netzweltler
Posts:
472
From:
Germany
Registered:
8/6/10


Re: It is a very bad idea and nothing less than stupid to define 1/3 = 0.333...
Posted:
Oct 3, 2017 3:25 PM


Am Dienstag, 3. Oktober 2017 16:20:25 UTC+2 schrieb Jim Burns: > On 10/3/2017 3:21 AM, netzweltler wrote: > > Am Dienstag, 3. Oktober 2017 03:22:11 UTC+2 > > schrieb Jim Burns: > >> On 10/2/2017 2:47 PM, netzweltler wrote: > >>> Am Montag, 2. Oktober 2017 20:35:56 UTC+2 > >>> schrieb Jim Burns: > >>>> On 10/2/2017 1:58 PM, netzweltler wrote: > >>>>> Am Montag, 2. Oktober 2017 17:59:21 UTC+2 > >>>>> schrieb Jim Burns: > >>>>>> On 10/1/2017 3:22 AM, netzweltler wrote: > > >>>>>>> Do you agree that 0.999... means infinitely many commands > >>>>>>> Add 0.9 + 0.09 > >>>>>>> Add 0.99 + 0.009 > >>>>>>> Add 0.999 + 0.0009 > >>>>>>> ...? > >>>>>> > >>>>>> 0.999... does not mean infinitely many commands. > >>>>> > >>>>> But that's exactly what it means. > >>>> > >>>> That's not the standard meaning. > >>> > >>> So, you disagree that > >>> 0.999... = 0.9 + 0.09 + 0.009 + ... ? > >> > >> Your '...' is not usable. If we say what we _really_ mean, > >> in a manner clear enough to reason about, then the '...' > >> disappears. Also, what we are left with are finitely many > >> statements of finite length. You will not find infinitely > >> many commands in those finitelymany, finitelength > >> statements. > >> > >> We sometimes write the set of natural numbers as > >> { 0, 1, 2, 3, ... } > >> The '...' is informal. We do not use '...' in our reasoning, > >> we use a correct description of what the '...' stands for. > >> > >> Do you see '...' anywhere in the following? > >> > >> The set N contains 0, and for every element x in N, its > >> successor Sx is in N. > >> > >> This is true of N but not true of any _proper_ subset of N. > >> > >> _Therefore_ , if we can prove that B is a subset of N > >> which contains 0 and which, for element x of B, contains Sx, > >> then B is not a _proper_ subset of N. > >> > >> B nonetheless is a subset of N, we just said so. The only subset > >> of N which B can be is N. Therefore, B = N. > >> > >> This is finite reasoning about the infinitely many elements > >> in N. Note that there is no '...' in it. > >> > >> I could continue and derive 0.999... = 1 from our definitions, > >> and nowhere in that derivation will be '...'. There will not be > >> infinitely many commands in it either. > > > Sorry, no. The meaning of "..." is absolutely clear in this > > context and > > Is it clear to you? Really? > > I ask because the basis for your whole complaint, in many > threads, is that '...' means "infinitely many commands" in > some way but then you're all "Whoa! that makes no sense, guys". > It does not look to me as though _what you think_ '...' > means in this context is at all clear _to you_ . > > ( _What you think_ it means is not what it means. This is > _my_ point.)
Sorry, I don't get what you are trying to teach me. > > we both know that there is a decimal place for each n ? N > > in 0.999... > > And what does that mean? Have you traded one thing that needs > explaining for another thing that needs explaining? It's not > very useful to do that. > > You refer to N here. What is N? Do you need to use the > successor operation infinitely many times to say what N is? > Is it clear to you what that means? > >  > N has a finite description. It is the minimal inductive set > with 0 and successor x > x+1. > > When we use N to describe something, for example, the set of all > finite initial expansions of 0.999..., > { 0.9, 0.99, 0.999, ... } > the use of N will not make that description infinite. > > The value that we assign to 0.999... is the least upper bound > of the set of all finite initial expansions of 0.999... > This is a _definition_ . > > And that assigned value is 1. Not "nearly 1", exactly 1.

