclude that > it is a location to the right of all those numbers.
Basically, yes. We prove 1 is an upper bound. But we also prove that there is no number to the left of 1 which is also an upper bound. Thus 1 is the unique _least_ upper bound.
> I could rather show that 0.999... is definitely not pointing > to a location coincident with point 1. Therefore we had to > agree that 0.999... can be represented by the line segments > {[0, 0.9], [0.9, 0.99], [0.99, 0.999], ...}.
So, if we changed the definition of 0.999... , you could show that 0.999... does not represent 1? I'm sure you could.
Why should anyone care about 0.999... with a different definition?
And why do you keep speaking as though the definition has been changed to what you suggest, when it hasn't been?
> But I guess  since you don't agree that 0.999... is the > result of infinitely many additions  you don't agree with > that notion either.
It's not actually a question of _my_ agreement. I could agree that there is an integer between 2 and 3, but there still would not be an integer there.
Now, we could _change the definitions_ so that I would be correct. Maybe we'd start counting 0, 1, 2, 4, 3, 5, 6, ... So, it's not inconceivable. But it's a useless point to make. When we talk about 2 and 3, we're talking about 2 and 3 _with the usual definitions_ . And when we talk about 0.999..., we talk about 0.999... _with the usual definitions_ . The usual definitions do not have infinitely many additions.
> You don't show how 0.999... is pointing to a location on > the number line.
The set { 0.9, 0.99, 0.999, ... } is bounded and nonempty. Proof:
For all x e { 0.9, 0.99, 0.999, ... }, x < 2.
0.9 e { 0.9, 0.99, 0.999, ... }
Done.
The infinite decimal 0.999... (a map from negative places to decimal digits, all 9s) points to a bounded nonempty set of finite partial sums. That set points to its least upper bound  a unique location on the number line.
It's just not true that I haven't showed you this, over and over and over.

