On Monday, 15 August 2022 at 23:45:59 UTC+8, Ben Bacarisse wrote:
> wij <wyni...@gmail.com> writes:
>
> > On Monday, 15 August 2022 at 20:02:58 UTC+8, richar...@gmail.com wrote:
> >> On 8/15/22 5:38 AM, wij wrote:
> >> > On Monday, 15 August 2022 at 08:34:39 UTC+8, richar...@gmail.com wrote:
> >> >> On 8/14/22 7:35 PM, wij wrote:
> >> >>> The vague, no-logic concept of infinity seems dominated people's mind.
> >> >>> What is infinity? What does "lim(x→∞) f(x)" mean?
> >> >>>
> >> >>> If infinity is merely a 'concept', not a number, what does x approach to?
> >> >>> If x is not getting "closer" to ∞? What does 'approach' mean?
> >> >>> Therefore, ∞-(x+1) < ∞-x must be valid inequality to mean x+1 is closer than x to infinity ∞.
> >> >>>
> >> >>> But valid what? Most people agree ∀n∈ℕ, n<∞.
> >> >>>
> >> >>> Is x+1 not closer than x to infinity?
> >> >>> So, infinity ∞ must have arithmetic meaning. Here is one:
> >> >>> The multiplicative inverse of ∞ is 1/∞, the additive inverse is -∞
> >> >>>
> >> >>> All in all, that is the definition of infinity (the symbol '∞') proposed.
> >> >>> All is that simple, the usage treating ∞ as if it is a unique number is
> >> >>> safe-guaranteed, what left is interpretation. Though I think I figured this
> >> >>> part (merely means a procedure never terminate), there may be lots more
> >> >>> instances to test its interpretation in various scenario.
> >> >> If we are talking the real number system, as implied by the limit
> >> >> operator, then the definition of what lim(x->inf) f(x) means
> >> >>
> >> >> is there a number L, such that for ANY error e > 0, no matter how small,
> >> >> can we find an X such that for all x > X that |f(x)-L| < e
> >> >>
> >> >> If L exists, then it is the value of lim(x->inf) f(x)
> >> >>
> >> >> Generally, we will find some bounding formula of some X(e) where we can
> >> >> prove that | F(x) - L | < e for all x > X(e),
> >> >
> >> > The issue has been discussed many times. This proposal is primarily about the
> >> > definition of infinity.
> >> >
> >> > Pythagorean's real number is Q, they could use the infinite-approaching argument
> >> > very validly deducing that all numbers are ratio number. Anyone can use Q to
> >> > approach any number and deduce that all real numbers are rational (sure modern
> >> > people won't do this).
> >> >
> >> Actually, the real Pythagorean's eventually realized (c 5th century BC)
> >> that the length of the hypotenuse of a right triangle with the two legs
> >> having length 1 was not a rational number, and this caused them problem.
> >>
> >> Yes, it took them a while, but that is the irrationality of Man when he
> >> sticks to wrong ideas.
> >> > Snippet from https://groups.google.com/g/comp.theory/c/DaybI0JY4Vc
> >> > ...
> >> > To add more material came up to me (not well ordered):
> >> >
> >> > ----------------------------
> >> > There are quite a number of proofs of "repeating decimals are irrational".
> >> > The basic is the correct equation of 1/3 and its decimal form from long
> >> > division (kids understand this 'infinity' with no problem) should be:
> >> >
> >> > 1/3= 0.333... + nonzero_remainder.
> >> >
> >> > ----------------------------
> >> > To translate the 0.999... problem to limit:
> >> >
> >> > Let A= lim(n->∞) 1-1/2^n = 0.999...
> >> > B= lim(n->∞) 1-1/10^n = 0.999...
> >> >
> >> > Assume A=B
> >> > <=> lim(n->∞) 1-1/2^n = lim(n->∞) 1-1/10^n
> >> > <=> lim(n->∞) 1/2^n = lim(n->∞) 1/10^n
> >> And this step is invalid. You either multiplied by a "non-number" or
> >> divided by zero depending on the steps you did to make that transition..
> >>
> >> This is the problem of assuming that "infinity" is a number.
> >> > <=> lim(n->∞) 1 = lim(n->∞) 1/5^n
> >> > <=> 1=0
> >> >
> >> > [Note] I just demonstrate an instance. The limit theory can evolve as it does
> >> > (e.g. one-sided limit... There are many slightly different versions of
> >> > interpretation of limit as it evolves). Readers might find different
> >> > authors use different rules.
> >> > Limit is a technic to find its 'limit', it cannot form a logically
> >> > consistent theory for real number, e.g. the result of limit in general
> >> > must be verified, e.g. numerically, one cannot absolutely trust the
> >> > result of limit arithmetic. And at final, lim(x->c) f(c)= L does not
> >> > 'deduce' f(c)=L (In text book, probably just reads "lim(x->c) f(c)= L, SO
> >> > WRITTEN as f(c)=L"). Limit theory only says the limit of 0.999... is 1,
> >> > the theory does not say 0.999...=1. There is no equality concept in the
> >> > ε-δ theory.
> >> > If one resorts to Dedekind-cut-like theories (I did not really read it),
> >> > from the knowledge that all the combinations of discrete symbols cannot
> >> > represent all the real numbers, I can conclude what those theories
> >> > claim are false, let alone I suspect there should be circular arguments
> >> > there, because many terms there must be well defined as a fundamental
> >> > theory, are undefined (prove me wrong).
> >> >
> >> > The limit example above demonstrated "0.999..." cannot denote a specific number,
> >> > which also means "repeating decimal" cannot specify a unique number (A!=B).
> >> > Using limit is invalid for me (for this question) but the result is correct,
> >> > see the provided reference (I found a typo there).
> >> >
> >> > -----------------------
> >> > Simple arithmetic (this should also be a valid way 2.718... is calculated):
> >> > (0.999....)^n approaches 1/e
> >> > (1.000...1)^n approaches e (or defined as e)
> >> > A possible rebuttal might be that the (1-1/n) in lim(n->∞) (1-1/n)^n is an invalid
> >> > number (approximated like 0.999...), or it is a 'concept' etc...
> >> > But if it is not a number, the whole equation is broken.
> >> >
> >> > -----------------------
> >> > A[0]=0
> >> > A[n]=(A[n-1]+1)/2
> >> >
> >> > The density property says (implicitly) n can enumerate infinitely (otherwise, it
> >> > won't be a rule) and A[∞] never be 1. A[n] infinitely approaches 1 in form
> >> > like 0.999.... This is like in the case of the interval [0,1), infinite numbers
> >> > of 0.999...s are located near the open end of [0,1).
> >> > Can we infinitely refine the scale of a ruler and the last scale never touches
> >> > the scale of 1? I think, yes, something like the √2 story, otherwise all numbers
> >> > can be 'proved' rational.
> >
> > Why do you point to where you seem not addressing.
> > Andy probably just missed a point, I provided a reference.
> > Ben made an error and (assume he saw my reply to Andy) made an error
> > again.
> I made no mistake.
> > Let A= lim(n->∞) 1-1/2^n = 0.999...
> > B= lim(n->∞) 1-1/10^n = 0.999...
> >
> > Assume A=B
> > <=> lim(n->∞) 1-1/2^n = lim(n->∞) 1-1/10^n
> > <=> lim(n->∞) 1/2^n = lim(n->∞) 1/10^n
> > <=> (lim(n->∞) 2^n)*(lim(n->∞) 1/2^n) = (lim(n->∞) 2^n)*(lim(n->∞) 1/10^n)
> No. This step is invalid. You can't multiply a limit by anything but a
> real number. lim(n->∞) 2^n is not a real number.
> > <=> lim(n->∞) 2^n/2^n = lim(n->∞) 2^n/10^n
> The product law for limits is only valid when both limits are real.
> > <=> lim(n->∞) 1 = lim(n->∞) 1/5^n
> > <=> 1=0
> >
> > I wonder how much does you guys really understand you are talking?
> Obviously if you don't understand the basics of real analysis, you will
> doubt anyone who points them out.
>
> --
> Ben.

Do you understand this?
https://tutorial.math.lamar.edu/classes/calci/limitsproperties.aspx